Simple harmonic motion

Class 12 Physics, unit 17, book pages 44–69

This unit is about oscillations, above all simple harmonic motion (SHM): motion in which the acceleration is proportional to the displacement from the mean position and always directed towards it, $a = - ω^{2} x$ . It shows that a mass on a spring, a simple pendulum and the projection of uniform circular motion all move in SHM, and derives their time period, velocity, acceleration and energy, together with phase and the graphs of SHM. It ends with real oscillators: damping (light, critical and heavy), forced oscillations and resonance, and where resonance is useful or dangerous.

Learning outcomes (19)

Describe simple examples of free oscillations.
Use the terms displacement, amplitude, period, frequency, angular frequency and phase difference in the context of oscillations.
Express the period of simple harmonic motion in terms of both frequency and angular frequency.
Explain that simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and in the opposite direction.
Use $a = - ω^{2} x$ to solve problems.
Use the equations $v = v_{0} cos ω t$ and $v = \pm ω x_{0}^{2} - x^{2}$ to solve problems.
Analyse graphical representations of the variations of displacement, velocity and acceleration for simple harmonic motion.
Analyse the interchange between kinetic and potential energy during simple harmonic motion.
Apply $\frac{1}{2} m ω^{2} x_{0}^{2}$ for the total energy of a system undergoing simple harmonic motion.
Describe that a resistive force acting on an oscillating system causes damping.
Use the terms light, critical and heavy damping.
Sketch displacement-time graphs to illustrate light, critical and heavy damping.
State that resonance involves a maximum amplitude of oscillations and that this occurs when an oscillating system is forced to oscillate at its natural frequency.
Describe practical examples of free and forced oscillations.
Describe practical examples of damped oscillations, with particular reference to the effects of the degree of damping and the importance of critical damping in cases such as a car suspension system.
Justify qualitatively the factors which determine the frequency response and sharpness of the resonance.
Identify the use of standing waves and resonance in applications such as Rubens tubes, Chladni plates and acoustic levitation (knowledge of wave harmonic modes is not required).
Justify the importance of critical damping in a car suspension system.
Justify that there are some circumstances in which resonance is useful (such as tuning a radio or a microwave oven) and other circumstances in which resonance should be avoided (such as an airplane's wing or a suspension bridge).

17.1Oscillations

A child on a swing, the pendulum of a wall clock and children bouncing on a trampoline all have one thing in common: they move back and forth between two points again and again. This kind of repeated motion is called oscillation. This unit studies the most important type of oscillation, simple harmonic motion (SHM). It then looks at what happens when friction takes energy away (damping) and when an outside force keeps pushing the system (forced oscillations and resonance).

Vibratory (oscillatory) motion: When a body moves to and fro about its mean position, then such motion is called vibratory or oscillatory motion.
Mean (equilibrium) position: The position where the body rests when it is not disturbed. The net force on the body is zero there, and displacement is measured from it.
Oscillation (vibration): The complete round trip of an oscillating or vibrating body about its mean position is called one oscillation or vibration.

Examples of oscillations

The motion of the Earth's surface during an earthquake.
The wings of a bird during flying.
A string of a guitar producing music.
Vibration of atoms about their fixed positions in a crystal.
The beating of the heart.
A mass attached to a compressed or stretched spring (horizontal or vertical): once it is released, it starts oscillating.

One complete oscillation

Look at the pendulum in the figure. It starts at the mean position O, swings to the extreme position A, comes back through O, swings to the other extreme position B and returns to O. This complete round trip is one oscillation. The motion then repeats in exactly the same way.

One oscillation of a pendulum: from the mean position O to the extreme A, back through O to the other extreme B, and back to O.

Terms related to oscillations

Displacement ( $x$ ): The distance of the vibrating body from the equilibrium (mean) position, on either side, at any instant of time. It is positive on one side and negative on the other. SI unit: metre (m).
Amplitude ( $x_{0}$ or $A$ ): The maximum displacement of a vibrating body from the mean position. SI unit: metre (m).
Time period ( $T$ ): The time taken to complete one oscillation or vibration. SI unit: second (s).
Frequency ( $f$ ): The number of vibrations or oscillations ( $n$ ) per unit time ( $t$ ): $f = n / t$ . SI unit: hertz (Hz).
Hertz (Hz): One oscillation per second: $1 Hz = 1 s^{- 1}$ .
Angular frequency ( $ω$ ): The angular displacement per unit time, $ω = 2 π f$ . SI unit: radian per second ( $rad s^{- 1}$ ).

Frequency and time period are reciprocals of each other. If one oscillation takes $T$ seconds, then $1/ T$ oscillations happen in each second.

f = \frac{1}{T} and T = \frac{1}{f}

One complete oscillation corresponds to $2 π$ radians (one full turn of the reference circle of section 17.3). So in one second the angle changes by $2 π$ times the number of oscillations per second:

ω = 2 π f = \frac{2 π}{T}

Rearranging gives the time period of SHM in terms of frequency and angular frequency:

T = \frac{1}{f} = \frac{2 π}{ω}

The terms used to describe an oscillation.
Quantity	Symbol	SI unit	Meaning in one line
Displacement	$x$	$m$	Where the body is now, measured from the mean position
Amplitude	$x_{0}$	$m$	The largest displacement
Time period	$T$	$s$	Time for one oscillation
Frequency	$f$	$Hz$	Oscillations per second
Angular frequency	$ω$	$rad s^{- 1}$	$2 π$ times the frequency

Heart beat as an oscillation

During strenuous exercise your heart beats 150 times per minute. Find the time period, the frequency and the angular frequency of the heart beat.

Frequency is the number of beats per second, so change the minute into seconds.
$f = \frac{n}{t} = \frac{150}{60 s} = 2.5 Hz$
The time period is the reciprocal of the frequency.
$T = \frac{1}{f} = \frac{1}{2.5} = 0.4 s$
Angular frequency.
$ω = 2 π f = 2 π (2.5) = 15.7 rad s^{- 1}$

T = 0.4 s, f = 2.5 Hz, ω = 15.7 rad s^{- 1}

17.2Simple harmonic motionOften examined

Every oscillating body has a mean position where it can stay at rest. When it is moved away from that position, a force appears that pulls it back. This force is the restoring force. Simple harmonic motion is a type of motion in which a system oscillates back and forth about a mean (equilibrium) position under a restoring force that grows in exact proportion to the displacement.

Simple harmonic motion (SHM): Such oscillatory motion in which the acceleration of a particle is directly proportional to its displacement from the mean position and is always directed towards the mean position.
Restoring force: The force that always acts towards the mean position and tries to bring a displaced body back to it.

a \propto - x or a = - ω^{2} x

$a$ : acceleration; $x$ : displacement from the mean position; $ω$ : angular frequency, a positive constant for the system.

Each part of the equation carries one condition of SHM:

Proportional: the acceleration is directly proportional to the displacement. Double the displacement and the acceleration doubles.
Minus sign: the acceleration is opposite to the displacement. When the body is on the right ( $x > 0$ ) it is pulled to the left, and when it is on the left ( $x < 0$ ) it is pulled to the right. So the acceleration always points towards the mean position.

Characteristics of SHM

The body oscillates about a fixed mean position, between two extreme positions $x = + x_{0}$ and $x = - x_{0}$ .
Acceleration and restoring force are largest at the extreme positions and zero at the mean position.
Speed is largest at the mean position and zero at the extreme positions.
The time period does not depend on the amplitude (for the mass-spring system and for a pendulum with small swings).
In ideal SHM there is no friction, so the total energy and the amplitude stay constant.

Testing an equation for SHM

An equation of the form $a = - (positive constant) \times x$ describes SHM, and the constant is $ω^{2}$ . Any other form does not.

Only a linear relation with a negative constant is SHM.
Relation	Result	Reason
$a = - 10 x$	SHM	Form $a = - ω^{2} x$ with $ω^{2} = 10$ , so $ω = 3.16 rad s^{- 1}$
$a = - 100 x$	SHM	$ω^{2} = 100$ , so $ω = 10 rad s^{- 1}$
$a = 0.7 x$	Not SHM	No minus sign: the acceleration points away from the mean position
$a = - 200 x^{2}$	Not SHM	Not proportional to $x$ ; $x^{2}$ never changes sign, so it cannot always point towards the mean position
$a = 100 x^{3}$	Not SHM	Not proportional to $x$ and no minus sign

17.2SHM of a mass-spring systemOften examined

A mass attached to a spring is the standard example of SHM. A block of mass $m$ is fixed to one end of a spring; the other end is attached to a rigid support. The block rests on a smooth (frictionless) horizontal surface, and the mass of the spring is ignored. With no outside force, the spring is neither stretched nor compressed, and the block stays at its equilibrium (mean) position, which we call $x = 0$ .

Mass

m

on a smooth horizontal surface. (a) Pulled right by

x

: the spring pulls left. (b) At

x = 0

the spring force is zero. (c) Pushed left: the spring pushes right. The force always points towards

x = 0

The same system in motion. The force

F

(coloured) always points towards

x = 0

and is largest at the extremes, where the velocity

v

is zero. At

x = 0

the force is zero and the speed is greatest.

Restoring force: Hooke's law

Pull the block to the right through a displacement $x$ with a force $F$ . Because the spring is elastic, it pulls back on the block with a force proportional to $x$ and opposite to it. This is the restoring force, and it obeys Hooke's law:

F = - k x (17.1)

$k$ is the spring constant: the force needed per unit extension (or compression) of the spring. Its SI unit is $N m^{- 1}$ , and a stiff spring has a large $k$ . The negative sign shows that the spring force is always opposite to the displacement: when the block is to the right ( $x > 0$ ), the force points left, towards the mean position.

Why the block keeps oscillating

Released at $x = + x_{0}$ , the block is pulled towards the mean position and speeds up.
At $x = 0$ the spring force is zero, but the block is moving at its greatest speed. Because of inertia it cannot stop there and moves past the mean position.
Now the spring is compressed. It pushes the block back, so the block slows down and stops at $x = - x_{0}$ , the same distance on the left (no energy is lost on a frictionless surface).
The compressed spring pushes the block back through the mean position and the cycle repeats. The block keeps changing its direction, so it is accelerating all the time. By Newton's second law this acceleration is caused by the spring force: $F = ma$ (17.2).

Derivation Motion of a mass attached to a spring is SHM

Aim. Show that the acceleration of the block is proportional to its displacement and directed towards the mean position, and find the angular frequency.

Assumptions.

The spring is massless and obeys Hooke's law.
The surface is frictionless and air resistance is negligible, so no energy is lost.
The block moves only along the line of the spring (the $x$ -axis).

The only horizontal force on the block is the spring force. By Hooke's law it is proportional to the displacement $x$ and opposite to it.
$F = - k x (17.1)$
The block has mass $m$ , so by Newton's second law the same force gives it an acceleration $a$ .
$F = ma (17.2)$
Both equations describe the same force, so equate them.
$ma = - k x$
Divide both sides by $m$ to get the acceleration.
$a = - \frac{k}{m} x (17.3)$
$k$ and $m$ are fixed positive quantities for a given spring and block, so $k / m$ is a positive constant. Write it as $ω^{2}$ (a square is always positive).
$\frac{k}{m} = ω^{2} ⟹ a = - ω^{2} x$
Because $ω^{2}$ is constant, the acceleration is directly proportional to the displacement. The minus sign shows that it is always opposite to the displacement, that is, towards the mean position. These are exactly the two conditions of SHM.
$a \propto - x$
Take the square root of $k / m = ω^{2}$ to get the angular frequency of the oscillating mass.
$ω = \frac{k}{m} (17.4)$

Result

a = - \frac{k}{m} x = - ω^{2} x, ω = \frac{k}{m}

The mass-spring system satisfies the definition of SHM. How quickly it oscillates depends only on the stiffness $k$ and the mass $m$ , not on how far the block was pulled.

Derivation Time period and frequency of a mass-spring system

Aim. Express the time period $T$ and frequency $f$ in terms of $m$ and $k$ .

Angular frequency is related to frequency and time period by its definition.
$ω = 2 π f = \frac{2 π}{T}$
Make $T$ the subject.
$T = \frac{2 π}{ω}$
Substitute $ω = k / m$ from (17.4). Dividing by $k / m$ is the same as multiplying by $m / k$ .
$T = \frac{2 π}{k / m} = 2 π \frac{m}{k}$
Frequency is the reciprocal of the time period.
$f = \frac{1}{T} = \frac{1}{2 π} \frac{k}{m}$

Result

T = 2 π \frac{m}{k}, f = \frac{1}{2 π} \frac{k}{m}

A heavier mass oscillates more slowly ( $T \propto m$ ) because it has more inertia. A stiffer spring oscillates faster ( $T \propto 1/ k$ ) because it pulls back harder. The amplitude does not appear, so the period is the same for large and small oscillations.

How the period of a mass-spring system responds to changes.
Change made	Effect on $T = 2 π m / k$
Mass doubled ( $2 m$ )	$T$ becomes $2 T = 1.41 T$
Mass made 4 times ( $4 m$ )	$T$ becomes $2 T$
Spring constant doubled ( $2 k$ )	$T$ becomes $T / 2 = 0.71 T$
Amplitude doubled	No change
Same system taken to the Moon	No change ( $g$ does not appear)

Example 17.1: a fish on a spring scale

A fish is hung on a spring scale. (a) What is the constant of the spring in such a scale if the spring stretches

8.0 cm

for a

10.0 kg

load? (b) What is the mass of the fish that stretches the spring

5.5 cm

Given

$x_{1} = 8.0 cm = 0.080 m$
$m_{1} = 10.0 kg$
$x_{2} = 5.5 cm = 0.055 m$
$g = 9.8 m s^{- 2}$

To find

(a) $k$
(b) mass of the fish $m_{2}$

(a) When the load hangs at rest, the spring force balances its weight, so in size $k x_{1} = m_{1} g$ .
$k = \frac{m _{1} g}{x _{1}} = \frac{10.0 \times 9.8}{0.080} = 1225 N m^{- 1}$
(b) The same spring now balances the weight of the fish: $k x_{2} = m_{2} g$ .
$m_{2} = \frac{k x _{2}}{g} = \frac{1225 \times 0.055}{9.8} = 6.875 kg \approx 6.9 kg$

k = 1225 N m^{- 1}, m_{2} \approx 6.9 kg

Assignment 17.1

An object with mass

500 g

is suspended from a spring. The spring is stretched by

9.8 cm

. Calculate the spring constant.

Change to SI units: $m = 0.500 kg$ and $x = 0.098 m$ . At rest the spring force balances the weight, so $k x = m g$ .
$k = \frac{m g}{x} = \frac{0.500 \times 9.8}{0.098} = 50 N m^{- 1}$

k = 50 N m^{- 1}

17.3Uniform circular motion and SHMOften examined

There is a close link between circular motion and SHM. Imagine a ball moving round a circle at a steady rate, lit from above so that its shadow falls on a straight line (a diameter). The ball goes round and round, but its shadow moves back and forth along the line, and the shadow's motion is exactly SHM. This picture lets us find the displacement, velocity and acceleration in SHM from what we already know about circular motion.

Let a point P move on a circle of radius x_0 in the xy-plane with constant angular velocity \omega. Let Q be the projection of P on the x-axis (the foot of the perpendicular from P). As P goes round, Q oscillates along the diameter between -x_0 and +x_0.

P moves round a circle of radius x_0 with constant angular velocity \omega. Q, the foot of the perpendicular from P on the X-axis, oscillates along the diameter. In triangle OPQ, OQ=x and OP=x_0.

The radius x_0 of the circle is the amplitude of Q.
One revolution of P takes the same time as one oscillation of Q, so both have the same time period T.
The angular velocity \omega of P is the angular frequency of Q's oscillation, because \omega=2\pi/T for both.

Derivation Instantaneous displacement of the projection Q

Aim. Find the displacement x of Q at time t.

Assumptions.

P moves anticlockwise with constant angular velocity \omega on a circle of radius x_0.
Time is counted from the moment P is on the x-axis at +x_0, so \theta=0 at t=0.

In time t the radius OP turns through angle \theta. For uniform circular motion, angle turned = angular velocity \times time.
\theta=\omega t
Triangle OPQ has a right angle at Q, hypotenuse OP=x_0 and base OQ=x (the displacement of Q). Use the cosine ratio.
\cos\theta=\dfrac{OQ}{OP}=\dfrac{x}{x_0}
Make x the subject and put \theta=\omega t.
x=x_0\cos\theta=x_0\cos\omega t\qquad(17.5)

Resultx=x_0\cos\omega t

The cosine swings between +1 and -1, so x swings between +x_0 and -x_0. It repeats every time \omega t grows by 2\pi, that is, every T=2\pi/\omega. So Q oscillates with amplitude x_0 and angular frequency \omega.

Derivation Instantaneous velocity of the projection Q

Aim. Find the velocity of Q in terms of time and in terms of displacement.

The velocity v_P=x_0\omega of P is along the tangent. Its horizontal component, v=v_P\sin\theta, is the velocity of Q.

P moves along the tangent with constant speed v=r\omega, and here r=x_0.
v_P=x_0\omega
Q always stays directly below (or above) P, so Q moves with the horizontal component of P's velocity. The tangent is perpendicular to OP and the vertical line through P is perpendicular to the x-axis, so the angle between v_P and the vertical is \theta, and between v_P and the horizontal it is 90^\circ-\theta.
v=v_P\cos(90^\circ-\theta)
Use \cos(90^\circ-\theta)=\sin\theta.
v=v_P\sin\theta=x_0\omega\sin\theta\qquad(17.6)
To write v in terms of x, use \sin^{2}\theta+\cos^{2}\theta=1.
v=x_0\omega\sqrt{1-\cos^{2}\theta}
From (17.5), \cos\theta=x/x_0. Substitute and take x_0 inside the root.
v=x_0\omega\sqrt{1-\dfrac{x^{2}}{x_0^{2}}}=\omega\sqrt{x_0^{2}-x^{2}}\qquad(17.7)
Direction: while P is in the upper half of the circle, Q moves towards -x, so its velocity is negative. Q passes each point once in each direction, hence the \pm sign.
v=-x_0\omega\sin\omega t,\qquad v=\pm\,\omega\sqrt{x_0^{2}-x^{2}}

Resultv=\omega\sqrt{x_0^{2}-x^{2}}\ \ (\text{speed}),\qquad v_{\max}=v_0=\omega x_0\ \text{at}\ x=0

The speed of Q is largest, \omega x_0, at the mean position (x=0) and zero at the extreme positions (x=\pm x_0), where Q stops and turns back.

Derivation Instantaneous acceleration of the projection Q

Aim. Show that the acceleration of Q is a=-\omega^{2}x.

The centripetal acceleration a_P=x_0\omega^2 of P points towards O. Its horizontal component, a=-a_P\cos\theta, is the acceleration of Q and also points towards O.

P moves in a circle at constant speed, so its only acceleration is the centripetal acceleration r\omega^{2}, directed from P towards the centre O.
a_P=x_0\omega^{2}
Q moves with the horizontal component of P's acceleration. The line PO makes angle \theta with the horizontal (alternate angles), so the component has size a_P\cos\theta. It points towards O, opposite to the displacement x of Q, so it carries a minus sign.
a=-a_P\cos\theta=-x_0\omega^{2}\cos\theta
Use x=x_0\cos\theta from (17.5).
a=-\omega^{2}\left(x_0\cos\theta\right)=-\omega^{2}x\qquad(17.8)

Resulta=-\omega^{2}x=-x_0\omega^{2}\cos\omega t

\omega^{2} is constant, so the acceleration of Q is proportional to its displacement and always directed towards the mean position O. Therefore the motion of Q is SHM: when a body moves in a circle, its projection on a diameter executes SHM.

Result. When a body moves in a circle with uniform speed, its projection on a diameter of the circle executes simple harmonic motion. The acceleration of any body in SHM can therefore be calculated from a=-\omega^{2}x (17.8).

Summary of the equations of SHM

Here v_0=\omega x_0 is the maximum speed. The book uses the first column; the SLO equation v=v_0\cos\omega t belongs to the second column.
Quantity	Timing from an extreme (x=+x_0 at t=0)	Timing from the mean position (x=0 at t=0, moving towards +x)
Displacement	x=x_0\cos\omega t	x=x_0\sin\omega t
Velocity	v=-v_0\sin\omega t	v=v_0\cos\omega t
Acceleration	a=-\omega^{2}x_0\cos\omega t	a=-\omega^{2}x_0\sin\omega t
In terms of x	v=\pm\omega\sqrt{x_0^{2}-x^{2}},\ a=-\omega^{2}x	The same

\begin{aligned}v_{\max}&=v_0=\omega x_0\ \ (\text{at the mean position})\\ a_{\max}&=\omega^{2}x_0\ \ (\text{at the extreme positions})\end{aligned}

Example 17.2: velocity and acceleration of a mass on a spring

A body with mass 800\,\mathrm{g} attached to a spring vibrates with amplitude 30\,\mathrm{cm}. The restoring force is 60\,\mathrm{N} when the displacement is 0.30\,\mathrm{m}. (a) Find out its angular frequency. (b) Also calculate the magnitude of its velocity and acceleration at x=12\,\mathrm{cm}.

Given

m = 800\,\mathrm{g}=0.8\,\mathrm{kg}
x_0 = 30\,\mathrm{cm}=0.30\,\mathrm{m}
F = 60\,\mathrm{N}\ \text{at}\ x=0.30\,\mathrm{m}
x = 12\,\mathrm{cm}=0.12\,\mathrm{m}

To find

(a) \omega
(b) v and a at x=0.12\,\mathrm{m}

(a) Find the spring constant from Hooke's law (size only).
k=\dfrac{F}{x}=\dfrac{60}{0.30}=200\,\mathrm{N\,m^{-1}}
Angular frequency of a mass-spring system.
\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{200}{0.8}}=\sqrt{250}=15.8\,\mathrm{rad\,s^{-1}}
(b) Speed at x=0.12\,\mathrm{m}.
v=\omega\sqrt{x_0^{2}-x^{2}}=15.81\sqrt{0.30^{2}-0.12^{2}}=15.81\sqrt{0.0756}=15.81\times0.275=4.35\,\mathrm{m\,s^{-1}}
Size of the acceleration at x=0.12\,\mathrm{m}. Note that \omega^{2}=k/m=250\,\mathrm{s^{-2}} exactly.
a=\omega^{2}x=250\times0.12=30\,\mathrm{m\,s^{-2}}

\omega=15.8\,\mathrm{rad\,s^{-1}},\quad v=4.35\,\mathrm{m\,s^{-1}},\quad a=30\,\mathrm{m\,s^{-2}}

Assignment 17.2

Time period of a mass attached at the end of a spring is 0.40\,\mathrm{s}. Find out the magnitude of acceleration when the displacement is 4\,\mathrm{cm}.

Find \omega from the time period.
\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.40}=15.7\,\mathrm{rad\,s^{-1}}
Use a=\omega^{2}x (size only) with x=0.04\,\mathrm{m}.
a=\omega^{2}x=(15.71)^{2}(0.04)=246.7\times0.04=9.87\,\mathrm{m\,s^{-2}}

a\approx9.9\,\mathrm{m\,s^{-2}}

17.4PhaseOften examined

The equation x=x_0\cos\theta tells us where Q is. The angle \theta=\omega t inside the cosine is called the phase. The phase tells us two things at once: the displacement of the body and the direction in which it is moving. The displacement alone is not enough. For example, x=0 at \theta=\pi/2 (moving towards -x) and again at \theta=3\pi/2 (moving towards +x); the phase tells these two moments apart.

Phase: The angle \theta=\omega t which specifies the displacement as well as the direction of a point executing SHM is called the phase of the motion.
Initial phase (\varphi): The phase at t=0. It tells where in its cycle the oscillation starts. It is also called the phase constant.
Phase difference: The difference between the phases of two oscillations of the same frequency. It shows how far one oscillation is ahead of the other.

In general, the displacement is written with the initial phase included:

x=x_0\cos(\omega t+\varphi)\qquad(17.9)

Here \omega t+\varphi is the phase at time t and \varphi is the initial phase. Phase is measured in radians (or degrees). One complete oscillation is a phase change of 2\pi\,\mathrm{rad}=360^\circ, after which everything repeats.

The phase fixes both the position and the direction of motion.
Phase \theta	Time (\varphi=0)	x=x_0\cos\theta	Position and motion
0	0	+x_0	Extreme position, momentarily at rest
\pi/2	T/4	0	Mean position, moving towards -x at maximum speed
\pi	T/2	-x_0	Other extreme, momentarily at rest
3\pi/2	3T/4	0	Mean position, moving towards +x at maximum speed
2\pi	T	+x_0	Back at the start: one oscillation complete

Physical significance of phase

To see what the initial phase does, draw displacement-time graphs for different values of \varphi. Since x is periodic, each graph repeats every time period T.

\varphi=0: x=x_0\cos\omega t. The body is at the extreme positions at t=0,\ T/2,\ T (\theta=0,\ \pi,\ 2\pi) and at the mean position at t=T/4,\ 3T/4 (\theta=\pi/2,\ 3\pi/2).
\varphi=90^\circ: x=x_0\cos(\omega t+90^\circ)=-x_0\sin\omega t. Now the body is at the mean position at t=0,\ T/2,\ T and at the extreme positions at t=T/4,\ 3T/4.
\varphi=180^\circ: x=x_0\cos(\omega t+180^\circ)=-x_0\cos\omega t. The body starts from the other extreme, -x_0.

Phase \varphi=0 and \varphi=90^\circ

The 90^\circ curve reaches every stage (for example -x_0) a quarter period (T/4) earlier, so it leads the \varphi=0 curve by 90^\circ.

Phase \varphi=0 and \varphi=180^\circ

With a phase difference of 180^\circ the two motions are always opposite: when one is at +x_0 the other is at -x_0 (out of phase).

The curve for \varphi=90^\circ leads the curve for \varphi=0 by 90^\circ: it reaches each stage of the motion a quarter period earlier. In the same way the curve for \varphi=180^\circ leads the curve for \varphi=0 by 180^\circ.

In phase: When the phase difference between two oscillating systems is 0^\circ or 360^\circ, they are oscillating in phase. They reach their extremes together and always move in the same direction.
Out of phase: When the phase difference between two oscillating systems is 180^\circ, they are oscillating out of phase. When one is at +x_0 the other is at -x_0; they always move in opposite directions.

17.5Graphs of displacement, velocity and accelerationOften examined

Graphs show at a glance how displacement, velocity and acceleration change during SHM. Use the book's timing: the body starts from the extreme position +x_0 at t=0. Then, from section 17.3,

\begin{aligned} x&=x_0\cos\omega t &&\text{(i)}\\ v&=-x_0\omega\sin\omega t &&\text{(ii)}\\ a&=-\omega^{2}x=-x_0\omega^{2}\cos\omega t &&\text{(iii)}\end{aligned}

Displacement against time

Displacement is a cosine curve between +x_0 and -x_0.

Velocity against time

Velocity is zero at the extremes (t=0,\ T/2,\ T) and has its largest size \omega x_0 at the mean position (t=T/4,\ 3T/4).

Acceleration against time

Acceleration is always opposite to displacement: its curve is the displacement curve turned upside down.

Values at each quarter period.
t	0	T/4	T/2	3T/4	T
Phase \omega t	0	\pi/2	\pi	3\pi/2	2\pi
x	+x_0	0	-x_0	0	+x_0
v	0	-\omega x_0	0	+\omega x_0	0
a	-\omega^{2}x_0	0	+\omega^{2}x_0	0	-\omega^{2}x_0

What the graphs show

Displacement varies between +x_0 and -x_0; its maximum value is the amplitude x_0.
Velocity is maximum (\pm\omega x_0) at the mean position and zero at the extreme positions.
Acceleration is maximum (\omega^{2}x_0 in size) at the extreme positions and zero at the mean position.
The **phase difference between velocity and displacement is \pi/2** (90^\circ): velocity is largest when the displacement is zero and zero when the displacement is largest. Velocity leads displacement by \pi/2.
The **phase difference between acceleration and displacement is \pi** (180^\circ): acceleration is always opposite to displacement.
So acceleration leads velocity by \pi/2, and velocity leads displacement by \pi/2.

x=x_0\cos\omega t,\qquad v=\omega x_0\cos\!\left(\omega t+\tfrac{\pi}{2}\right),\qquad a=\omega^{2}x_0\cos(\omega t+\pi)

Written as cosines, the three equations show the phase leads directly: 0, \pi/2 and \pi.

Acceleration against displacement

A straight line through the origin with negative slope (-\omega^2): acceleration is proportional to displacement and opposite to it. This graph is the signature of SHM.

17.6Simple pendulumOften examined

A simple pendulum is a small heavy bob hanging from a string. Pull the bob a little to one side and let it go: it swings to and fro along the arc of a circle whose centre is the point of suspension. We now show that, for small swings, this motion is SHM, and find its time period.

Ideal simple pendulum: A small but heavy bob of mass m suspended by a light and inextensible string of length l, the other end of which is attached to a fixed frictionless support.
Length of the pendulum (l): The distance from the point of suspension to the centre of the bob.

Forces on the bob at A: tension T along the string and weight mg. The weight splits into mg\cos\theta along the string and mg\sin\theta along the arc towards O. Only mg\sin\theta is unbalanced; it is the restoring force.

Forces on the bob

Let the bob be at A, at a displacement x (measured along the arc) from the mean position O, with the string at angle \theta to the vertical (\angle OPA=\theta). Two forces act on the bob: the tension T in the string and its weight mg acting vertically downwards. Resolve the weight into two rectangular components:

mg\cos\theta along the string, away from the support. The tension balances this component, so there is no motion along the string.
mg\sin\theta along the tangent to the arc, towards O. Nothing balances it, so it is the net force on the bob: the restoring force that accelerates it towards the mean position.

Derivation Simple pendulum executes SHM; its time period

Aim. Show that the acceleration of the bob is proportional to its displacement and directed towards O, and find T.

Assumptions.

The bob is a point mass and the string is light and inextensible.
The support is frictionless and air resistance is negligible.
The angle \theta is small (less than about 10^\circ or 0.2\,\mathrm{rad}), so \sin\theta\approx\theta and the arc OA is almost a straight line.

The restoring force on the bob is the tangential component of its weight. The minus sign shows that it acts towards O, opposite to the displacement.
F=-mg\sin\theta\qquad\text{(i)}
For a small angle, \sin\theta\approx\theta when \theta is in radians (see the table below).
F=-mg\,\theta\qquad\text{(ii)}
An angle in radians equals arc length divided by radius. The arc OA is the displacement x and the radius is the length l.
\theta=\dfrac{\text{arc }OA}{l}=\dfrac{x}{l}
Substitute this \theta into (ii).
F=-mg\,\dfrac{x}{l}\qquad\text{(iii)}
By Newton's second law the same force gives the bob an acceleration a.
F=ma\qquad\text{(iv)}
Equate (iii) and (iv) and cancel m.
ma=-mg\,\dfrac{x}{l}\ \Longrightarrow\ a=-\dfrac{g}{l}\,x\qquad\text{(v)}
g and l are constants, so the acceleration is proportional to the displacement and directed towards O. The motion is SHM.
a\propto -x\qquad\text{(vi)}
Compare (v) with the SHM equation a=-\omega^{2}x (vii) to read off \omega.
\omega^{2}=\dfrac{g}{l}\ \Longrightarrow\ \omega=\sqrt{\dfrac{g}{l}}\qquad\text{(viii)}
Put (viii) into T=2\pi/\omega.
T=\dfrac{2\pi}{\sqrt{g/l}}=2\pi\sqrt{\dfrac{l}{g}}\qquad(17.10)

ResultT=2\pi\sqrt{\dfrac{l}{g}},\qquad f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

The time period of a simple pendulum depends only on its length and on g. It does not depend on the mass of the bob, and for small swings it does not depend on the amplitude.

Why the angle must be small

Up to about 10^\circ the error is only about 0.5\%, so a pendulum with small swings moves in SHM to a very good approximation. At larger angles the formula for T becomes less accurate.
\theta (degree)	\theta (radian)	\sin\theta	Error in \sin\theta\approx\theta
0	0.0000	0.0000	0\%
2	0.0349	0.0349	0.02\%
4	0.0698	0.0698	0.08\%
6	0.1047	0.1045	0.18\%
8	0.1396	0.1392	0.32\%
10	0.1745	0.1736	0.51\%
20	0.3491	0.3420	2.0\%
30	0.5236	0.5000	4.5\%

What the period depends on

T\propto\sqrt{l}: a longer pendulum swings more slowly. Four times the length gives twice the period.
T\propto 1/\sqrt{g}: where g is smaller (on the Moon, or high on a mountain) the pendulum swings more slowly.
T does not depend on the mass m: a heavier bob feels a larger restoring force, mg\sin\theta, but it also has more inertia, and the two effects cancel.
For small angles, T does not depend on the amplitude.

Example 17.3: length of a pendulum

A simple pendulum completes one vibration in one second. Calculate its length when g=9.8\,\mathrm{m\,s^{-2}}.

One vibration in one second means T=1\,\mathrm{s}. Square both sides of T=2\pi\sqrt{l/g} to remove the root, then make l the subject.
T^{2}=4\pi^{2}\dfrac{l}{g}\ \Longrightarrow\ l=\dfrac{gT^{2}}{4\pi^{2}}
Substitute the values.
l=\dfrac{9.8\times(1)^{2}}{4\times(3.14)^{2}}=\dfrac{9.8}{39.4}=0.248\,\mathrm{m}=24.8\,\mathrm{cm}

l=0.248\,\mathrm{m}\ \ (24.8\,\mathrm{cm})

Assignment 17.3

Find the time periods of a simple pendulum with length 1\,\mathrm{m}, placed on Earth and on Moon. The value of g on the surface of Moon is 1/6th of its value on Earth.

On Earth, g=9.8\,\mathrm{m\,s^{-2}}.
T_E=2\pi\sqrt{\dfrac{1}{9.8}}=2\pi(0.319)=2.01\,\mathrm{s}
On the Moon, g_M=9.8/6=1.63\,\mathrm{m\,s^{-2}}. Since T\propto1/\sqrt{g}, the period becomes \sqrt{6}=2.45 times longer.
T_M=2\pi\sqrt{\dfrac{1}{1.63}}=\sqrt{6}\,T_E=2.45\times2.01=4.92\,\mathrm{s}

T_E\approx2.0\,\mathrm{s},\qquad T_M\approx4.9\,\mathrm{s}

17.7Energy conservation in SHMOften examined

In SHM the body keeps changing its speed and position, but its total energy stays constant. The energy only changes form: it is stored as elastic potential energy when the spring is stretched or compressed, and appears as kinetic energy when the mass moves. We work this out for the mass-spring system; the same conclusion holds for every body in SHM.

Potential energy at displacement x

Derivation Elastic potential energy of a stretched spring

Aim. Find the potential energy stored when the spring is stretched (or compressed) through x.

Assumptions.

The spring obeys Hooke's law and is stretched slowly from 0 to x.

Work done in stretching a spring

The work done is the area under the force-extension line: a triangle of base x and height kx, so W=\tfrac12(x)(kx)=\tfrac12 kx^2.

The work done by the applied force equals the average force times the displacement.
W=F_{\text{av}}\,x\qquad\text{(i)}
By Hooke's law the applied force needed at extension x is
F=kx\qquad\text{(ii)}
The force increases steadily (linearly) from 0 at the start to kx at the end, so its average is half the sum of the first and last values.
F_{\text{av}}=\dfrac{0+kx}{2}=\dfrac12 kx\qquad\text{(iii)}
Substitute (iii) in (i).
W=\left(\dfrac12 kx\right)(x)=\dfrac12 kx^{2}
This work is stored in the spring as elastic potential energy.
\text{P.E.}=\dfrac12 kx^{2}\qquad(17.11)

Result\text{P.E.}=\dfrac12 kx^{2}

P.E. is zero at the mean position and grows as the square of the displacement. It is largest at the extreme positions, where \text{P.E.}_{\max}=\tfrac12 kx_0^{2}.

Kinetic energy at displacement x

Derivation Kinetic energy of the oscillating mass

Aim. Find the kinetic energy of the mass when its displacement is x.

The speed of a body in SHM at displacement x is
v=\omega\sqrt{x_0^{2}-x^{2}}\qquad\text{(iv)}
For the mass-spring system \omega=\sqrt{k/m}, so
v=\sqrt{\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)}\qquad\text{(v)}
At the mean position (x=0) the speed is maximum.
v_0=\sqrt{\dfrac{k}{m}}\,x_0\qquad\text{(vi)}
Put v^{2} from (v) into \text{K.E.}=\tfrac12 mv^{2}. The mass m cancels.
\text{K.E.}=\dfrac12 m\cdot\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)=\dfrac12 k\left(x_0^{2}-x^{2}\right)\qquad(17.13)

Result\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 k\left(x_0^{2}-x^{2}\right)

K.E. is largest at the mean position, \tfrac12 kx_0^{2}, and zero at the extreme positions, where the body stops for an instant.

Total energy is constant

Derivation Total energy of a body in SHM is constant

Aim. Show that \text{K.E.}+\text{P.E.} has the same value at every displacement.

At an extreme position (x=x_0) the block is at rest, so K.E. is zero and all the energy is elastic P.E.
\text{T.E.}=\text{P.E.}_{\max}=\dfrac12 kx_0^{2}\qquad(17.12)
At the mean position (x=0) the spring is neither stretched nor compressed, so P.E. is zero and all the energy is kinetic.
\text{T.E.}=\text{K.E.}_{\max}=\dfrac12 mv_{\max}^{2}=\dfrac12 kx_0^{2}\qquad(17.14)
**At any displacement x**, add the energies from (17.11) and (17.13).
\text{T.E.}=\text{K.E.}+\text{P.E.}=\dfrac12 k\left(x_0^{2}-x^{2}\right)+\dfrac12 kx^{2}
The x^{2} terms cancel, leaving a quantity that does not depend on x.
\text{T.E.}=\dfrac12 kx_0^{2}\qquad(17.15)
Since \omega=\sqrt{k/m}, k=m\omega^{2}. This gives the total energy in terms of \omega.
\text{T.E.}=\dfrac12 m\omega^{2}x_0^{2}\qquad(17.16)

Result\text{T.E.}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}=\text{constant}

The total energy of the mass-spring system is constant and is proportional to the square of the amplitude. This statement of conservation of energy is equally valid for all bodies executing SHM.

Interchange of kinetic and potential energy

The energy is shared differently at each position, but the total never changes.
Position	x	P.E.	K.E.	T.E.
Extreme	\pm x_0	\tfrac12 kx_0^{2} (maximum)	0	\tfrac12 kx_0^{2}
Halfway	\pm\tfrac12 x_0	\tfrac14 of T.E.	\tfrac34 of T.E.	\tfrac12 kx_0^{2}
Equal sharing	\pm x_0/\sqrt{2}	\tfrac12 of T.E.	\tfrac12 of T.E.	\tfrac12 kx_0^{2}
Mean	0	0	\tfrac12 kx_0^{2} (maximum)	\tfrac12 kx_0^{2}

Energy against displacement

P.E. is zero at the mean position and largest at the extremes; K.E. is the opposite. The two are equal at x=\pm x_0/\sqrt{2}, and at every point K.E. + P.E. equals the same total energy.

Going from an extreme position to the mean position, P.E. turns into K.E.; from the mean position to the other extreme, K.E. turns back into P.E. The energy oscillates between the two forms, but their sum stays constant.

Example 17.4: energy of a mass-spring oscillator

A 0.025\,\mathrm{kg} mass is attached to a spring which is displaced through 0.10\,\mathrm{m} to the right of its mean position and then released. Time period of its oscillation is 1.57\,\mathrm{s}. Calculate its (a) angular speed, (b) the total energy, (c) the maximum acceleration.

Given

m = 0.025\,\mathrm{kg}
x_0 = 0.10\,\mathrm{m} released from rest, so this is the amplitude
T = 1.57\,\mathrm{s}

To find

(a) \omega
(b) T.E.
(c) a_{\max}

(a) Angular speed from the time period.
\omega=\dfrac{2\pi}{T}=\dfrac{2(3.14)}{1.57}=4.0\,\mathrm{rad\,s^{-1}}
(b) Total energy.
\text{T.E.}=\dfrac12 m\omega^{2}x_0^{2}=\dfrac12(0.025)(4)^{2}(0.10)^{2}=2.0\times10^{-3}\,\mathrm{J}
(c) The acceleration is largest at the extreme positions, where x=x_0.
a_{\max}=\omega^{2}x_0=(4)^{2}(0.10)=1.6\,\mathrm{m\,s^{-2}}

\omega=4.0\,\mathrm{rad\,s^{-1}},\quad \text{T.E.}=2.0\times10^{-3}\,\mathrm{J},\quad a_{\max}=1.6\,\mathrm{m\,s^{-2}}

Assignment 17.4

Find the amplitude, frequency and time period of an object oscillating at the end of a spring, if the equation for its position at any instant t is given by x=0.25\cos\left(\dfrac{\pi}{8}t\right). Also find the displacement of the object after 2.0\,\mathrm{s}.

Compare with x=x_0\cos\omega t.
x_0=0.25\,\mathrm{m},\qquad \omega=\dfrac{\pi}{8}\,\mathrm{rad\,s^{-1}}=0.393\,\mathrm{rad\,s^{-1}}
Frequency and time period.
f=\dfrac{\omega}{2\pi}=\dfrac{\pi/8}{2\pi}=\dfrac{1}{16}=0.0625\,\mathrm{Hz},\qquad T=\dfrac{1}{f}=16\,\mathrm{s}
Displacement at t=2.0\,\mathrm{s}. The angle is in radians: \tfrac{\pi}{8}\times2=\tfrac{\pi}{4}.
x=0.25\cos\dfrac{\pi}{4}=0.25\times0.707=0.177\,\mathrm{m}

x_0=0.25\,\mathrm{m},\ \ f=0.0625\,\mathrm{Hz},\ \ T=16\,\mathrm{s},\ \ x(2\,\mathrm{s})\approx0.18\,\mathrm{m}

17.8Free and forced oscillationsOften examined

Depending on the situation, oscillations may be free, forced or damped. Free and forced oscillations differ in whether an outside periodic force keeps acting on the oscillator. Damped oscillations, in the next section, are oscillations that lose energy.

Free oscillations

Natural frequency: Every oscillator has a natural frequency of vibration with which it vibrates freely after an initial disturbance. It depends only on the oscillator itself (its mass and stiffness, or its length).
Free oscillations: A body is said to be executing free oscillations when it oscillates under the influence of a restoring force without any external force acting on it.

An ideal simple pendulum, slightly displaced and released, oscillates freely with its natural frequency f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}; change its length and you change its natural frequency. A plucked guitar string keeps vibrating after it is released, at its natural frequency, and this gives the particular note you hear. The natural frequency of the string depends on its length.

Free oscillations have a constant amplitude and a constant period because no external force acts on the oscillator. Ideally, free oscillations do not undergo damping.

A simple pendulum displaced slightly and released.
A mass on a spring pulled aside and released.
A plucked guitar string or a struck tuning fork.
A swing given one push and then left alone.

Forced oscillations

If an external periodic force acts on an oscillator, it can change the amplitude of its oscillations. The external force transfers energy to the oscillator at its own frequency, which is not necessarily the natural frequency of the oscillator. This frequency is called the driving (driven) frequency. After a short time the oscillator vibrates at the driving frequency.

Driving frequency: The frequency of the external periodic force that acts on an oscillator.
Forced oscillations: If the external forces make the object oscillate at the frequency of the applied force rather than at its natural frequency, then such oscillations are called forced oscillations.

A swing that you keep pushing periodically so that it does not lose amplitude.
The pendulum of a clock, driven by a small motor (or by the clock's spring).
A bridge vibrating when soldiers march across it or when wind blows on it.

Difference between free and forced oscillations.
Free oscillations	Forced oscillations
No external periodic force acts after the start	An external periodic force acts all the time
Oscillate at the natural frequency f_0	Oscillate at the driving frequency f
Energy is given once, at the start	Energy is supplied continuously by the driver
Amplitude constant (ideal case)	Amplitude depends on how close f is to f_0; largest when f=f_0
Example: a plucked guitar string	Example: a clock pendulum driven by a motor

17.8Damped oscillationsOften examined

A real pendulum does not swing for ever. After some time it stops oscillating because of air resistance and friction at the support. All real oscillating systems feel such resistive forces, known as damping forces. They take energy away, so the amplitude becomes smaller from one oscillation to the next until the motion stops.

Damping: The effect of resistive forces which dissipate energy from a vibrating object.
Damping forces: Resistive forces that oppose the motion of an oscillator, for example friction between moving parts, air resistance, and internal forces in a spring. They turn the oscillator's mechanical energy into heat.
Damped oscillations: If the amplitude of oscillations decreases under damping forces, then such oscillations are called damped oscillations. Equivalently: when an oscillator undergoes oscillations before coming to rest under the action of a damping force, such oscillations are called damped oscillations.

As an oscillator vibrates, it does work against the force of friction. This work comes out of the oscillator's energy, which gradually decreases. Since the total energy is \tfrac12 kx_0^{2}, a smaller energy means a smaller amplitude.

Undamped and damped oscillations

Damping does not change the time between swings much, but each swing is smaller than the one before, until the motion dies out.

Light, heavy and critical damping

How an oscillator comes to rest depends on how strong the damping force is compared with the restoring force.

Light damping: The damping is light when the amplitude of oscillations decreases gradually with time. The system completes many oscillations before it stops. Example: a swing in a playground left to oscillate freely, or a pendulum in air.
Heavy damping: A very large resistive force makes the oscillator move back to its equilibrium position gradually, without completing a single oscillation. It takes a long time to come to rest. Example: a mass on a spring placed in a thick, viscous liquid such as oil or honey.
Critical damping: When an oscillator comes to rest without any oscillation in the shortest time under a damping force, the damping is called critical damping. Example: the shock absorbers of a car.

Light, critical and heavy damping

All three start from the same displacement x_0 at rest. T is the natural period of the undamped oscillator.

Comparison of the three kinds of damping.
Type	Resistive force	Oscillation	Time to settle
Light	Small	Yes, with slowly decreasing amplitude	Long (many oscillations)
Critical	Just enough	None	Shortest possible
Heavy	Very large	None	Long (creeps back slowly)

Importance of critical damping in a car suspension

A car has springs between the wheels and the frame (body). When a wheel goes over a bump, the spring is compressed and then pushes back. Without damping the car would move up and down violently for some time, which is uncomfortable, may injure the passengers and makes the car hard to control.

Car suspension: the spring carries the load and absorbs the jolt from a bump; the shock absorber (a piston moving through oil) removes the energy so the body does not keep bouncing.

Energy-absorbing devices called shock absorbers are fitted parallel to the springs. They are fluid-filled tubes in which a piston moves through oil, and they turn the simple harmonic motion of the springs into damped motion.
With critical damping the car body returns to its normal level in the shortest time without oscillating, so the ride is smooth and the tyres stay in contact with the road.
With light damping (for example worn-out shock absorbers) the car keeps bouncing after every bump.
With heavy damping the suspension is too stiff and slow: it has not recovered from one bump before the next one arrives, so the jolts pass straight to the passengers.

Modern suspensions are set up so that the shock absorbers absorb the spring's energy and remove the vibration within a single oscillation. This prevents the car from continually bouncing.

17.9ResonanceOften examined

A driving force is always needed to start the oscillations of any oscillator, damped or undamped, by supplying the initial energy for the motion. The driving force has its own frequency, and it makes the oscillator vibrate at this driving frequency rather than at its natural frequency. The size of the resulting oscillation depends on how close the driving frequency is to the natural frequency.

Start with a driving frequency near zero and increase it slowly. At first the oscillator vibrates with a small amplitude. As the driving frequency comes closer to the natural frequency, the driving force transfers energy more efficiently and the amplitude grows. When the driving frequency equals the natural frequency, the amplitude reaches its maximum value. This situation is called resonance. If the driving frequency is increased further, the amplitude falls again.

Resonance: Resonance is the phenomenon in which the amplitude of vibration of an oscillator attains its maximum value when the driving frequency becomes equal to the natural frequency of the oscillator.
Resonance frequency: The natural frequency of the oscillator, at which resonance occurs.

\text{Resonance:}\qquad f_{\text{driving}}=f_0\quad\Longrightarrow\quad\text{amplitude is maximum}

At the resonance frequency the efficiency of energy transfer from the driving force to the oscillator is maximum. The force is always in step with the motion: it pushes in the direction in which the body is already moving, so it does positive work in every cycle and energy keeps building up.

Demonstration with pendulums

Five pendulums hang from one rod. A and B have length l, C and D have length L. When the driver E has length l and is set swinging, A and B pick up large swings (resonance); C and D hardly move.

Five pendulums A, B, C, D and E hang from the same horizontal rod. A and B have the same length l; C and D have the same length L. E is a heavy driver pendulum whose length can be changed.
Make the length of E equal to that of A and B, and set E vibrating in a direction perpendicular to the plane of the paper. It shakes the rod at its own frequency. After some time A and B start large vibrations by themselves, but C and D remain almost at rest, because E has the same length, and so the same frequency, as A and B.
Now make E equal in length to C and D and set it vibrating again. This time C and D start vibrating strongly while A and B remain nearly at rest.

Effect of damping on resonance

Damping reduces the maximum amplitude of an oscillator at its resonance frequency and broadens the resonance curve. A graph of the amplitude of a damped oscillator against the frequency of the driving force shows this.

Amplitude against driving frequency

Resonance: the amplitude is largest when the driving frequency f equals the natural frequency f_0. Less damping gives a taller and narrower (sharper) peak.

Small damping: a tall, narrow peak. The amplitude is very large at f_0 and falls quickly on either side; the resonance is sharp.
Medium damping: a lower and wider peak.
Heavy damping: a low, broad peak. The amplitude at f_0 is only a little larger than at other frequencies; the resonance is flat (not sharp).

Frequency response: The way the amplitude of a forced oscillator changes with the driving frequency, that is, the shape of the amplitude-frequency curve.
Sharpness of resonance: How quickly the amplitude falls as the driving frequency moves away from the natural frequency. A sharp resonance has a tall, narrow peak.

Factors that decide the frequency response and sharpness

Natural frequency of the oscillator. It fixes where the peak lies. It depends on the oscillator itself: \sqrt{k/m} for a spring, \sqrt{g/l} for a pendulum, and the coil and capacitor for a radio circuit.
Amount of damping. This is the main factor for sharpness. With little damping only a little energy is lost in each cycle, so energy builds up to a large amplitude near f_0 and the peak is tall and narrow. With more damping, energy is lost quickly and the peak is low and broad.
Strength of the driving force. A larger driving force raises the whole curve (a larger amplitude at every frequency) but does not change the position of the peak.

17.9Resonance in daily lifeOften examined

Resonance plays a very important role in daily life. Sometimes we use it on purpose; sometimes things must be designed to avoid it.

Where resonance is useful

Tuning a radio (electrical resonance). When we turn the tuning knob of a radio, we change the natural frequency of the electrical circuit of the receiver. When it becomes equal to the transmission frequency of a radio station, resonance occurs. The energy absorbed from that station is then maximum, and it is the only station we hear.
Magnetic resonance imaging (MRI). In this medical diagnostic tool, atomic nuclei (mostly hydrogen nuclei) are made to oscillate by radio waves of the order of 100\,\mathrm{MHz}. At resonance the nuclei absorb maximum energy, and the pattern of energy absorbed is used to produce computer-enhanced images of the inside of the body.
Heating food in a microwave oven. The oven uses microwaves with a frequency similar to the natural frequency of vibration of water (or fat) molecules. The water molecules in the food absorb maximum energy from the microwaves and oscillate strongly, which heats the food. Plastic or glass containers do not heat up in the oven, since they do not contain water molecules.
Resonance in a guitar. When a guitarist strikes the strings, the vibration passes to the hollow wooden box. The box resonates and the sound gets amplified.
Pushing a swing. Small pushes given at the natural frequency of the swing build up a large amplitude.

Where resonance must be avoided

Bridges. Soldiers marching on a bridge are ordered to break their steps. The rhythmic march applies a periodic force; if its frequency equals the natural frequency of the bridge, the amplitude of vibration grows (resonance) and the bridge may collapse. The most studied example is the collapse of the Tacoma Narrows Bridge (USA) in 1940: a strong, continuous wind drove oscillations of the bridge deck that increased in amplitude until it broke apart.
Airplane wings. The wing is a very flexible part of an airplane. If gusts of wind vibrate it at a frequency equal to its natural structural frequency, resonance occurs and the amplified vibrations can destroy the wing. Wings are therefore designed so that their natural frequency does not match the frequencies of external vibrations.

17.9Applications of resonance and standing waves

When a wave meets its own reflection, the two overlap and form a standing wave. A standing wave has points where there is no vibration, called nodes, and points where the vibration is largest, called antinodes. A standing wave builds up to a large amplitude only at particular resonant frequencies that suit the size of the system. Several devices use such resonant standing waves. (Knowledge of wave harmonic modes is not required.)

Node: A point of a standing wave where the oscillation (of the particles, or of the pressure in a sound wave) is always zero.
Antinode: A point of a standing wave where the oscillation has its largest amplitude.

Rubens tube

A Rubens tube (also known as a standing wave flame tube) is used to demonstrate acoustic standing waves. It was invented by the German physicist Heinrich Rubens in 1905. It consists of a metal pipe with holes drilled along the top and sealed at both ends. One sealed end is attached to a small speaker driven by a frequency generator, while the other end is connected to a supply of a flammable gas. The pipe is filled with the gas, and the gas leaking from the holes is lit to give a row of flames.

Rubens tube: a standing sound wave in the gas-filled tube makes the flames tall where the pressure swings most and short where it hardly changes. Neighbouring tall flames are \lambda/2 apart.

The frequency generator is set to a resonance frequency of the tube (based on its dimensions). A standing sound wave forms in the gas inside the tube.
The standing wave creates points where the pressure oscillates strongly (higher and lower than normal) and points where it hardly changes. More gas escapes from the holes where the pressure swings are large, so the flames are high there; less gas escapes where the pressure hardly changes, so the flames are low.
The row of flames therefore shows the shape of the standing wave. It shows graphically the relationship between sound waves and sound pressure, acting as a primitive oscilloscope.
The wavelength can be measured with a ruler: two neighbouring high flames are \lambda/2 apart (a high flame and the next low flame are \lambda/4 apart). Knowing the frequency f, the speed of sound in the gas is v=f\lambda.

Acoustic levitation

Acoustic levitation is a method for suspending matter in air against gravity by using acoustic pressure from high-intensity sound waves. A sound source (transducer) sends sound towards a reflector; the incident and reflected waves create a standing wave, and its nodes can trap small particles or droplets, which then float without any support.

Acoustic levitation: sound from the source reflects back and forms a standing wave. Small beads are held at the pressure nodes, against gravity.

Acoustic levitation provides a container-less environment for some experiments: a sample can be studied, heated or mixed without touching, and being contaminated by, the walls of a container.

Chladni plates

A Chladni plate consists of a flat metal sheet (usually circular or square) mounted on a central stalk attached to a strong base. The plate is made to vibrate, for example by drawing a violin bow along its edge or with a speaker. When it oscillates in a particular mode of vibration, its nodes and antinodes form a complex but symmetrical pattern. Sand sprinkled on the plate vibrates away from the antinodes and gathers at the nodes, so the pattern becomes visible. Different frequencies give different patterns.

Chladni plate: a metal plate fixed at its centre is made to vibrate (with a bow or a speaker). Sand is thrown off the vibrating antinodes and settles on the still nodal lines, drawing a symmetric pattern.

Resonance and standing waves at work.
Device	What resonates	What we see or use
Rubens tube	Standing sound wave in a gas-filled tube	Flame heights show where the pressure swings most and least; gives \lambda
Chladni plate	Metal plate vibrating in one of its modes	Sand collects on the nodal lines and shows the pattern
Acoustic levitation	Standing sound wave between a source and a reflector	Small objects held at the nodes without contact

Formula sheet

Frequency

f=\dfrac{n}{t}=\dfrac{1}{T}

n: number of oscillations in time t; T: time period

Unit: \mathrm{Hz} (\mathrm{s^{-1}})

Angular frequency

\omega=2\pi f=\dfrac{2\pi}{T}

f: frequency; T: time period

Unit: \mathrm{rad\,s^{-1}}

Time period in terms of f and \omega

T=\dfrac{1}{f}=\dfrac{2\pi}{\omega}

f: frequency; \omega: angular frequency

Unit: \mathrm{s}

Hooke's law (restoring force)

F=-kx

k: spring constant; x: displacement from the mean position; minus sign: force opposite to x

Units: F in \mathrm{N}; k in \mathrm{N\,m^{-1}}

Condition for SHM

a=-\omega^{2}x

a: acceleration; x: displacement; \omega: angular frequency

Unit: \mathrm{m\,s^{-2}}

Acceleration of a mass on a spring

a=-\dfrac{k}{m}\,x

k: spring constant; m: mass; x: displacement

Unit: \mathrm{m\,s^{-2}}

Angular frequency of a mass-spring system

\omega=\sqrt{\dfrac{k}{m}}

k: spring constant; m: mass

Unit: \mathrm{rad\,s^{-1}}

Time period of a mass-spring system

T=2\pi\sqrt{\dfrac{m}{k}}

m: oscillating mass; k: spring constant

Unit: \mathrm{s}

Frequency of a mass-spring system

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k: spring constant; m: mass

Unit: \mathrm{Hz}

Displacement in SHM

x=x_0\cos\omega t

x_0: amplitude; \omega t=\theta: phase (timing from an extreme position)

Unit: \mathrm{m}

Displacement with initial phase

x=x_0\cos(\omega t+\varphi)

\varphi: initial phase (phase at t=0)

Unit: \mathrm{m}

Velocity in terms of displacement

v=\pm\,\omega\sqrt{x_0^{2}-x^{2}}

x_0: amplitude; x: displacement; sign gives the direction

Unit: \mathrm{m\,s^{-1}}

Velocity in terms of time

v=-v_0\sin\omega t\quad\text{or}\quad v=v_0\cos\omega t

v_0=\omega x_0; first form when timing starts at an extreme, second when it starts at the mean position

Unit: \mathrm{m\,s^{-1}}

Maximum speed

v_{\max}=v_0=\omega x_0

at the mean position

Unit: \mathrm{m\,s^{-1}}

Maximum acceleration

a_{\max}=\omega^{2}x_0

at the extreme positions

Unit: \mathrm{m\,s^{-2}}

Acceleration of a simple pendulum

a=-\dfrac{g}{l}\,x,\qquad \omega=\sqrt{\dfrac{g}{l}}

g: acceleration due to gravity; l: length; x: arc displacement (small angles)

Units: \mathrm{m\,s^{-2}}; \mathrm{rad\,s^{-1}}

Time period of a simple pendulum

T=2\pi\sqrt{\dfrac{l}{g}}

l: length from support to centre of bob; g: acceleration due to gravity

Unit: \mathrm{s}

Elastic potential energy

\text{P.E.}=\dfrac12 kx^{2}

k: spring constant; x: displacement

Unit: \mathrm{J}

Kinetic energy in SHM

\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 k\left(x_0^{2}-x^{2}\right)

m: mass; v: speed; x_0: amplitude; x: displacement

Unit: \mathrm{J}

Total energy in SHM

\text{T.E.}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}

constant; proportional to the square of the amplitude

Unit: \mathrm{J}

Condition for resonance

f_{\text{driving}}=f_0

f_0: natural frequency of the oscillator

Unit: \mathrm{Hz}

Wavelength from a Rubens tube

\lambda=2\times(\text{distance between neighbouring high flames}),\qquad v=f\lambda

f: frequency of the generator; v: speed of sound in the gas

Unit: \lambda in \mathrm{m}

Key terms

Vibratory (oscillatory) motion: To and fro motion of a body about its mean position.
Mean (equilibrium) position: The position where the net force on the body is zero; displacement is measured from it.
Oscillation (vibration): One complete round trip of a vibrating body about its mean position.
Displacement: The distance of the vibrating body from the mean position, on either side, at any instant. SI unit: m.
Amplitude: The maximum displacement from the mean position, x_0. SI unit: m.
Time period: The time taken to complete one oscillation, T. SI unit: s.
Frequency: The number of oscillations per unit time, f=n/t=1/T. SI unit: Hz.
Hertz: One oscillation per second.
Angular frequency: Angular displacement per unit time, \omega=2\pi f=2\pi/T. SI unit: \mathrm{rad\,s^{-1}}.
Simple harmonic motion: Oscillatory motion in which the acceleration is directly proportional to the displacement from the mean position and always directed towards the mean position, a=-\omega^{2}x.
Restoring force: The force that always acts towards the mean position and tries to bring a displaced body back to it.
Hooke's law: The restoring force of a spring is proportional to its extension or compression and opposite to it, F=-kx.
Spring constant: The force needed per unit extension of a spring, k=F/x. SI unit: \mathrm{N\,m^{-1}}.
Projection of circular motion: When a body moves in a circle with uniform speed, its projection on a diameter executes SHM.
Phase: The angle \theta=\omega t which specifies the displacement as well as the direction of motion of a point executing SHM.
Initial phase: The phase at t=0, \varphi in x=x_0\cos(\omega t+\varphi).
Phase difference: The difference between the phases of two oscillations of the same frequency.
In phase: Phase difference 0^\circ or 360^\circ: the oscillations move together.
Out of phase: Phase difference 180^\circ: the oscillations always move in opposite directions.
Ideal simple pendulum: A small heavy bob suspended by a light, inextensible string from a fixed frictionless support.
Elastic potential energy: Energy stored in a stretched or compressed spring, \tfrac12 kx^{2}.
Natural frequency: The frequency with which an oscillator vibrates freely after an initial disturbance.
Free oscillations: Oscillations under the restoring force alone, with no external force acting; they occur at the natural frequency.
Driving frequency: The frequency of the external periodic force acting on an oscillator.
Forced oscillations: Oscillations in which an external force makes the object oscillate at the frequency of the applied force rather than at its natural frequency.
Damping: The effect of resistive forces which dissipate energy from a vibrating object.
Damping forces: Resistive forces such as friction and air resistance that take energy from an oscillator.
Damped oscillations: Oscillations whose amplitude decreases under damping forces.
Light damping: Damping in which the amplitude decreases gradually with time while the body keeps oscillating.
Heavy damping: Damping so large that the body returns slowly to equilibrium without completing a single oscillation.
Critical damping: Damping with which the body comes to rest without any oscillation in the shortest possible time.
Shock absorber: A fluid-filled tube with a piston, fitted parallel to a car's springs, that damps their oscillations.
Resonance: The phenomenon in which the amplitude of an oscillator becomes maximum when the driving frequency equals its natural frequency.
Resonance frequency: The natural frequency of an oscillator, at which resonance occurs.
Frequency response: How the amplitude of a forced oscillator varies with the driving frequency.
Sharpness of resonance: How quickly the amplitude falls on either side of the resonance frequency; small damping gives a sharp resonance.
Standing wave: The wave pattern formed when a wave overlaps its reflection, with fixed nodes and antinodes.
Node: A point of a standing wave where the oscillation is always zero.
Antinode: A point of a standing wave where the oscillation has its largest amplitude.
Rubens tube: A gas-filled metal tube with a row of flames, used to demonstrate acoustic standing waves.
Acoustic levitation: Suspending matter in air against gravity using the acoustic pressure of high-intensity standing sound waves.
Chladni plate: A vibrating metal plate on which sand collects along the nodal lines, showing its mode of vibration.

Multiple choice questionsp. 66–67

Which of the following statements is true regarding the acceleration of the particle executing simple harmonic motion?

AAcceleration is zero at the mean positionAnswer
BAcceleration is maximum at the mean position
CAcceleration is zero at the extreme position
DAcceleration is a constant

Why

In SHM a=-\omega^{2}x. At the mean position x=0, so a=0. The acceleration is largest at the extreme positions (x=\pm x_0), so B and C have it the wrong way round, and it changes with x, so it is not constant (D).

If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in seconds?

A150
B2.5
C0.4Answer
D0.0067

Why

The time per beat is the time period: T=\dfrac{60\,\mathrm{s}}{150}=0.4\,\mathrm{s}. The value 2.5 is the frequency (beats per second), not the time per beat.

Which of the following relationship between the acceleration a and the displacement x of a particle involve SHM?

Aa=0.7x
Ba=-200x^{2}
Ca=-10xAnswer
Da=100x^{3}

Why

SHM needs a=-(\text{constant})\,x: proportional to x and opposite to it. Only a=-10x has this form, with \omega^{2}=10. Option A has no minus sign, and B and D are not proportional to x.

A mass-spring system undergoes simple harmonic motion has amplitude A. When the kinetic energy of the object equals twice the potential energy stored in the spring, what is the position x of the object?

AA
B\dfrac{A}{3}
C\dfrac{A}{\sqrt{3}}Answer
D0

Why

Total energy \tfrac12 kA^{2}=\text{K.E.}+\text{P.E.}=2\,\text{P.E.}+\text{P.E.}=3\,\text{P.E.}, so the P.E. is one third of the total:

\tfrac12 kx^{2}=\tfrac13\cdot\tfrac12 kA^{2}\ \Longrightarrow\ x^{2}=\dfrac{A^{2}}{3}\ \Longrightarrow\ x=\dfrac{A}{\sqrt{3}}

Option B forgets the square root: energy depends on x^{2}, not on x.

Suppose we replace the spring in a simple harmonic oscillator with a stronger spring, having twice the spring constant. What is the ratio of the new period of oscillation to the original period?

A\dfrac{1}{2}
B\dfrac{1}{\sqrt{2}}Answer
C1
D\sqrt{2}

Why

T=2\pi\sqrt{m/k}, so T\propto1/\sqrt{k}. Doubling k gives \dfrac{T'}{T}=\sqrt{\dfrac{k}{2k}}=\dfrac{1}{\sqrt{2}}. The stiffer spring oscillates faster, but the period does not halve because of the square root.

A pinball machine uses a spring that is compressed 4.0\,\mathrm{cm} to launch a ball. If the spring constant is 13\,\mathrm{N/m}, what is the force on the ball at the moment the spring is released?

A52\,\mathrm{N}
B0.52\,\mathrm{N}Answer
C0.52\,\mathrm{N\,m}
D5.2\,\mathrm{N\,m}

Why

By Hooke's law, F=kx=13\times0.040=0.52\,\mathrm{N}. Option A forgets to change centimetres into metres, and \mathrm{N\,m} is not a unit of force.

A particle is executing simple harmonic motion with period T. At time t=0 it is at the equilibrium point. At which time is it furthest from the equilibrium point?

A0.50\,T
B0.70\,T
C0.25\,TAnswer
D1.4\,T

Why

Starting from the mean position, x=x_0\sin(2\pi t/T). The particle first reaches an extreme (furthest point) after a quarter period, t=0.25\,T, and again at 0.75\,T,\ 1.25\,T,\dots At 0.50\,T it is back at equilibrium, and 0.70\,T and 1.4\,T are not quarter-period points.

The acceleration of a body executing simple harmonic motion leads the velocity by what phase?

A0\,\mathrm{rad}
B\pi/8\,\mathrm{rad}
C\pi/4\,\mathrm{rad}
D\pi/2\,\mathrm{rad}Answer

Why

With x=x_0\cos\omega t: v=\omega x_0\cos(\omega t+\pi/2) and a=\omega^{2}x_0\cos(\omega t+\pi). The phase of a is ahead of that of v by \pi-\pi/2=\pi/2. Physically, v is largest at the mean position and a is largest at the extremes, a quarter cycle apart.

92026 Annual I, Set B, MCQ 8

A simple pendulum has a period of 2.5\,\mathrm{s}. What is its period if its length is made four times larger?

A0.625\,\mathrm{s}
B1.25\,\mathrm{s}
C2.5\,\mathrm{s}
D5\,\mathrm{s}Answer

Why

T=2\pi\sqrt{l/g}, so T\propto\sqrt{l}. Four times the length gives \sqrt{4}=2 times the period: 2\times2.5=5\,\mathrm{s}.

What is the frequency of a pendulum that swings at the rate of 45 cycles per minute?

A0.75\,\mathrm{Hz}Answer
B1.3\,\mathrm{Hz}
C2700\,\mathrm{Hz}
D60\,\mathrm{Hz}

Why

Frequency is cycles per second: f=\dfrac{45}{60\,\mathrm{s}}=0.75\,\mathrm{Hz}. The number 1.3 is the period in seconds, not a frequency, and 2700 comes from multiplying instead of dividing.

A block at the end of a horizontal spring is pulled from equilibrium at x=0 to x=A, and then released. Through what total distance does it travel in one full cycle of its motion?

A\dfrac{A}{2}
BA
C2A
D4AAnswer

Why

In one cycle the block goes from +A to 0, from 0 to -A, from -A back to 0 and from 0 to +A: four stretches of length A, so 4A. Its displacement after a full cycle is zero, but the distance travelled is 4A.

A particle is vibrating simple harmonically with an acceleration of 16\,\mathrm{cm\,s^{-2}} when it is at a distance of 4\,\mathrm{cm} from the mean position. Its time period is:

A1\,\mathrm{s}
B2.572\,\mathrm{s}
C3.142\,\mathrm{s}Answer
D6.028\,\mathrm{s}

Why

\omega^{2}=\dfrac{a}{x}=\dfrac{16}{4}=4\,\mathrm{s^{-2}}\ \Longrightarrow\ \omega=2\,\mathrm{rad\,s^{-1}},\qquad T=\dfrac{2\pi}{\omega}=\pi=3.142\,\mathrm{s}

The centimetres cancel in a/x, so they need not be changed into metres here.

Short questionsp. 67

13 marks

If we halve the length of a simple pendulum to its original length, what is the alteration in the period of this pendulum? What is its new frequency?

Solution

The time period of a simple pendulum is T=2\pi\sqrt{l/g}, so T\propto\sqrt{l}. Let the new length be l'=l/2.

T'=2\pi\sqrt{\dfrac{l/2}{g}}=\dfrac{1}{\sqrt{2}}\left(2\pi\sqrt{\dfrac{l}{g}}\right)=\dfrac{T}{\sqrt{2}}=0.707\,T

So the period decreases to 0.707 of its original value (about 29\% shorter): the shorter pendulum swings faster. The frequency is the reciprocal of the period, so it increases by the same factor:

f'=\dfrac{1}{T'}=\sqrt{2}\,f=1.41\,f

For example, a pendulum of period 2.0\,\mathrm{s} (f=0.50\,\mathrm{Hz}) would have T'=1.41\,\mathrm{s} and f'=0.71\,\mathrm{Hz} after its length is halved.

23 marks

If the amplitude of vibration of a body executing SHM is doubled, what will happen to the maximum kinetic energy?

Solution

The kinetic energy is maximum at the mean position, where all the energy of the body is kinetic:

\text{K.E.}_{\max}=\dfrac12 mv_0^{2}=\dfrac12 m\omega^{2}x_0^{2}=\dfrac12 kx_0^{2}

The angular frequency \omega=\sqrt{k/m} does not depend on the amplitude, so \text{K.E.}_{\max}\propto x_0^{2}. If the amplitude is doubled, the maximum speed v_0=\omega x_0 doubles, and

\text{K.E.}'_{\max}=\dfrac12 k(2x_0)^{2}=4\left(\dfrac12 kx_0^{2}\right)=4\,\text{K.E.}_{\max}

The maximum kinetic energy becomes four times as large. The total energy of the oscillator also becomes four times, since it equals \text{K.E.}_{\max}.

33 marks

When marching soldiers are about to cross a bridge, they break steps. Why?

Solution

Soldiers marching in step strike the bridge together at regular intervals. This applies a periodic force to the bridge at the frequency of their steps. The bridge, like every structure, has its own natural frequency of vibration.

If the frequency of the steps equals the natural frequency of the bridge, resonance occurs: every step adds energy in step with the bridge's motion, so the amplitude of vibration keeps growing. The large vibrations can damage the bridge or even make it collapse. The Broughton Suspension Bridge in England collapsed in 1831 while soldiers marched across it in step.

When the soldiers break step, their feet land at random times. The forces no longer act in a regular rhythm, so there is no single driving frequency and resonance cannot build up.

43 marks

Suppose that a driving force has half frequency as compared to the frequency of an oscillator. Will it produce resonance? Similarly, if the driving frequency is twice the frequency of the oscillator, will it produce resonance?

Solution

No, in both cases. Resonance occurs only when the driving frequency equals the natural frequency of the oscillator (f=f_0). Only then is the force always in step with the motion, so that it does positive work in every cycle and the amplitude builds up to its maximum.

When f=f_0/2 or f=2f_0, the driving force is in step with the motion for part of the time and against it for the rest. The energy fed in during one part of the cycle is taken back during another part, so little energy is stored. The oscillator is forced to vibrate at the driving frequency with a small amplitude, as the resonance curve shows at f_0/2 and 2f_0.

53 marks

Pendulum clocks are made to run at the correct rate by adjusting the pendulum's length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you. Will you have to lengthen or shorten the pendulum to keep the correct time, other factors are remaining constant? Explain your answer.

Solution

The period of the pendulum is T=2\pi\sqrt{l/g}. In the new city g is slightly greater, so with the same length the period becomes slightly shorter. The pendulum swings faster and the clock gains time (runs fast).

To keep the correct time the period must not change, so the ratio l/g must stay the same:

\dfrac{l'}{g'}=\dfrac{l}{g}\quad\Longrightarrow\quad l'=l\,\dfrac{g'}{g}>l

Since g'>g, the new length must be greater. So the pendulum must be lengthened, in the same proportion as the increase in g.

63 marks

Two mass-spring systems vibrate with simple harmonic motion. If the spring constants are equal and the mass of one system is twice that of the other, which system has a greater period?

Solution

The period of a mass-spring system is T=2\pi\sqrt{m/k}. With the same k, T\propto\sqrt{m}:

\dfrac{T_2}{T_1}=\sqrt{\dfrac{m_2}{m_1}}=\sqrt{\dfrac{2m}{m}}=\sqrt{2}\quad\Longrightarrow\quad T_2=1.41\,T_1

The system with the larger mass has the greater period, about 1.41 times that of the other. Reason: both springs exert the same restoring force kx at the same displacement, but the heavier mass has more inertia, so its acceleration a=kx/m is smaller and it takes longer to complete each oscillation.

73 marks

Give some applications in which resonance plays an important role.

Solution

Resonance (maximum amplitude when the driving frequency equals the natural frequency) is used in many devices:

Radio and TV tuning: the natural frequency of the receiver circuit is adjusted to match the frequency of one station, which is then received with maximum energy.
Microwave oven: microwaves with a frequency close to the natural frequency of water molecules make them absorb maximum energy and heat the food.
MRI: hydrogen nuclei in the body absorb maximum energy from radio waves at resonance; the pattern of absorption forms a medical image.
Musical instruments: the hollow box of a guitar resonates and amplifies the sound of the strings.
Standing-wave devices: the Rubens tube, Chladni plates and acoustic levitation all use resonant standing waves.
Pushing a swing: pushes at its natural frequency build up a large amplitude.

83 marks

A simple pendulum is set into vibrations and left untouched, eventually stops, why?

Solution

A real pendulum is not an ideal free oscillator; its oscillations are damped. Two resistive (damping) forces act on it: air resistance on the moving bob and string, and friction at the point of support.

In every swing the pendulum does work against these forces. This work comes out of its mechanical energy, which is converted into heat in the air and the support. As the energy \tfrac12 m\omega^{2}x_0^{2} decreases, the amplitude decreases gradually (light damping). When all the mechanical energy has been dissipated, the pendulum comes to rest at its mean position. The energy is not destroyed; it is transferred to the surroundings as heat.

93 marks

Under what condition(s) the motion of a simple pendulum be simple harmonic motion?

Solution

The motion of a simple pendulum is SHM only when its acceleration is proportional to its displacement, a=-(g/l)x. This requires:

Small amplitude: the angle of swing must be small (less than about 10^\circ, or 0.2\,\mathrm{rad}), so that \sin\theta\approx\theta and the restoring force mg\sin\theta\approx mg\,x/l is proportional to the displacement.
Ideal string: the string is light (massless) and inextensible, so that the length l stays fixed.
Small heavy bob: the mass of the bob is concentrated at one point.
No damping: the support is frictionless and air resistance is negligible, so that the amplitude does not decrease.

103 marks

At what position is the velocity of a particle executing simple harmonic motion a) maximum b) minimum?

Solution

The velocity of a particle in SHM at displacement x is

v=\pm\,\omega\sqrt{x_0^{2}-x^{2}}

(a) Maximum at the mean position. At x=0 the square root is largest, so v_{\max}=\omega x_0. Here the restoring force is zero, the particle has stopped speeding up, and all its energy is kinetic.

(b) Minimum (zero) at the extreme positions. At x=\pm x_0, v=\omega\sqrt{x_0^{2}-x_0^{2}}=0. The particle stops for an instant and reverses its direction; all its energy is potential.

113 marks

Show that the motion of projection of a body revolving in a circle describes S.H.M.

Solution

Let a body P move round a circle of radius x_0 with constant angular velocity \omega, and let Q be its projection (the foot of the perpendicular) on the diameter along the x-axis. At time t, OP makes angle \theta=\omega t with the x-axis.

The centripetal acceleration a_P=x_0\omega^2 of P points towards O. Its horizontal component, a=-a_P\cos\theta, is the acceleration of Q and also points towards O.

Displacement of Q, from the right triangle OPQ:
x=x_0\cos\theta
P has a centripetal acceleration directed towards O:
a_P=x_0\omega^{2}
Q moves with the horizontal component of this acceleration, which points towards O (opposite to x):
a=-a_P\cos\theta=-x_0\omega^{2}\cos\theta
Substitute x=x_0\cos\theta:
a=-\omega^{2}x

Since \omega^{2} is constant, the acceleration of Q is proportional to its displacement and always directed towards the mean position O. Hence the projection Q executes SHM.

122026 Annual I, Set B, Q.2 (xii) OR3 marks

Sketch displacement-time graphs to illustrate light, critical and heavy damping.

Solution

All three graphs start from the same displacement x_0 with the oscillator at rest:

Light, critical and heavy damping

Light damping: oscillations with decreasing amplitude. Critical damping: fastest return to x=0 without oscillating. Heavy damping: slow return without oscillating.

Light damping: the body oscillates about x=0 with a gradually decreasing amplitude.
Critical damping: the body returns to x=0 in the shortest possible time without oscillating (no overshoot).
Heavy damping: the body creeps back to x=0 slowly, without completing even one oscillation; it takes longer than with critical damping.

133 marks

Justify the importance of critical damping in a car suspension system.

Solution

A car's body rests on springs. When a wheel hits a bump, the spring is compressed and the body then starts to oscillate up and down. Shock absorbers (pistons moving in oil, fitted parallel to the springs) provide the damping, and the best amount is critical damping:

With critical damping the body returns to its normal position in the shortest possible time without oscillating, so the passengers feel one smooth movement.
If the damping were light, the car would keep bouncing after every bump; this is uncomfortable, can injure passengers and makes the tyres lose grip on the road.
If the damping were heavy, the suspension would return so slowly that it would still be compressed when the next bump arrived, so the jolts would pass straight to the passengers.

So critical damping gives both comfort and safety.

143 marks

Differentiate free and forced oscillations.

Solution

Free oscillations: a body oscillates under its own restoring force, with no external force acting after the start, at its natural frequency. Forced oscillations: an external periodic force makes the body oscillate at the frequency of the applied force rather than at its natural frequency.

Free oscillations	Forced oscillations
No external periodic force acts after the start	An external periodic force acts all the time
Frequency equals the natural frequency f_0	Frequency equals the driving frequency f
Energy is given once, at the start	Energy is supplied continuously by the driver
Amplitude constant (ideal case)	Amplitude depends on how close f is to f_0; maximum at resonance
Example: plucked guitar string, pendulum released once	Example: clock pendulum driven by a motor, a swing pushed again and again

153 marks

How the time period of a simple pendulum changes if mass of its bob is doubled? What change arises in time period of mass attached to a spring if same is done here?

Solution

Simple pendulum: no change. T=2\pi\sqrt{l/g} contains no mass. Doubling the mass doubles the restoring force mg\sin\theta, but it also doubles the inertia, so the acceleration a=-(g/l)x and the period stay the same.

**Mass-spring system: the period increases by a factor \sqrt{2}.** From T=2\pi\sqrt{m/k},

T'=2\pi\sqrt{\dfrac{2m}{k}}=\sqrt{2}\,T=1.41\,T

Here the restoring force kx depends only on the spring, not on the mass. A doubled mass therefore has half the acceleration at each displacement, so it oscillates more slowly and its period becomes about 41\% longer.

Comprehensive questionsp. 68

Define and explain phase in simple harmonic motion.

Solution

Definition

The angle \theta=\omega t which specifies the displacement as well as the direction of motion of a point executing SHM is called the phase of the motion.

Explanation

The displacement of a body executing SHM is

x=x_0\cos(\omega t+\varphi)

Here x_0 is the amplitude, \omega the angular frequency and \omega t+\varphi the phase at time t. The constant \varphi is the initial phase: the phase at t=0. Phase is measured in radians (or degrees); one complete oscillation is a phase change of 2\pi\,\mathrm{rad}=360^\circ.

The displacement alone does not describe the state of the motion completely, because the body passes through every point twice in each oscillation, once in each direction. The phase gives both the displacement and the direction of motion:

For \varphi=0, the phase fixes both the position and the direction of motion.
Phase \theta	Time	x=x_0\cos\theta	State of motion
0	0	+x_0	At one extreme, momentarily at rest
\pi/2	T/4	0	At the mean position, moving towards -x
\pi	T/2	-x_0	At the other extreme, momentarily at rest
3\pi/2	3T/4	0	At the mean position, moving towards +x
2\pi	T	+x_0	One oscillation completed

Physical significance of the initial phase

The initial phase tells where the oscillation starts. Displacement-time graphs for different values of \varphi show this:

\varphi=0: x=x_0\cos\omega t. The body starts from the extreme position +x_0; it is at the extremes at t=0,\ T/2,\ T and at the mean position at t=T/4,\ 3T/4.
\varphi=90^\circ: x=x_0\cos(\omega t+90^\circ)=-x_0\sin\omega t. The body starts from the mean position; it is at the extremes at t=T/4,\ 3T/4.
\varphi=180^\circ: x=-x_0\cos\omega t. The body starts from the other extreme, -x_0.

Phase \varphi=0 and \varphi=90^\circ

The 90^\circ curve reaches every stage (for example -x_0) a quarter period (T/4) earlier, so it leads the \varphi=0 curve by 90^\circ.

Phase \varphi=0 and \varphi=180^\circ

With a phase difference of 180^\circ the two motions are always opposite: when one is at +x_0 the other is at -x_0 (out of phase).

Phase difference

The phase difference between two oscillations of the same frequency shows how far one is ahead of the other. In the graphs, the curve for \varphi=90^\circ reaches every stage a quarter period earlier than the curve for \varphi=0, so it leads by 90^\circ. The curve for \varphi=180^\circ leads the curve for \varphi=0 by 180^\circ.

In phase: phase difference 0^\circ or 360^\circ. The two bodies reach their extremes together and always move in the same direction.
Out of phase: phase difference 180^\circ. When one is at +x_0 the other is at -x_0; they always move in opposite directions.

Within a single SHM, the velocity leads the displacement by \pi/2 and the acceleration leads the displacement by \pi.

Derive the expressions for instantaneous displacement, instantaneous velocity and acceleration of the projection of a particle moving in a circle.

Solution

Set-up

Let a particle P move anticlockwise on a circle of radius x_0 with constant angular velocity \omega. Let Q be the projection of P on the x-axis, that is, the foot of the perpendicular from P. As P goes round the circle, Q moves to and fro along the diameter between -x_0 and +x_0. Time is counted from the moment P is at +x_0 on the x-axis.

Instantaneous displacement

Derivation Instantaneous displacement of the projection Q

Aim. Find the displacement x of Q at time t.

Assumptions.

P moves anticlockwise with constant angular velocity \omega on a circle of radius x_0.
Time is counted from the moment P is on the x-axis at +x_0, so \theta=0 at t=0.

P moves round a circle of radius x_0 with constant angular velocity \omega. Q, the foot of the perpendicular from P on the X-axis, oscillates along the diameter. In triangle OPQ, OQ=x and OP=x_0.

In time t the radius OP turns through angle \theta. For uniform circular motion, angle turned = angular velocity \times time.
\theta=\omega t
Triangle OPQ has a right angle at Q, hypotenuse OP=x_0 and base OQ=x (the displacement of Q). Use the cosine ratio.
\cos\theta=\dfrac{OQ}{OP}=\dfrac{x}{x_0}
Make x the subject and put \theta=\omega t.
x=x_0\cos\theta=x_0\cos\omega t\qquad(17.5)

Resultx=x_0\cos\omega t

Instantaneous velocity

Derivation Instantaneous velocity of the projection Q

Aim. Find the velocity of Q in terms of time and in terms of displacement.

The velocity v_P=x_0\omega of P is along the tangent. Its horizontal component, v=v_P\sin\theta, is the velocity of Q.

P moves along the tangent with constant speed v=r\omega, and here r=x_0.
v_P=x_0\omega
Q always stays directly below (or above) P, so Q moves with the horizontal component of P's velocity. The tangent is perpendicular to OP and the vertical line through P is perpendicular to the x-axis, so the angle between v_P and the vertical is \theta, and between v_P and the horizontal it is 90^\circ-\theta.
v=v_P\cos(90^\circ-\theta)
Use \cos(90^\circ-\theta)=\sin\theta.
v=v_P\sin\theta=x_0\omega\sin\theta\qquad(17.6)
To write v in terms of x, use \sin^{2}\theta+\cos^{2}\theta=1.
v=x_0\omega\sqrt{1-\cos^{2}\theta}
From (17.5), \cos\theta=x/x_0. Substitute and take x_0 inside the root.
v=x_0\omega\sqrt{1-\dfrac{x^{2}}{x_0^{2}}}=\omega\sqrt{x_0^{2}-x^{2}}\qquad(17.7)
Direction: while P is in the upper half of the circle, Q moves towards -x, so its velocity is negative. Q passes each point once in each direction, hence the \pm sign.
v=-x_0\omega\sin\omega t,\qquad v=\pm\,\omega\sqrt{x_0^{2}-x^{2}}

Resultv=\omega\sqrt{x_0^{2}-x^{2}}\ \ (\text{speed}),\qquad v_{\max}=v_0=\omega x_0\ \text{at}\ x=0

The speed of Q is largest, \omega x_0, at the mean position (x=0) and zero at the extreme positions (x=\pm x_0), where Q stops and turns back.

Instantaneous acceleration

Derivation Instantaneous acceleration of the projection Q

Aim. Show that the acceleration of Q is a=-\omega^{2}x.

The centripetal acceleration a_P=x_0\omega^2 of P points towards O. Its horizontal component, a=-a_P\cos\theta, is the acceleration of Q and also points towards O.

P moves in a circle at constant speed, so its only acceleration is the centripetal acceleration r\omega^{2}, directed from P towards the centre O.
a_P=x_0\omega^{2}
Q moves with the horizontal component of P's acceleration. The line PO makes angle \theta with the horizontal (alternate angles), so the component has size a_P\cos\theta. It points towards O, opposite to the displacement x of Q, so it carries a minus sign.
a=-a_P\cos\theta=-x_0\omega^{2}\cos\theta
Use x=x_0\cos\theta from (17.5).
a=-\omega^{2}\left(x_0\cos\theta\right)=-\omega^{2}x\qquad(17.8)

Resulta=-\omega^{2}x=-x_0\omega^{2}\cos\omega t

Conclusion

The projection Q has displacement x=x_0\cos\omega t, velocity v=-x_0\omega\sin\omega t=\pm\omega\sqrt{x_0^{2}-x^{2}} and acceleration a=-\omega^{2}x. Its acceleration is proportional to its displacement and directed towards the mean position, so the projection of uniform circular motion on a diameter is SHM. Its speed is greatest (\omega x_0) at the mean position and its acceleration is greatest (\omega^{2}x_0) at the extreme positions.

32026 Annual I, Set B, Q.3 OR

Show that motion of a mass attached to a spring executes S.H.M.

Solution

Set-up and diagram

A block of mass m is attached to one end of a spring whose other end is fixed to a rigid support. The block lies on a smooth horizontal surface and the mass of the spring is negligible. At the equilibrium (mean) position x=0 the spring is neither stretched nor compressed.

Mass m on a smooth horizontal surface. (a) Pulled right by x: the spring pulls left. (b) At x=0 the spring force is zero. (c) Pushed left: the spring pushes right. The force always points towards x=0.

Restoring force

When the block is pulled through a displacement x to the right, the stretched spring pulls it to the left; when it is pushed to the left, the compressed spring pushes it to the right. In both cases the spring force points towards the mean position and, by Hooke's law, is proportional to the displacement: F=-kx, where k is the spring constant (\mathrm{N\,m^{-1}}).

When released, the block accelerates towards x=0. There the force is zero, but the block is moving fast and its inertia carries it past the mean position. The spring then slows it, stops it at -x_0 and sends it back, so the block oscillates about x=0.

Proof that the motion is SHM

Derivation Motion of a mass attached to a spring is SHM

Aim. Show that the acceleration of the block is proportional to its displacement and directed towards the mean position, and find the angular frequency.

Assumptions.

The spring is massless and obeys Hooke's law.
The surface is frictionless and air resistance is negligible, so no energy is lost.
The block moves only along the line of the spring (the x-axis).

The only horizontal force on the block is the spring force. By Hooke's law it is proportional to the displacement x and opposite to it.
F=-kx\qquad(17.1)
The block has mass m, so by Newton's second law the same force gives it an acceleration a.
F=ma\qquad(17.2)
Both equations describe the same force, so equate them.
ma=-kx
Divide both sides by m to get the acceleration.
a=-\dfrac{k}{m}\,x\qquad(17.3)
k and m are fixed positive quantities for a given spring and block, so k/m is a positive constant. Write it as \omega^{2} (a square is always positive).
\dfrac{k}{m}=\omega^{2}\quad\Longrightarrow\quad a=-\omega^{2}x
Because \omega^{2} is constant, the acceleration is directly proportional to the displacement. The minus sign shows that it is always opposite to the displacement, that is, towards the mean position. These are exactly the two conditions of SHM.
a\propto -x
Take the square root of k/m=\omega^{2} to get the angular frequency of the oscillating mass.
\omega=\sqrt{\dfrac{k}{m}}\qquad(17.4)

Resulta=-\dfrac{k}{m}\,x=-\omega^{2}x,\qquad \omega=\sqrt{\dfrac{k}{m}}

The mass-spring system satisfies the definition of SHM. How quickly it oscillates depends only on the stiffness k and the mass m, not on how far the block was pulled.

Time period and frequency

Derivation Time period and frequency of a mass-spring system

Aim. Express the time period T and frequency f in terms of m and k.

Angular frequency is related to frequency and time period by its definition.
\omega=2\pi f=\dfrac{2\pi}{T}
Make T the subject.
T=\dfrac{2\pi}{\omega}
Substitute \omega=\sqrt{k/m} from (17.4). Dividing by \sqrt{k/m} is the same as multiplying by \sqrt{m/k}.
T=\dfrac{2\pi}{\sqrt{k/m}}=2\pi\sqrt{\dfrac{m}{k}}
Frequency is the reciprocal of the time period.
f=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

ResultT=2\pi\sqrt{\dfrac{m}{k}},\qquad f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

A heavier mass oscillates more slowly (T\propto\sqrt{m}) because it has more inertia. A stiffer spring oscillates faster (T\propto 1/\sqrt{k}) because it pulls back harder. The amplitude does not appear, so the period is the same for large and small oscillations.

Conclusion

The acceleration of the block is directly proportional to its displacement (a=-\omega^{2}x) and always directed towards the mean position, which is the definition of SHM. The motion repeats with time period T=2\pi\sqrt{m/k}, which depends only on the mass and the spring constant, not on the amplitude.

42026 Annual I, Set D, Q.3 OR

Analyse the interchange between kinetic and potential energy during simple harmonic motion.

Solution

Idea

A body in SHM has two kinds of mechanical energy. The spring (or, for a pendulum, the raised bob) stores potential energy, which depends on the displacement. The moving mass has kinetic energy, which depends on its speed. As the body oscillates, energy passes from one form to the other, but the total stays constant (no friction).

Potential energy

The work done in stretching a spring through x is the average force \tfrac12 kx times the displacement x. It is stored as elastic potential energy:

\text{P.E.}=\dfrac12 kx^{2}

Kinetic energy

The speed at displacement x is v=\omega\sqrt{x_0^{2}-x^{2}} with \omega^{2}=k/m, so

\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 m\,\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)=\dfrac12 k\left(x_0^{2}-x^{2}\right)

Total energy

\text{T.E.}=\text{K.E.}+\text{P.E.}=\dfrac12 k\left(x_0^{2}-x^{2}\right)+\dfrac12 kx^{2}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}

The result does not contain x: the total energy is the same at every position and is proportional to the square of the amplitude.

Interchange during one oscillation

At the extreme position x=+x_0: the body is momentarily at rest. K.E. =0 and P.E. =\tfrac12 kx_0^{2} is maximum: all the energy is potential.
Moving towards the mean position: the speed rises and the stretch falls, so P.E. is converted into K.E.
At the mean position x=0: P.E. =0 and K.E. =\tfrac12 kx_0^{2} is maximum: all the energy is kinetic, and the speed is greatest (\omega x_0).
Moving towards the other extreme: the spring is compressed, the body slows down, and K.E. is converted back into P.E.
At x=-x_0: all the energy is potential again. The cycle then repeats in the reverse direction.

Position	P.E.	K.E.
x=\pm x_0	\tfrac12 kx_0^{2} (maximum)	0
x=\pm x_0/2	\tfrac14 of T.E.	\tfrac34 of T.E.
x=\pm x_0/\sqrt{2}	\tfrac12 of T.E.	\tfrac12 of T.E.
x=0	0	\tfrac12 kx_0^{2} (maximum)

Energy against displacement

P.E. is zero at the mean position and largest at the extremes; K.E. is the opposite. The two are equal at x=\pm x_0/\sqrt{2}, and at every point K.E. + P.E. equals the same total energy.

Conclusion

In SHM the energy oscillates between kinetic and potential forms, twice in every period, but their sum remains constant at \tfrac12 kx_0^{2}=\tfrac12 m\omega^{2}x_0^{2}. This is the law of conservation of energy applied to an oscillator.

What is resonance? Give three of its applications in our daily life?

Solution

Definition

Resonance is the phenomenon in which the amplitude of vibration of an oscillator attains its maximum value when the driving frequency becomes equal to the natural frequency of the oscillator. The natural frequency at which this happens is called the resonance frequency.

f_{\text{driving}}=f_0\quad\Longrightarrow\quad\text{maximum amplitude}

Explanation

Every oscillator has a natural frequency f_0 at which it vibrates freely. When an external periodic force acts on it, it is forced to vibrate at the driving frequency f. If f is far from f_0, the force is sometimes in step with the motion and sometimes against it, so little energy is stored and the amplitude stays small. As f approaches f_0, the energy transfer becomes more efficient and the amplitude grows. When f=f_0, the force always pushes in the direction of motion, the energy transfer is maximum, and the amplitude becomes maximum.

A familiar example is pushing a swing: small pushes given once in every swing, at the right moment, make the swing go higher and higher.

Amplitude against driving frequency

The amplitude of a forced oscillator peaks when the driving frequency equals the natural frequency f_0.

A simple demonstration uses several pendulums hanging from one rod: when a driver pendulum is set swinging, only the pendulums of the same length (same natural frequency) pick up large swings.

Five pendulums hang from one rod. A and B have length l, C and D have length L. When the driver E has length l and is set swinging, A and B pick up large swings (resonance); C and D hardly move.

Applications in daily life

Tuning a radio. Turning the knob changes the natural frequency of the receiver's electrical circuit. When it equals the frequency of a station, resonance occurs, the circuit absorbs maximum energy from that station, and only that station is heard.
Microwave oven. The microwaves have a frequency close to the natural frequency of vibration of water molecules. The water in the food absorbs maximum energy, its molecules vibrate strongly, and the food is heated. Glass and plastic containers, which contain no water, stay cool.
Magnetic resonance imaging (MRI). Hydrogen nuclei in the body are made to oscillate by radio waves (about 100\,\mathrm{MHz}). At resonance they absorb maximum energy, and the pattern of absorption is used to make detailed medical images.
Musical instruments (a fourth example): the hollow wooden box of a guitar resonates with the strings and amplifies the sound.

Justify some circumstances in which resonance should be avoided.

Solution

Why resonance can be harmful

At resonance the driving force feeds energy into a system in every cycle, so the amplitude of vibration can become very large. Structures such as bridges, buildings and aircraft wings have natural frequencies. If a periodic force (wind, footsteps, engines, earthquakes) matches one of them, the large vibrations can crack or destroy the structure. Such resonance must therefore be avoided.

Bridges

Soldiers marching on a bridge are ordered to break their steps. Marching in step applies a periodic force at the step frequency; if it equals the natural frequency of the bridge, resonance occurs, the amplitude of vibration increases and the bridge may collapse. The most studied example is the Tacoma Narrows Bridge (USA), which collapsed in 1940 during a windstorm: the strong, continuous wind drove oscillations of the bridge deck that grew in amplitude until the deck broke apart.

Airplane wings

The wing is a very flexible part of an airplane. If gusts of wind vibrate it with a frequency equal to its natural structural frequency, resonance occurs and the vibrations of the wing become too large, which can eventually destroy it. Engineers therefore design the wing so that its natural frequency does not match the frequencies of the external vibrations it will meet.

Buildings and machines

Tall buildings are designed so that their natural frequencies differ from those of earthquake waves and wind gusts; otherwise the building would sway with a growing amplitude. In machines, unbalanced rotating parts (for example a washing machine drum or a car wheel) can shake the frame violently at the speed where the rotation frequency equals a natural frequency of the frame.

How resonance is avoided

Design the structure so that its natural frequency is far from the frequencies of the forces it will face.
Add damping (for example dampers in bridges and tall buildings) so that the resonance peak is low and broad.
Remove the regular driving force, for example by breaking step on a bridge or balancing rotating parts.

72026 Annual I, Set D, Q.3 OR

Derive equations for kinetic and potential energy of a body of mass m executing S.H.M.

Solution

Consider a body of mass m attached to a spring of spring constant k, oscillating in SHM with amplitude x_0 on a frictionless horizontal surface. Let its displacement from the mean position be x at some instant.

Potential energy

Derivation Elastic potential energy of a stretched spring

Aim. Find the potential energy stored when the spring is stretched (or compressed) through x.

Assumptions.

The spring obeys Hooke's law and is stretched slowly from 0 to x.

Work done in stretching a spring

The work done is the area under the force-extension line: a triangle of base x and height kx, so W=\tfrac12(x)(kx)=\tfrac12 kx^2.

The work done by the applied force equals the average force times the displacement.
W=F_{\text{av}}\,x\qquad\text{(i)}
By Hooke's law the applied force needed at extension x is
F=kx\qquad\text{(ii)}
The force increases steadily (linearly) from 0 at the start to kx at the end, so its average is half the sum of the first and last values.
F_{\text{av}}=\dfrac{0+kx}{2}=\dfrac12 kx\qquad\text{(iii)}
Substitute (iii) in (i).
W=\left(\dfrac12 kx\right)(x)=\dfrac12 kx^{2}
This work is stored in the spring as elastic potential energy.
\text{P.E.}=\dfrac12 kx^{2}\qquad(17.11)

Result\text{P.E.}=\dfrac12 kx^{2}

P.E. is zero at the mean position and grows as the square of the displacement. It is largest at the extreme positions, where \text{P.E.}_{\max}=\tfrac12 kx_0^{2}.

Kinetic energy

Derivation Kinetic energy of the oscillating mass

Aim. Find the kinetic energy of the mass when its displacement is x.

The speed of a body in SHM at displacement x is
v=\omega\sqrt{x_0^{2}-x^{2}}\qquad\text{(iv)}
For the mass-spring system \omega=\sqrt{k/m}, so
v=\sqrt{\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)}\qquad\text{(v)}
At the mean position (x=0) the speed is maximum.
v_0=\sqrt{\dfrac{k}{m}}\,x_0\qquad\text{(vi)}
Put v^{2} from (v) into \text{K.E.}=\tfrac12 mv^{2}. The mass m cancels.
\text{K.E.}=\dfrac12 m\cdot\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)=\dfrac12 k\left(x_0^{2}-x^{2}\right)\qquad(17.13)

Result\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 k\left(x_0^{2}-x^{2}\right)

K.E. is largest at the mean position, \tfrac12 kx_0^{2}, and zero at the extreme positions, where the body stops for an instant.

Total energy

Derivation Total energy of a body in SHM is constant

Aim. Show that \text{K.E.}+\text{P.E.} has the same value at every displacement.

At an extreme position (x=x_0) the block is at rest, so K.E. is zero and all the energy is elastic P.E.
\text{T.E.}=\text{P.E.}_{\max}=\dfrac12 kx_0^{2}\qquad(17.12)
At the mean position (x=0) the spring is neither stretched nor compressed, so P.E. is zero and all the energy is kinetic.
\text{T.E.}=\text{K.E.}_{\max}=\dfrac12 mv_{\max}^{2}=\dfrac12 kx_0^{2}\qquad(17.14)
**At any displacement x**, add the energies from (17.11) and (17.13).
\text{T.E.}=\text{K.E.}+\text{P.E.}=\dfrac12 k\left(x_0^{2}-x^{2}\right)+\dfrac12 kx^{2}
The x^{2} terms cancel, leaving a quantity that does not depend on x.
\text{T.E.}=\dfrac12 kx_0^{2}\qquad(17.15)
Since \omega=\sqrt{k/m}, k=m\omega^{2}. This gives the total energy in terms of \omega.
\text{T.E.}=\dfrac12 m\omega^{2}x_0^{2}\qquad(17.16)

Result\text{T.E.}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}=\text{constant}

The total energy of the mass-spring system is constant and is proportional to the square of the amplitude. This statement of conservation of energy is equally valid for all bodies executing SHM.

Graph

Energy against displacement

P.E. is zero at the mean position and largest at the extremes; K.E. is the opposite. The two are equal at x=\pm x_0/\sqrt{2}, and at every point K.E. + P.E. equals the same total energy.

The graph shows that P.E. is maximum and K.E. zero at the extreme positions, K.E. is maximum and P.E. zero at the mean position, and their sum (the horizontal line) is the same everywhere.

Explain what is meant by damped oscillations?

Solution

Definition

If the amplitude of oscillations decreases under damping forces, then such oscillations are called damped oscillations. In other words, when an oscillator keeps oscillating for some time before coming to rest under the action of a damping force, its oscillations are damped.

Cause of damping

Damping is the effect of resistive forces which dissipate energy from a vibrating object. Every real oscillator feels such forces, called damping forces:

friction between moving parts (for example at the support of a pendulum),
air resistance on the moving body,
internal forces in a spring, which turn some energy into heat.

Energy and amplitude

As the oscillator vibrates, it does work against these forces. The work comes out of its mechanical energy, which is gradually converted into heat. Since the energy of an oscillator is \tfrac12 kx_0^{2}, the loss of energy shows up as a steady decrease of the amplitude, until the motion stops. The period is hardly changed by light damping.

Undamped and damped oscillations

Damping does not change the time between swings much, but each swing is smaller than the one before, until the motion dies out.

Degrees of damping

Light damping: the amplitude decreases gradually and the body makes many oscillations before stopping (a swing, a pendulum in air).
Critical damping: the body returns to its mean position in the shortest possible time without oscillating (car shock absorbers).
Heavy damping: a very large resistive force makes the body creep back to the mean position slowly without completing even one oscillation (a mass on a spring in thick oil).

Examples and uses

A simple pendulum in air slowly comes to rest.
Car shock absorbers damp the oscillations of the car springs after a bump.
The pointer of a moving-coil meter is damped so that it settles quickly on the reading.

92026 Annual I, Set B, Q.2 (iii)

Explain the relation between damping and sharpness of resonance.

Solution

Resonance and sharpness

When an oscillator of natural frequency f_0 is driven by a periodic force, its amplitude depends on the driving frequency f and is maximum at f=f_0 (resonance). The graph of amplitude against driving frequency is the resonance curve. The sharpness of resonance describes how quickly the amplitude falls as f moves away from f_0: a sharp resonance has a tall, narrow peak, a flat resonance a low, broad one.

Amplitude against driving frequency

Resonance: the amplitude is largest when the driving frequency f equals the natural frequency f_0. Less damping gives a taller and narrower (sharper) peak.

Effect of damping

Small damping: very large amplitude at f_0 and a narrow peak. The resonance is sharp: the oscillator responds strongly only when f is very close to f_0.
Medium damping: smaller maximum amplitude and a wider peak.
Heavy damping: a small maximum amplitude and a broad, flat curve. The oscillator responds almost equally over a wide range of frequencies, so the resonance is hardly noticeable.

So the smaller the damping, the sharper the resonance; damping reduces the maximum amplitude at the resonance frequency and broadens the resonance curve.

Why damping controls sharpness

At resonance the driving force feeds energy in during every cycle. The amplitude grows until the energy lost per cycle to the damping forces equals the energy supplied. With little damping, very little energy is lost, so the energy, and with it the amplitude, can build up to a large value, but only when the force stays exactly in step with the motion, that is, very near f_0. With heavy damping, energy is lost quickly, so the amplitude stays small even at f_0, and the difference between f_0 and nearby frequencies is small.

Practical importance

A radio receiver needs sharp resonance (low damping in its tuning circuit) so that it picks up one station and rejects stations of nearby frequencies.
A car suspension is given strong damping so that the car never builds up large vibrations at any speed or on any road.
The other factors in the frequency response are the natural frequency of the oscillator, which fixes where the peak lies, and the strength of the driving force, which scales the whole curve.

Discuss the use of standing waves and resonance in (a) rubens tubes (b) chladni plates (c) acoustic levitation.

Solution

When a wave overlaps its own reflection, a standing wave forms, with nodes (no vibration) and antinodes (maximum vibration). At particular resonant frequencies, set by the size of the system, the standing wave builds up to a large amplitude. The following devices make resonant standing waves visible or useful.

(a) Rubens tube

A Rubens tube (standing wave flame tube), invented by Heinrich Rubens in 1905, is a metal pipe with a row of holes along the top, sealed at both ends. A loudspeaker driven by a frequency generator is attached to one end, and a flammable gas is fed in at the other. The gas escaping from the holes is lit.

Rubens tube: a standing sound wave in the gas-filled tube makes the flames tall where the pressure swings most and short where it hardly changes. Neighbouring tall flames are \lambda/2 apart.

When the generator is set to a resonance frequency of the tube, a standing sound wave forms in the gas. Where the pressure oscillates strongly, more gas is pushed out and the flames are tall; where the pressure hardly changes, less gas escapes and the flames are short. The flames thus draw the standing wave, acting as a primitive oscilloscope. The distance between two neighbouring tall flames is \lambda/2, so the wavelength can be measured with a ruler, and with the known frequency the speed of sound follows from v=f\lambda.

(b) Chladni plates

A Chladni plate is a flat metal sheet (usually circular or square) fixed at its centre on a stalk attached to a strong base. It is set vibrating by a violin bow drawn along its edge or by a loudspeaker. At resonant frequencies the plate vibrates in a particular mode, with a symmetrical pattern of nodal lines (no vibration) and antinodes (strong vibration).

Sand sprinkled on the plate is thrown away from the vibrating antinodes and collects along the still nodal lines, making the pattern visible. Changing the frequency to another resonance gives a different, often more complex pattern. The plates are used to study and demonstrate the modes of vibration of plates, such as the bodies of violins and guitars.

Acoustic levitation is a method of suspending matter in air against gravity using the acoustic pressure of high-intensity sound waves. A sound source (transducer) sends sound towards a reflector; the incident and reflected waves form a standing wave between them.

Acoustic levitation: sound from the source reflects back and forms a standing wave. Small beads are held at the pressure nodes, against gravity.

The standing wave has fixed nodes where the pressure forces push small objects inwards from both sides. Small particles or droplets placed there are trapped and float without any support. Acoustic levitation provides a container-less environment for experiments: samples can be studied, heated or mixed without touching (and being contaminated by) the walls of a container.

Analyze graphical representations of the variations of displacement, velocity and acceleration for simple harmonic motion.

Solution

Equations

For a body that starts from the extreme position +x_0 at t=0:

\begin{aligned} x&=x_0\cos\omega t\\ v&=-x_0\omega\sin\omega t\\ a&=-\omega^{2}x=-x_0\omega^{2}\cos\omega t\end{aligned}

Graphs

Displacement against time

Displacement is a cosine curve between +x_0 and -x_0.

Velocity against time

Velocity is zero at the extremes (t=0,\ T/2,\ T) and has its largest size \omega x_0 at the mean position (t=T/4,\ 3T/4).

Acceleration against time

Acceleration is always opposite to displacement: its curve is the displacement curve turned upside down.

Values at each quarter period

t	0	T/4	T/2	3T/4	T
x	+x_0	0	-x_0	0	+x_0
v	0	-\omega x_0	0	+\omega x_0	0
a	-\omega^{2}x_0	0	+\omega^{2}x_0	0	-\omega^{2}x_0

t	x	v	a
0	+x_0	0	-\omega^{2}x_0
T/4	0	-\omega x_0	0
T/2	-x_0	0	+\omega^{2}x_0
3T/4	0	+\omega x_0	0
T	+x_0	0	-\omega^{2}x_0

Analysis

Displacement is a cosine curve between +x_0 and -x_0; its maximum value is the amplitude.
Velocity is zero at the extreme positions (t=0,\ T/2,\ T) and has its maximum size \omega x_0 at the mean position (t=T/4,\ 3T/4).
Acceleration is zero at the mean position and has its maximum size \omega^{2}x_0 at the extreme positions.
The phase difference between velocity and displacement is \pi/2: the velocity curve is shifted a quarter period from the displacement curve (velocity leads).
The phase difference between acceleration and displacement is \pi: the acceleration curve is the displacement curve turned upside down, because a=-\omega^{2}x.
All three quantities repeat with the same time period T=2\pi/\omega.

Plotting acceleration directly against displacement gives a straight line through the origin with negative slope -\omega^{2}. This graph is the defining property of SHM.

Acceleration against displacement

A straight line through the origin with negative slope (-\omega^2): acceleration is proportional to displacement and opposite to it. This graph is the signature of SHM.

122026 Annual I, Set B, Q.2 (xii) OR

Explain the terms light, critical and heavy damping, with the help of examples.

Solution

Damping

Damping is the effect of resistive forces (friction, air resistance, viscous forces of a liquid) which dissipate energy from an oscillating system. How the oscillations die away depends on the size of the damping force compared with the restoring force.

Light damping

The damping is light when the amplitude of oscillations decreases gradually with time. The damping force is small, so the system completes many oscillations, each a little smaller than the one before, before it stops. Examples: a swing in a playground that is left to oscillate freely; a simple pendulum swinging in air; a plucked guitar string.

Critical damping

When an oscillator comes to rest without any oscillation in the shortest possible time under a damping force, the damping is critical. The body returns to the mean position once, without overshooting. Examples: the shock absorbers of a car, which bring the car body back to its level quickly without bouncing after a bump; the pointer of a moving-coil meter; automatic door closers.

Heavy damping

In heavy damping the resistive force is very large. The oscillator merely moves back to its equilibrium position gradually, without completing a single oscillation, and takes a long time to come to rest (longer than with critical damping). Example: a mass attached to a spring placed in a thick, viscous liquid such as oil or honey.

Displacement-time graphs

Light, critical and heavy damping

All three start from the same displacement x_0 at rest. T is the natural period of the undamped oscillator.

Type	Resistive force	Oscillation	Time to settle	Example
Light	Small	Yes, decreasing amplitude	Long	Swing, pendulum in air
Critical	Just enough	None	Shortest	Car shock absorbers
Heavy	Very large	None	Long (slow creep)	Mass on a spring in oil

Numerical problemsp. 68–69

Change every quantity to SI units first. Most problems need only \omega=2\pi f=2\pi/T, \omega=\sqrt{k/m} or \sqrt{g/l}, v=\omega\sqrt{x_0^{2}-x^{2}}, a=-\omega^{2}x and E=\tfrac12 m\omega^{2}x_0^{2}.

The amplitude of the motion of a mass attached to a spring is 2.48\,\mathrm{m}, while maximum speed of its mass is 4.36\,\mathrm{m\,s^{-1}}. What is the period of the motion?

Solution

Given

x_0 = 2.48\,\mathrm{m} amplitude
v_0 = 4.36\,\mathrm{m\,s^{-1}} maximum speed

To find

The maximum speed occurs at the mean position: v_0=\omega x_0. Solve for \omega.
\omega=\dfrac{v_0}{x_0}=\dfrac{4.36}{2.48}=1.758\,\mathrm{rad\,s^{-1}}
Find the period from \omega.
T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{1.758}=3.57\,\mathrm{s}

T=3.57\,\mathrm{s}

Book answer. 3.57\,\mathrm{s}

A particle moves, whose displacement as a function of time is: x=3.0\cos(2t), where distance is measured in meter and time in second.

Solution

Given

x = 3.0\cos(2t)\ \mathrm{m}

To find

(a) x_0, f, \omega, T
(b) times to reach x=0 and a turning point

(a) Calculate the amplitude, the frequency, the angular frequency and the period of this particle.

Compare x=3.0\cos(2t) with x=x_0\cos\omega t.
x_0=3.0\,\mathrm{m},\qquad \omega=2.0\,\mathrm{rad\,s^{-1}}
Frequency from \omega=2\pi f.
f=\dfrac{\omega}{2\pi}=\dfrac{2.0}{2\pi}=0.318\,\mathrm{Hz}
Period.
T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2.0}=\pi=3.14\,\mathrm{s}

x_0=3.0\,\mathrm{m},\ f=0.318\,\mathrm{Hz},\ \omega=2.0\,\mathrm{rad\,s^{-1}},\ T=3.14\,\mathrm{s}

(b) Calculate the time at which the particle reaches the midpoint (i.e., x=0) and the turning point.

At t=0, x=3.0\cos0=+3.0\,\mathrm{m}, so the particle starts at a turning point. It reaches the midpoint when the cosine first becomes zero.
\cos2t=0\ \Longrightarrow\ 2t=\dfrac{\pi}{2}\ \Longrightarrow\ t=\dfrac{\pi}{4}=0.785\,\mathrm{s}\ \ \left(=\dfrac{T}{4}\right)
The next turning point is the other extreme, x=-3.0\,\mathrm{m}, where \cos2t=-1.
2t=\pi\ \Longrightarrow\ t=\dfrac{\pi}{2}=1.57\,\mathrm{s}\ \ \left(=\dfrac{T}{2}\right)

t_{\text{midpoint}}=0.785\,\mathrm{s},\qquad t_{\text{turning point}}=1.57\,\mathrm{s}

x_0=3.0\,\mathrm{m},\ f=0.318\,\mathrm{Hz},\ \omega=2.0\,\mathrm{rad\,s^{-1}},\ T=3.14\,\mathrm{s};\ \ 0.785\,\mathrm{s},\ 1.57\,\mathrm{s}

Book answer. (a) 3.0\,\mathrm{m}, 0.318\,\mathrm{Hz}, 2.0\,\mathrm{rad\,s^{-1}}, 3.14\,\mathrm{s} (b) 0.785\,\mathrm{s}, 1.57\,\mathrm{s}

A particle executes simple harmonic motion, that moves back and forth along x-axis between points -0.20\,\mathrm{m} and +0.20\,\mathrm{m}. The period of the motion is 1.2\,\mathrm{s}. At the time t=0, the particle is at +0.20\,\mathrm{m} and its velocity is zero.

Solution

Given

x = -0.20\,\mathrm{m}\ \text{to}\ +0.20\,\mathrm{m}
T = 1.2\,\mathrm{s}
x(0) = +0.20\,\mathrm{m},\ v(0)=0

To find

(a) f, \omega
(b) x_0
(c) times to reach x=0 and x=0.10\,\mathrm{m}
(d) speeds at x=0 and x=0.10\,\mathrm{m}

The particle starts from rest at +0.20\,\mathrm{m} (an extreme), so its motion is x=x_0\cos\omega t.

(a) What is the frequency of the motion and the angular frequency?

Frequency is the reciprocal of the period.
f=\dfrac{1}{T}=\dfrac{1}{1.2}=0.833\,\mathrm{Hz}
Angular frequency.
\omega=2\pi f=\dfrac{2\pi}{1.2}=5.24\,\mathrm{rad\,s^{-1}}

f=0.83\,\mathrm{Hz},\qquad \omega=5.24\,\mathrm{rad\,s^{-1}}

(b) What is the amplitude of the motion?

The amplitude is the maximum displacement from the mean position x=0, which is half the distance between the two extreme points: x_0=\tfrac12\left(0.20-(-0.20)\right)=0.20\,\mathrm{m}.

x_0=0.20\,\mathrm{m}

(c) At what time will the particle reach the point x=0? At what time will the particle reach the point x=0.10\,\mathrm{m}?

Starting from an extreme, the particle reaches the mean position after a quarter period.
x=0:\quad \cos\omega t=0\ \Rightarrow\ \omega t=\dfrac{\pi}{2}\ \Rightarrow\ t=\dfrac{T}{4}=\dfrac{1.2}{4}=0.30\,\mathrm{s}
For x=0.10\,\mathrm{m}: 0.10=0.20\cos\omega t.
\cos\omega t=0.5\ \Rightarrow\ \omega t=\dfrac{\pi}{3}\ \Rightarrow\ t=\dfrac{\pi/3}{2\pi/1.2}=\dfrac{1.2}{6}=0.20\,\mathrm{s}

t=0.30\,\mathrm{s}\ \text{at}\ x=0,\qquad t=0.20\,\mathrm{s}\ \text{at}\ x=0.10\,\mathrm{m}

(d) What is the speed of the particle when it is at x=0? What is the speed of the particle when it reaches the point x=0.10\,\mathrm{m}?

At x=0 the speed is maximum.
v_0=\omega x_0=5.236\times0.20=1.05\,\mathrm{m\,s^{-1}}
At x=0.10\,\mathrm{m} use v=\omega\sqrt{x_0^{2}-x^{2}}.
v=5.236\sqrt{0.20^{2}-0.10^{2}}=5.236\sqrt{0.030}=5.236\times0.1732=0.91\,\mathrm{m\,s^{-1}}

v=1.05\,\mathrm{m\,s^{-1}}\ \text{at}\ x=0,\qquad v=0.91\,\mathrm{m\,s^{-1}}\ \text{at}\ x=0.10\,\mathrm{m}

0.83\,\mathrm{Hz},\ 5.24\,\mathrm{rad\,s^{-1}};\ 0.20\,\mathrm{m};\ 0.30\,\mathrm{s},\ 0.20\,\mathrm{s};\ 1.05\,\mathrm{m\,s^{-1}},\ 0.91\,\mathrm{m\,s^{-1}}

Book answer. (a) 0.83\,\mathrm{Hz}, 5.2\,\mathrm{rad\,s^{-1}} (b) 0.20\,\mathrm{m} (c) 0.30\,\mathrm{s}, 0.20\,\mathrm{s} (d) 1.05\,\mathrm{m\,s^{-1}}, 0.91\,\mathrm{m\,s^{-1}}

A mass of 8.0\,\mathrm{kg} is attached to a spring and oscillates with amplitude of 0.25\,\mathrm{m} and has a frequency of 0.60\,\mathrm{Hz}. What is the energy of the motion?

Solution

Given

m = 8.0\,\mathrm{kg}
x_0 = 0.25\,\mathrm{m}
f = 0.60\,\mathrm{Hz}

To find

total energy E

Angular frequency.
\omega=2\pi f=2\pi(0.60)=3.77\,\mathrm{rad\,s^{-1}}
The total energy of SHM.
E=\dfrac12 m\omega^{2}x_0^{2}=\dfrac12(8.0)(3.77)^{2}(0.25)^{2}
Evaluate.
E=4.0\times14.21\times0.0625=3.55\,\mathrm{J}\approx3.6\,\mathrm{J}

E\approx3.6\,\mathrm{J}

Book answer. 3.6\,\mathrm{J}

Calculate the period and frequency of a 3.5\,\mathrm{m} long pendulum at the following locations:

Solution

Given

l = 3.5\,\mathrm{m}
g_{\text{Karachi}} = 9.832\,\mathrm{m\,s^{-2}}
g_{\text{K-2}} = 9.782\,\mathrm{m\,s^{-2}}

To find

T and f at each place

Use T=2\pi\sqrt{l/g} and f=1/T.

(a) at Karachi, where g=9.832\,\mathrm{m\,s^{-2}}.

Period.
T=2\pi\sqrt{\dfrac{3.5}{9.832}}=2\pi(0.5966)=3.749\,\mathrm{s}
Frequency.
f=\dfrac{1}{T}=\dfrac{1}{3.749}=0.2668\,\mathrm{Hz}

T=3.749\,\mathrm{s},\qquad f=0.2668\,\mathrm{Hz}

(b) at K-2, where g=9.782\,\mathrm{m\,s^{-2}}.

Period. g is smaller on K-2, so the period is slightly longer.
T=2\pi\sqrt{\dfrac{3.5}{9.782}}=2\pi(0.5982)=3.758\,\mathrm{s}
Frequency.
f=\dfrac{1}{T}=\dfrac{1}{3.758}=0.2661\,\mathrm{Hz}

T=3.758\,\mathrm{s},\qquad f=0.2661\,\mathrm{Hz}

T_{\text{Karachi}}=3.749\,\mathrm{s},\ f=0.2668\,\mathrm{Hz};\quad T_{\text{K-2}}=3.758\,\mathrm{s},\ f=0.2661\,\mathrm{Hz}

Book answer. (a) 3.747\,\mathrm{s}, 0.2669\,\mathrm{Hz} (b) 3.757\,\mathrm{s}, 0.2662\,\mathrm{Hz}

Note. The book's answers (3.747\,\mathrm{s} and 3.757\,\mathrm{s}) come from taking \pi=3.14; with \pi=3.1416 the periods are 3.749\,\mathrm{s} and 3.758\,\mathrm{s}. The difference is only in the fourth significant figure, and either set earns full marks.

In a car engine, a piston executes SHM with amplitude of 0.41\,\mathrm{m}. The engine is running at angular frequency of 2400\,\mathrm{rpm} (251\,\mathrm{rad\,s^{-1}}). What is the maximum speed of the piston?

Solution

Given

x_0 = 0.41\,\mathrm{m}
\omega = 2400\,\mathrm{rev\,min^{-1}}

To find

v_{\max}

Change revolutions per minute into radians per second: each revolution is 2\pi\,\mathrm{rad} and a minute is 60\,\mathrm{s}.
\omega=\dfrac{2400\times2\pi}{60}=80\pi=251\,\mathrm{rad\,s^{-1}}
The maximum speed of SHM occurs at the mean position.
v_{\max}=\omega x_0=251\times0.41=103\,\mathrm{m\,s^{-1}}

v_{\max}\approx103\,\mathrm{m\,s^{-1}}

Book answer. 103\,\mathrm{m\,s^{-1}}

A ball connected to a spring executes SHM. At t=0, its displacement is 0.50\,\mathrm{m} and its acceleration is -0.72\,\mathrm{m\,s^{-2}}. The phase constant for its motion is 0.84\,\mathrm{rad}. What is the ball's displacement at t=3.4\,\mathrm{s}?

Solution

Given

x(0) = 0.50\,\mathrm{m}
a(0) = -0.72\,\mathrm{m\,s^{-2}}
\varphi = 0.84\,\mathrm{rad} phase constant (initial phase)
t = 3.4\,\mathrm{s}

To find

x at t=3.4\,\mathrm{s}

Write the displacement with the phase constant.
x=x_0\cos(\omega t+\varphi)
Find \omega from a=-\omega^{2}x at t=0.
\omega^{2}=-\dfrac{a}{x}=-\dfrac{-0.72}{0.50}=1.44\,\mathrm{s^{-2}}\ \Longrightarrow\ \omega=1.2\,\mathrm{rad\,s^{-1}}
Find the amplitude from the displacement at t=0: x(0)=x_0\cos\varphi.
x_0=\dfrac{0.50}{\cos0.84}=\dfrac{0.50}{0.667}=0.749\,\mathrm{m}
Phase at t=3.4\,\mathrm{s}, in radians.
\omega t+\varphi=1.2(3.4)+0.84=4.92\,\mathrm{rad}
Displacement at t=3.4\,\mathrm{s} (calculator in radian mode).
x=0.749\cos(4.92)=0.749\times0.206=0.154\,\mathrm{m}

x\approx0.15\,\mathrm{m}

Book answer. 0.15\,\mathrm{m}

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 41 mark

In SHM, at which value of \theta, the instantaneous velocity (v) is maximum (v_0)?

A\theta=\dfrac{3\pi}{4}
B\theta=\dfrac{\pi}{2}Answer
C\theta=\dfrac{\pi}{4}
D\theta=0

Why

The formula list printed on the paper gives v=v_0\sin\theta. Since \sin\theta has its largest value, 1, at \theta=\pi/2, the velocity is maximum (v=v_0) there. This is the moment the body passes through the mean position; at \theta=0 it is at an extreme and v=0.

2026 Annual I, Set B, MCQ 51 mark

Total energy of the mass spring system in SHM, is proportional to:

A\dfrac{m\omega}{x_0^{2}}
B\dfrac{mx_0}{\omega^{2}}
Cm\omega^{2}x_0^{2}Answer
Dm\omega x_0

Why

\text{T.E.}=\tfrac12 m\omega^{2}x_0^{2}, so the total energy is proportional to m\omega^{2}x_0^{2}. The other options do not have this form (energy grows with the square of both \omega and the amplitude).

2026 Annual I, Set B, MCQ 81 mark

A simple pendulum has a period of 2.5\,\mathrm{s}. What will be its period if its length becomes four times large?

A2.5
B1.25\,\mathrm{s}
C0.625\,\mathrm{s}
D5\,\mathrm{s}Answer

Why

T=2\pi\sqrt{l/g}, so T\propto\sqrt{l}. Four times the length doubles the period: 2\times2.5=5\,\mathrm{s}.

2026 Annual I, Set B, Q.2 (i) OR3 marks

Time period of a mass attached at the end of a spring is 0.4\,\mathrm{s}. Find out the magnitude of acceleration when the displacement is 4\,\mathrm{cm}.

Solution

Marks: formula (1), substitution in SI units (1), answer with unit (1). This is Assignment 17.2 of the book.

Given

T = 0.4\,\mathrm{s}
x = 4\,\mathrm{cm}=0.04\,\mathrm{m}

To find

Angular frequency from the time period.
\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.4}=15.7\,\mathrm{rad\,s^{-1}}
Magnitude of the acceleration in SHM, a=\omega^{2}x.
a=(15.71)^{2}\times0.04=246.7\times0.04=9.87\,\mathrm{m\,s^{-2}}

a\approx9.9\,\mathrm{m\,s^{-2}}

2026 Annual I, Set B, Q.2 (iii)3 marks

Justify qualitatively the factors which determine the frequency response and sharpness of resonance.

Solution

The frequency response of a forced oscillator is the way its amplitude changes with the driving frequency. The amplitude is largest (resonance) when the driving frequency equals the natural frequency f_0. The sharpness of resonance is how quickly the amplitude falls on either side of f_0. The factors are:

Natural frequency of the oscillator. It decides where the peak lies. It is set by the oscillator itself, for example f_0=\frac{1}{2\pi}\sqrt{k/m} for a spring or \frac{1}{2\pi}\sqrt{g/l} for a pendulum.
Damping (the main factor for sharpness). With small damping little energy is lost per cycle, so near f_0 energy builds up to a large amplitude and the peak is tall and narrow: a sharp resonance. With heavy damping energy is lost quickly, so the peak is low and broad: a flat resonance.
Strength of the driving force. A stronger driving force raises the whole curve but does not move the peak.

Amplitude against driving frequency

Smaller damping gives a higher and sharper resonance peak; heavy damping reduces the amplitude and broadens the curve.

2026 Annual I, Set B, Q.2 (xii) OR1x3 = 3 marks

What is meant by light damping and heavy damping? Show these graphically. (Displacement-time graph)

Solution

Light damping (1 mark). The damping force is small, so the body keeps oscillating while its amplitude decreases gradually with time. Example: a swing, or a pendulum swinging in air.

Heavy damping (1 mark). The damping force is very large, so the body does not complete even one oscillation; it moves back slowly to its equilibrium position and takes a long time to come to rest. Example: a mass on a spring in thick oil.

Graph (1 mark).

Light and heavy damping

Light damping: the curve crosses the time axis again and again with shrinking peaks. Heavy damping: the curve never crosses the axis and approaches it slowly.

2026 Annual I, Set B, Q.3 OR1+3+2+1 = 7 marks

What is meant by simple harmonic motion (SHM)? Show that mass attached to a spring executes SHM. Derive the expressions for frequency 'f' and period 'T'. Draw diagram as well.

Solution

Simple harmonic motion (1 mark)

SHM is the oscillatory motion in which the acceleration of a particle is directly proportional to its displacement from the mean position and is always directed towards the mean position:

a\propto -x\qquad\text{or}\qquad a=-\omega^{2}x

Diagram (1 mark)

A block of mass m on a smooth horizontal surface is attached to a spring of spring constant k whose other end is fixed. At x=0 the spring is unstretched. When the block is displaced by x and released, the spring exerts a restoring force towards x=0, and the block oscillates about this mean position.

Mass-spring system executes SHM (3 marks)

Derivation Motion of a mass attached to a spring is SHM

Aim. Show that the acceleration of the block is proportional to its displacement and directed towards the mean position, and find the angular frequency.

Assumptions.

The spring is massless and obeys Hooke's law.
The surface is frictionless and air resistance is negligible, so no energy is lost.
The block moves only along the line of the spring (the x-axis).

The only horizontal force on the block is the spring force. By Hooke's law it is proportional to the displacement x and opposite to it.
F=-kx\qquad(17.1)
The block has mass m, so by Newton's second law the same force gives it an acceleration a.
F=ma\qquad(17.2)
Both equations describe the same force, so equate them.
ma=-kx
Divide both sides by m to get the acceleration.
a=-\dfrac{k}{m}\,x\qquad(17.3)
k and m are fixed positive quantities for a given spring and block, so k/m is a positive constant. Write it as \omega^{2} (a square is always positive).
\dfrac{k}{m}=\omega^{2}\quad\Longrightarrow\quad a=-\omega^{2}x
Because \omega^{2} is constant, the acceleration is directly proportional to the displacement. The minus sign shows that it is always opposite to the displacement, that is, towards the mean position. These are exactly the two conditions of SHM.
a\propto -x
Take the square root of k/m=\omega^{2} to get the angular frequency of the oscillating mass.
\omega=\sqrt{\dfrac{k}{m}}\qquad(17.4)

Resulta=-\dfrac{k}{m}\,x=-\omega^{2}x,\qquad \omega=\sqrt{\dfrac{k}{m}}

The mass-spring system satisfies the definition of SHM. How quickly it oscillates depends only on the stiffness k and the mass m, not on how far the block was pulled.

Frequency and time period (2 marks)

Derivation Time period and frequency of a mass-spring system

Aim. Express the time period T and frequency f in terms of m and k.

Angular frequency is related to frequency and time period by its definition.
\omega=2\pi f=\dfrac{2\pi}{T}
Make T the subject.
T=\dfrac{2\pi}{\omega}
Substitute \omega=\sqrt{k/m} from (17.4). Dividing by \sqrt{k/m} is the same as multiplying by \sqrt{m/k}.
T=\dfrac{2\pi}{\sqrt{k/m}}=2\pi\sqrt{\dfrac{m}{k}}
Frequency is the reciprocal of the time period.
f=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

ResultT=2\pi\sqrt{\dfrac{m}{k}},\qquad f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Since the acceleration is proportional to the displacement and directed towards the mean position, the mass-spring system executes SHM with T=2\pi\sqrt{m/k} and f=\frac{1}{2\pi}\sqrt{k/m}, independent of the amplitude.

2026 Annual I, Set D, MCQ 61 mark

In SHM, the relationship between acceleration and displacement of a particle when \omega=10 is:

Aa=-100xAnswer
Ba=100x^{2}
Ca=10x
Da=-10x^{2}

Why

In SHM a=-\omega^{2}x. With \omega=10\,\mathrm{rad\,s^{-1}}, \omega^{2}=100, so a=-100x. Options B and D are not proportional to x, and C has the wrong sign and does not square \omega.

2026 Annual I, Set D, MCQ 71 mark

What is the frequency of the body when its time period is 2\,\mathrm{s}?

A0.5\,\mathrm{Hz}Answer
B0.2\,\mathrm{Hz}
C1\,\mathrm{Hz}
D2\,\mathrm{Hz}

Why

f=\dfrac{1}{T}=\dfrac{1}{2\,\mathrm{s}}=0.5\,\mathrm{Hz}.

2026 Annual I, Set D, Q.2 (i) OR1+1+1 = 3 marks

A body with mass 800\,\mathrm{g} attached to a spring executes SHM. The restoring force is 60\,\mathrm{N} when the displacement is 0.3\,\mathrm{m}.

Solution

Given

m = 800\,\mathrm{g}=0.8\,\mathrm{kg}
F = 60\,\mathrm{N}\ \text{at}\ x=0.3\,\mathrm{m}
x = 12\,\mathrm{cm}=0.12\,\mathrm{m}\ \text{in part (b)}

To find

(a) \omega
(b) a

Marks: spring constant (1), angular frequency (1), acceleration (1). These are the data of Example 17.2 of the book.

a. Find out its angular frequency

Spring constant from Hooke's law (size only).
k=\dfrac{F}{x}=\dfrac{60}{0.3}=200\,\mathrm{N\,m^{-1}}
Angular frequency of the mass-spring system.
\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{200}{0.8}}=\sqrt{250}=15.8\,\mathrm{rad\,s^{-1}}

\omega=15.8\,\mathrm{rad\,s^{-1}}

b. Also find the magnitude of acceleration when the displacement is 12\,\mathrm{cm}.

Use a=\omega^{2}x with \omega^{2}=250\,\mathrm{s^{-2}} and x=0.12\,\mathrm{m}.
a=250\times0.12=30\,\mathrm{m\,s^{-2}}

a=30\,\mathrm{m\,s^{-2}}

Note. The book's Example 17.2 prints a=3\,\mathrm{m\,s^{-2}} for these data; that is an arithmetic slip. The correct value is 30\,\mathrm{m\,s^{-2}}.

2026 Annual I, Set D, Q.2 (iii)3 marks

How does resonance play an important role in tuning of radio station?

Solution

Many radio stations broadcast at the same time, each at its own frequency. All their waves reach the aerial and try to drive small electrical oscillations in the receiver.

The tuning circuit of the receiver (a coil and a variable capacitor) is an electrical oscillator with its own natural frequency. When we turn the tuning knob, we change the capacitance and so change this natural frequency.

When the natural frequency of the circuit becomes equal to the transmission frequency of a particular station, resonance occurs. The circuit then oscillates with maximum amplitude at that frequency and absorbs maximum energy from that station only. Signals of other stations are far from resonance and produce very small responses, so we hear only the tuned station. For good selection the tuning circuit has little damping, which makes the resonance sharp.

2026 Annual I, Set D, Q.2 (xii) OR1+2 = 3 marks

What is meant by free oscillations? Also give its two examples.

Solution

Free oscillations (1 mark). A body is said to be executing free oscillations when it oscillates under the influence of its restoring force alone, without any external force acting on it. It then vibrates at its own natural frequency, and (ideally, with no damping) its amplitude and period stay constant.

Examples (2 marks):

A simple pendulum displaced slightly from its mean position and released oscillates freely at its natural frequency f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}.
A mass attached to a spring, pulled aside and released, oscillates freely at f_0=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}.
(Also acceptable: a plucked guitar string, a struck tuning fork, a swing given a single push.)

2026 Annual I, Set D, Q.3 OR2+2+2+1 = 7 marks

Derive expressions for Kinetic and Potential energies of a mass attached to a spring executing SHM. Show that total energy of the spring system remains constant. Also make neat sketch.

Solution

Consider a mass m attached to a spring of spring constant k, oscillating on a smooth horizontal surface with amplitude x_0. Let its displacement be x at some instant.

Kinetic energy (2 marks)

Derivation Kinetic energy of the oscillating mass

Aim. Find the kinetic energy of the mass when its displacement is x.

The speed of a body in SHM at displacement x is
v=\omega\sqrt{x_0^{2}-x^{2}}\qquad\text{(iv)}
For the mass-spring system \omega=\sqrt{k/m}, so
v=\sqrt{\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)}\qquad\text{(v)}
At the mean position (x=0) the speed is maximum.
v_0=\sqrt{\dfrac{k}{m}}\,x_0\qquad\text{(vi)}
Put v^{2} from (v) into \text{K.E.}=\tfrac12 mv^{2}. The mass m cancels.
\text{K.E.}=\dfrac12 m\cdot\dfrac{k}{m}\left(x_0^{2}-x^{2}\right)=\dfrac12 k\left(x_0^{2}-x^{2}\right)\qquad(17.13)

Result\text{K.E.}=\dfrac12 mv^{2}=\dfrac12 k\left(x_0^{2}-x^{2}\right)

K.E. is largest at the mean position, \tfrac12 kx_0^{2}, and zero at the extreme positions, where the body stops for an instant.

Potential energy (2 marks)

Derivation Elastic potential energy of a stretched spring

Aim. Find the potential energy stored when the spring is stretched (or compressed) through x.

Assumptions.

The spring obeys Hooke's law and is stretched slowly from 0 to x.

The work done by the applied force equals the average force times the displacement.
W=F_{\text{av}}\,x\qquad\text{(i)}
By Hooke's law the applied force needed at extension x is
F=kx\qquad\text{(ii)}
The force increases steadily (linearly) from 0 at the start to kx at the end, so its average is half the sum of the first and last values.
F_{\text{av}}=\dfrac{0+kx}{2}=\dfrac12 kx\qquad\text{(iii)}
Substitute (iii) in (i).
W=\left(\dfrac12 kx\right)(x)=\dfrac12 kx^{2}
This work is stored in the spring as elastic potential energy.
\text{P.E.}=\dfrac12 kx^{2}\qquad(17.11)

Result\text{P.E.}=\dfrac12 kx^{2}

P.E. is zero at the mean position and grows as the square of the displacement. It is largest at the extreme positions, where \text{P.E.}_{\max}=\tfrac12 kx_0^{2}.

Total energy remains constant (2 marks)

Derivation Total energy of a body in SHM is constant

Aim. Show that \text{K.E.}+\text{P.E.} has the same value at every displacement.

At an extreme position (x=x_0) the block is at rest, so K.E. is zero and all the energy is elastic P.E.
\text{T.E.}=\text{P.E.}_{\max}=\dfrac12 kx_0^{2}\qquad(17.12)
At the mean position (x=0) the spring is neither stretched nor compressed, so P.E. is zero and all the energy is kinetic.
\text{T.E.}=\text{K.E.}_{\max}=\dfrac12 mv_{\max}^{2}=\dfrac12 kx_0^{2}\qquad(17.14)
**At any displacement x**, add the energies from (17.11) and (17.13).
\text{T.E.}=\text{K.E.}+\text{P.E.}=\dfrac12 k\left(x_0^{2}-x^{2}\right)+\dfrac12 kx^{2}
The x^{2} terms cancel, leaving a quantity that does not depend on x.
\text{T.E.}=\dfrac12 kx_0^{2}\qquad(17.15)
Since \omega=\sqrt{k/m}, k=m\omega^{2}. This gives the total energy in terms of \omega.
\text{T.E.}=\dfrac12 m\omega^{2}x_0^{2}\qquad(17.16)

Result\text{T.E.}=\dfrac12 kx_0^{2}=\dfrac12 m\omega^{2}x_0^{2}=\text{constant}

The total energy of the mass-spring system is constant and is proportional to the square of the amplitude. This statement of conservation of energy is equally valid for all bodies executing SHM.

Sketch (1 mark)

Energy against displacement

P.E. is zero at the mean position and largest at the extremes; K.E. is the opposite. The two are equal at x=\pm x_0/\sqrt{2}, and at every point K.E. + P.E. equals the same total energy.

K.E. and P.E. change continuously, one increasing as the other decreases, but their sum is the horizontal line \tfrac12 kx_0^{2}: the total energy of the spring system remains constant.

Exam focus

Define SHM. Show that a mass attached to a spring executes SHM and derive T=2\pi\sqrt{m/k} and f=\frac{1}{2\pi}\sqrt{k/m}, with a diagram. (long)
Came as the full 7-mark Section C question in 2026 (Set B, Q.3 OR); the most likely long question of this unit.
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Derive expressions for K.E. and P.E. of a mass-spring system in SHM and show that the total energy \tfrac12 kx_0^{2}=\tfrac12 m\omega^{2}x_0^{2} is constant, with a sketch. (long)
The other 7-mark Section C question in 2026 (Set D, Q.3 OR); the energy graph carries a separate mark.
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Derive the displacement, velocity and acceleration of the projection of a particle moving in a circle, and show that its motion is SHM. (long)
A standard derivation that also gives v=\pm\omega\sqrt{x_0^{2}-x^{2}} and a=-\omega^{2}x, the formulas used in most numericals.
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Show that a simple pendulum executes SHM for small angles and derive T=2\pi\sqrt{l/g}. (long)
A core derivation; its conditions and scaling (T\propto\sqrt{l}, independent of mass) appear in MCQs and short questions every year.
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Explain light, critical and heavy damping and sketch their displacement-time graphs; justify critical damping in a car suspension. (short)
Asked in 2026 (Set B, Q.2 (xii) OR); the graph earns a mark on its own.
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Explain the factors that determine the frequency response and sharpness of resonance (effect of damping), with the resonance curves. (short)
Asked in 2026 (Set B, Q.2 (iii)) and matches Comprehensive Question 9.
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How does resonance help in tuning a radio? Give other uses of resonance and cases where it must be avoided. (short)
Radio tuning was asked in 2026 (Set D, Q.2 (iii)); bridges and aircraft wings are the standard 'avoid resonance' examples.
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What are free oscillations? Give two examples, and distinguish free from forced oscillations. (short)
Asked in 2026 (Set D, Q.2 (xii) OR) and in the book's short questions.
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Numericals with \omega=2\pi/T, \omega=\sqrt{k/m}, a=\omega^{2}x and v=\omega\sqrt{x_0^{2}-x^{2}} (Example 17.2 and Assignment 17.2). (numerical)
Both 2026 sets had a 3-mark numerical of exactly this type (Set B and Set D, Q.2 (i) OR). Remember the book's printed answer to Example 17.2 is wrong: a=30\,\mathrm{m\,s^{-2}}.
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Energy numericals using E=\tfrac12 m\omega^{2}x_0^{2} and positions where K.E. and P.E. are in a given ratio. (numerical)
Total energy appeared as a 2026 MCQ (Set B, MCQ 5); the book's MCQ 4 and Numerical 4 practise it.
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Define phase and explain it with displacement-time graphs for \varphi=0,\ 90^\circ,\ 180^\circ; analyse the x, v, a graphs of SHM. (long)
Comprehensive Questions 1 and 11; phase relations (v leads x by \pi/2, a leads v by \pi/2) are frequent MCQs.
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Quick MCQ rules: T\propto\sqrt{l} for a pendulum, T\propto\sqrt{m/k} for a spring, a=-\omega^{2}x, v_{\max} at the mean position, a_{\max} at the extremes. (mcq)
Five 2026 MCQs (Set B 4, 5, 8; Set D 6, 7) came from these facts.
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