Waves

Class 11 Physics, unit 9, book pages 154–183

Practise 76 MCQs Take the unit quiz Review 42 cards

This unit studies what waves do when they carry energy, move relative to an observer, meet each other or meet obstacles. It covers the intensity of a wave, Doppler's effect, the superposition principle with interference, beats and stationary waves on strings and in air columns, and then polarisation, diffraction and gravitational waves. These ideas explain musical instruments, noise-cancelling headphones, radar and medical ultrasound, polaroid sunglasses and the detection of gravitational waves.

Learning outcomes (21)

Use intensity = power/area to solve problems. Use intensity $\propto$ (amplitude) $^{2}$ for a progressive wave to solve problems.
Explain that when a source of sound waves moves relative to a stationary observer, the observed frequency is different from the source frequency [describing the Doppler effect for a stationary source and a moving observer is not required].
Use the expression $f' = \frac{v}{v \pm v _{s}} f$ for the observed frequency when a source of sound waves moves relative to a stationary observer.
Explain the applications of the Doppler effect [such as radar, sonar, astronomy, satellite, radar speed traps and studying cardiac problems in humans].
Explain that polarisation is a phenomenon associated with transverse waves.
Define and apply Malus's law [ $I = I_{0} cos^{2} θ$ ] to calculate the intensity of a plane-polarised electromagnetic wave after transmission through a polarising filter or a series of polarising filters [calculation of the effect of a polarising filter on the intensity of an un-polarised wave is not required].
Use the principle of superposition of waves to solve problems.
Differentiate between constructive and destructive interference.
Apply the principle of superposition to explain the working of noise-cancelling headphones.
Illustrate experiments that demonstrate stationary waves [using microwaves, stretched strings and air columns (it will be assumed that end corrections are negligible; knowledge of the concept of end corrections is not required)].
Explain the formation of a stationary wave using graphical representation.
Explain the formation of harmonics in stationary waves.
Analyse experiments that demonstrate diffraction [including the qualitative effect of the gap width relative to the wavelength of the wave; for example, diffraction of water waves in a ripple tank].
Explain the term coherence.
Explain beats [as the pulsation caused by two waves of slightly different frequencies interfering with each other].
Illustrate examples of how beats are generated in musical instruments.
Explain the use of polaroids in sky photography and stress analysis of materials.
Describe qualitatively gravitational waves [as waves of the intensity of gravity generated by the accelerated masses of an orbital binary system that propagate as waves outward from their source at the speed of light].
State that when a gravitational wave passes a body with mass, the distortion in space-time can cause the body to stretch and compress periodically.
State that gravitational waves pass through the Earth due to far-off celestial events, but they have very minute amplitude.
Describe the use of interferometers in detecting gravitational waves [interferometers are very sensitive detection devices that make use of the interference of laser beams (working and set-up details are not required) and were used to first detect the existence of gravitational waves].

Introduction

The world is full of waves: sound waves, waves on a string, seismic waves, and electromagnetic waves such as visible light, radio waves, television signals and X-rays. All of these waves have a source: a vibrating object. Radio and television signals and other forms of electromagnetic waves are the backbone of communication in modern civilisation.

Waves transfer energy from one location to another. In this unit we study several wave phenomena: the intensity of a wave, Doppler's effect, superposition and interference, beats, stationary (standing) waves in strings and air columns, polarisation, diffraction, and gravitational waves.

9.1Intensity of waves

All waves carry energy, and sometimes their energy can be observed directly. Earthquakes can shake whole cities, doing the work of thousands of wrecking balls. Loud sounds can damage nerve cells in the inner ear and cause permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves wash away beaches.

Intensity: The energy carried by a wave per unit area in unit time.

The intensity of a progressive wave is directly related to the square of its amplitude:

I \propto (amplitude)^{2}

So an earthquake of large amplitude produces large ground displacements. The energy of a wave also depends on time: the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight can be focused to burn wood, while earthquakes spread out and do less damage the farther they get from the source. All these factors are included in the definition of intensity $I$ as power per unit area:

I = \frac{P}{A} (9.1 a)

$P$ : power carried by the wave through the area $A$ (area perpendicular to the direction of travel).

Power is energy per unit time, $P = E / t$ , so equation (9.1 a) can be written as:

I = \frac{E}{A \times t} (9.1 b)

$E$ : energy passing through area $A$ in time $t$ .

The definition is valid for any energy in transit, including that carried by progressive waves. The SI unit of intensity is watt per square metre ( $W m^{- 2}$ ).

Example 9.1 (1): energy falling on a solar collector

The average intensity of sunlight on Earth\'s surface is about

500 W m^{- 2}

. Calculate the amount of energy that falls on a solar collector with an area of

0.50 m^{2}

4.0 h

Given

$I = 500 W m^{- 2}$
$A = 0.50 m^{2}$
$t = 4.0 h = 4.0 \times 3600 s = 14400 s$

To find

energy $E$ falling on the collector

Start from the definition of intensity and make $E$ the subject.
$I = \frac{E}{A \times t} ⟹ E = I \times A \times t$
Convert the time to seconds (SI unit) and substitute.
$E = (500 W m^{- 2}) (0.50 m^{2}) (4.0 \times 3600 s)$
Multiply: $500 \times 0.50 = 250 W$ is the power received, and $250 \times 14400$ gives the energy.
$E = 250 \times 14400 = 3.6 \times 1 0^{6} J$

E = 3.6 \times 1 0^{6} J

Example 9.1 (2): doubling the amplitude

If the amplitude of a wave is doubled, how much does the energy increase?

Energy depends on intensity, and intensity is proportional to the square of the amplitude.
$I \propto (amplitude)^{2}$
Put the new amplitude equal to twice the old one and compare the intensities.
$\frac{I _{2}}{I _{1}} = (\frac{2 a}{a})^{2} = 4$

the energy becomes 4 times

Assignment 9.1: amplitude for 50 times the intensity

To increase the intensity of a wave by a factor of 50, by what factor should the amplitude of the wave be increased?

Intensity is proportional to the square of the amplitude, so the ratio of intensities equals the square of the ratio of amplitudes.
$\frac{I _{2}}{I _{1}} = (\frac{a _{2}}{a _{1}})^{2} = 50$
Take the square root.
$\frac{a _{2}}{a _{1}} = 50 = 7.07$

amplitude must be increased by a factor of 50 \approx 7.1

9.2Doppler's effectOften examined

Possibly you have noticed how the sound of a vehicle's horn changes as the vehicle moves away from you. The frequency of the sound you hear appears higher as the vehicle approaches you and appears lower as it moves away. This phenomenon is known as Doppler's effect. It is named after the Austrian physicist Christian Doppler (1803–1853), who described it in 1842.

Doppler's effect: The apparent change in the frequency of a wave due to the relative motion between the observer and the source.

Doppler's effect is most often associated with sound, but it is common to all waves, including water waves and light. In deriving the effect we assume that the air is stationary and that all speed measurements are made relative to this stationary medium.

Why the frequency changes

Motion of a source of sound towards an observer increases the rate at which he or she receives the vibrations. The velocity of each vibration is the speed of sound whether the source is moving or not. Each vibration from an approaching source has a shorter distance to travel, so the wavelength is shortened when the source moves towards the observer and lengthened when it moves away. The vibrations are therefore received at a higher frequency than they are sent. Similarly, sound waves from a receding source are received at a lower frequency than they are sent.

Consider a source of sound $S$ emitting a wave of speed $v$ , frequency $f$ and wavelength $λ$ . When the source $S$ and the listener $L$ are both at rest, the listener receives $f$ waves per second, according to

v = f λ ⟹ f = \frac{v}{λ}

There are four possible situations. In the symbols below $v$ is the speed of sound, $f$ the true frequency of the source, $f'$ the apparent frequency heard, $v_{L}$ the speed of the listener and $v_{s}$ the speed of the source.

(1) The source is at rest and the listener is moving

Derivation Listener moves towards the stationary source

Aim. Find the apparent frequency $f'$ heard by a listener moving towards a stationary source.

Figure 9.1 (a): The listener

L

moves with speed

v_{L}

towards the stationary source

S

. The wavelength is unchanged but the listener meets the waves faster, so

f' > f

The source is at rest, so the wavefronts are concentric circles and the wavelength in the air is unchanged.
$λ = \frac{v}{f}$
The listener moves towards the source with speed $v_{L}$ . Relative to the listener the waves approach faster: their speed is the speed of sound plus the speed of the listener.
$v_{relative} = v + v_{L}$
The number of waves received per second is the relative speed divided by the wavelength. This is the apparent frequency.
$f' = \frac{v + v _{L}}{λ}$
Put $λ = v / f$ from the source.
$f' = \frac{v + v _{L}}{v} f (9.2)$
Since $v + v_{L}$ is larger than $v$ , the factor is greater than one, so the listener hears a higher frequency.
$\frac{v + v _{L}}{v} > 1 ⟹ f' > f$

Result

f' = \frac{v + v _{L}}{v} f

The frequency of the sound increases, and as a result its pitch also increases.

Derivation Listener moves away from the stationary source

Aim. Find $f'$ for a listener moving away from a stationary source.

Figure 9.1 (b): The listener moves away from the stationary source. The waves overtake the listener more slowly, so

f' < f

The wavelength is again unchanged because the source is at rest.
$λ = \frac{v}{f}$
The listener moves away with speed $v_{L}$ , so the waves catch up with the listener more slowly: the relative speed is the speed of sound minus $v_{L}$ .
$v_{relative} = v - v_{L}$
Apparent frequency is relative speed divided by wavelength.
$f' = \frac{v - v _{L}}{λ}$
Put $λ = v / f$ .
$f' = \frac{v - v _{L}}{v} f (9.3)$
Since $v - v_{L}$ is smaller than $v$ , the factor is less than one.
$\frac{v - v _{L}}{v} < 1 ⟹ f' < f$

Result

f' = \frac{v - v _{L}}{v} f

The frequency of the sound decreases, and as a result its pitch also decreases.

(2) The source is moving and the listener is at rest

Derivation Source moves towards the stationary listener

Aim. Find the apparent frequency $f'$ for a source moving with speed $v_{s}$ towards a stationary listener $A$ .

Figure 9.1 (c): The source moves to the right and the listener is at rest. Each wavefront is centred on the point where it was emitted, so the wavefronts are crowded in front of the source (listener A) and spread out behind it (listener B).

Each new wave is emitted from a position closer to the listener than the previous one. The wavefronts reaching the listener are therefore closer together, and the wavelength $λ'$ measured by the listener is shorter than the wavelength $λ$ of the source.
During one vibration, which lasts one period $T$ , the source moves a distance $v_{s} T = v_{s} / f = Δ λ$ . This is the amount by which the wavelength is shortened.
$Δ λ = v_{s} T = \frac{v _{s}}{f}$
The apparent wavelength is the original wavelength minus this shortening. Use $λ = v / f$ .
$λ' = λ - Δ λ = \frac{v}{f} - \frac{v _{s}}{f}$
Combine the fractions.
$λ' = \frac{v - v _{s}}{f} (9.4)$
The waves still travel at speed $v$ through the air, so the apparent frequency is $v$ divided by the apparent wavelength.
$f' = \frac{v}{λ '} (9.5)$
Substitute (9.4) into (9.5).
$f' = \frac{v}{v - v _{s}} f (9.6)$
Since $v - v_{s}$ is smaller than $v$ , the factor is greater than one.
$\frac{v}{v - v _{s}} > 1 ⟹ f' > f$

Result

f' = \frac{v}{v - v _{s}} f

When the source moves towards a stationary listener, the frequency of the sound increases and so does its pitch.

Derivation Source moves away from the stationary listener

Aim. Find $f'$ for a source moving away from a stationary listener.

The observer now measures a wavelength $λ'$ greater than $λ$ , because each wave is emitted from a position farther away than the last one.
The wavelength is lengthened by the distance the source moves in one period.
$λ' = λ + Δ λ = \frac{v}{f} + \frac{v _{s}}{f}$
Combine the fractions.
$λ' = \frac{v + v _{s}}{f} (9.7)$
The apparent frequency is again $v$ divided by the apparent wavelength.
$f' = \frac{v}{λ '} (9.8)$
Substitute (9.7) into (9.8).
$f' = \frac{v}{v + v _{s}} f (9.9)$
Since $v + v_{s}$ is larger than $v$ , the factor is less than one.
$\frac{v}{v + v _{s}} < 1 ⟹ f' < f$

Result

f' = \frac{v}{v + v _{s}} f

When the source moves away from a stationary listener, the frequency of the sound decreases and so does its pitch.

The four cases together

Situation	Apparent frequency	Result	Equation
Source at rest, listener moves towards source	$f' = \frac{v + v _{L}}{v} f$	$f' > f$	(9.2)
Source at rest, listener moves away	$f' = \frac{v - v _{L}}{v} f$	$f' < f$	(9.3)
Listener at rest, source moves towards listener	$f' = \frac{v}{v - v _{s}} f$	$f' > f$	(9.6)
Listener at rest, source moves away	$f' = \frac{v}{v + v _{s}} f$	$f' < f$	(9.9)

Example 9.2: a receding sound source

A sound source is receding from a stationary observer at

22 m s^{- 1}

and has frequency

354 Hz

. If the speed of sound is

332 m s^{- 1}

, what frequency does the observer hear?

Given

$f = 354 Hz$
$v_{s} = 22 m s^{- 1}$
$v = 332 m s^{- 1}$

To find

apparent frequency $f'$

The source moves away from a stationary observer, so use equation (9.9).
$f' = \frac{v}{v + v _{s}} f$
Substitute the values: the denominator is $332 + 22 = 354$ .
$f' = \frac{332}{332 + 22} \times 354 = \frac{332}{354} \times 354$
The factor $354$ cancels.
$f' = 332 Hz$

f' = 332 Hz

Assignment 9.2: a bus with its horn sounding

A bus is moving at

20 m s^{- 1}

along a straight road with its

500 Hz

horn sounding. You are standing at the roadside. What frequency do you hear as the bus is (a) approaching you, (b) receding from you? (Take the speed of sound as

340 m s^{- 1}

Given

$f = 500 Hz$
$v_{s} = 20 m s^{- 1}$
$v = 340 m s^{- 1}$

To find

(a) $f'$ for approach
(b) $f'$ for recession

(a) The source approaches a listener at rest: use (9.6).
$f' = \frac{v}{v - v _{s}} f = \frac{340}{340 - 20} \times 500 = \frac{340}{320} \times 500 = 531 Hz$
(b) The source recedes: use (9.9).
$f' = \frac{v}{v + v _{s}} f = \frac{340}{340 + 20} \times 500 = \frac{340}{360} \times 500 = 472 Hz$

(a) 531 Hz, (b) 472 Hz

Worked example: a siren passing an observer

An ambulance siren has a frequency of

700 Hz

. The ambulance passes a stationary pedestrian at

30 m s^{- 1}

. Take the speed of sound as

340 m s^{- 1}

. Find the frequency heard (a) while the ambulance approaches and (b) while it moves away, and (c) the fall in pitch as it passes.

Given

$f = 700 Hz$
$v_{s} = 30 m s^{- 1}$
$v = 340 m s^{- 1}$

To find

(a) $f'$ approaching
(b) $f'$ receding
(c) the fall in frequency

(a) The source moves towards the listener, so the waves are crowded together and the pitch rises: use $v - v_{s}$ in the denominator.
$f' = \frac{v}{v - v _{s}} f = \frac{340}{340 - 30} \times 700 = \frac{238000}{310} = 767.7 Hz$
(b) The source moves away, so the waves are stretched and the pitch falls: use $v + v_{s}$ in the denominator.
$f' = \frac{v}{v + v _{s}} f = \frac{340}{340 + 30} \times 700 = \frac{238000}{370} = 643.2 Hz$
(c) The listener hears the pitch drop by the difference of the two values as the ambulance goes by.
$767.7 - 643.2 = 124.5 Hz \approx 1.2 \times 1 0^{2} Hz$
Check the signs: the approaching value is above $700 Hz$ and the receding value is below it. If your answers do not fall on those sides of $f$ , the signs are swapped.

f' \approx 768 Hz (approaching), f' \approx 643 Hz (receding), Δ f \approx 1.2 \times 1 0^{2} Hz

9.2.1Applications of Doppler's effect

Doppler's effect has many interesting applications. It is applied in weather observation to characterise cloud movement and weather patterns, and has other applications in aviation and radiology. Some of the most significant ones are listed below.

SONAR

SONAR (Sound Navigation and Ranging) helps to detect the speed of ships, aeroplanes and submarines using Doppler's effect. When sound waves are reflected from a moving body such as a submarine, their frequency changes. This change in frequency allows us to calculate the speed and direction of the submarine.

Satellites

The velocity of the Earth's satellites is determined from the Doppler shift in the frequency of the radio waves they transmit.

Radar

Doppler's effect is used in a radar system to find the speed and direction of aeroplanes. A radar system sends out radio waves. If the frequency of the reflected radio waves from an aeroplane decreases, the aeroplane is moving away from the radar. If the frequency of the reflected waves increases, the aeroplane is moving towards the radar.

Astronomy

Doppler's effect is of great interest to astronomers, who use the shift in the frequency of the electromagnetic waves produced by moving stars in our galaxy and beyond to gather information about those stars and galaxies. In 1929, Edwin Hubble observed that all galaxies appeared to be red-shifted (that is, moving away from us and from each other), which led him to propose that the universe is expanding. Specific information about stars within galaxies can be determined by using Doppler's effect.

Figure 9.2: Blue shift and red shift. Light from a star moving towards the Earth is shifted to a higher frequency (blue shift); light from a star moving away is shifted to a lower frequency (red shift).

Fig. 9.2 shows that electromagnetic radiation emitted by stars in a distant galaxy appears shifted to lower frequencies (a red shift) if the star is moving away from the Earth. On the other hand, there is an upward shift in frequency (a blue shift) of such observed radiation if the star is moving towards the Earth.

Radar speed traps

Doppler's effect is used to measure speed in RADAR sensors, known as radar speed traps. When a fixed-frequency radio wave sent by the sender continuously strikes an object moving towards or away from the sender, the frequency of the reflected radio wave is changed by the object's motion. Analysing the frequency of the reflected wave gives the speed of the moving object.

Diagnosis of cardiac diseases

Doppler's effect is used in the diagnosis of cardiac diseases. Doppler ultrasound (or Doppler echocardiography) is a test in which very high frequency sound waves are bounced off the heart and blood vessels. The returning sound waves are picked up and turned into pictures showing blood flow through the arteries or the heart. Doppler ultrasound allows doctors to see clearly the flow of blood through the heart and blood vessels. It also allows them to see and measure obstructions in arteries and to measure the degree of narrowing or leakage of heart valves.

9.3Superposition of wavesOften examined

Various wave phenomena in nature require two or more waves passing through the same region of space at the same time. Two travelling waves can meet and pass through each other without being destroyed or even altered.

When two waves of the same nature pass through the same medium, their interaction may cause the formation of a new wave. In the region of overlap, the resultant wave is found by adding the displacements of the individual waves. For such analysis the superposition principle applies:

Superposition principle: When two or more waves are passing through the same region at the same time, the total displacement at the point where they interact is equal to the vector sum of the individual displacements due to each pulse at that point.

If a particle of the medium is acted upon simultaneously by n waves, such that its displacement due to each of the individual waves is y_1,\ y_2,\ y_3,\ \dots,\ y_n, then the resultant displacement y of the particle is:

y=y_1+y_2+y_3+\dots+y_n

In Fig. 9.3 two waves (X and Y) superpose to form the resultant wave (Z). The diagram shows only three points where it displays how the displacement of the resultant wave is calculated. For example, when time t=0 the displacement of the Y wave is -0.9 and the displacement of the X wave is -2.1. Signs must be taken into account, because displacement is a vector quantity. So the displacement of the resultant wave is the vector sum (-0.9)+(-2.1)=-3. This is simply the principle of superposition.

Figure 9.3: Two waves (X and Y) superposing to form the resultant wave (Z). At every instant the displacement of Z is the sum of the displacements of X and Y; at t=0 it is (-0.9)+(-2.1)=-3.

The superposition of waves can lead to the following three effects:

When two waves having the same frequency travel with the same speed along the same direction in a specific medium, they superpose and create an effect known as the interference of waves.
When two waves having the same frequency travel with the same speed along opposite directions in a specific medium, they superpose to produce stationary waves.
When two waves having slightly different frequencies travel with the same speed along the same direction in a specific medium, they superpose to produce beats.

Coherent waves

When two or more light waves travel together in such a way that their phase difference is constant, such waves are known as coherent waves. This is possible if the waves have the same frequency and wavelength.

Coherent source: A source which emits light waves with the same frequency, wavelength and phase (or with a constant phase difference).

The laser is a highly coherent light source. Coherent light waves can also be produced by Young's double slit experiment.

9.4InterferenceOften examined

Waves which have a constant phase difference and the same frequency are called coherent waves.

Interference: The effect produced due to the superposition of waves from two coherent sources.

There are two types of interference: constructive interference and destructive interference.

Constructive interference

During constructive interference the crest of one wave falls on the crest of the other wave, and the trough of one wave falls on the trough of the other, as shown in Fig. 9.4 (a). Both of the superposing waves have amplitude X, and the amplitude of the resultant wave is increased to 2X.

Figure 9.4 (a): Constructive interference. Two waves of amplitude X in phase give a resultant of amplitude 2X.

In the case of longitudinal waves, constructive interference occurs when the compressions of one wave overlap with the compressions of the other, and the rarefactions of one wave overlap with the rarefactions of the other.

Constructive interference: When two waves overlap at a point in the same phase, the two waves reinforce each other.

Destructive interference

During destructive interference the crest of one wave falls on the trough of the other. The amplitude of the resultant wave is equal to the difference between the amplitudes of the individual waves, that is X-X=0, as shown in Fig. 9.4 (b).

Figure 9.4 (b): Destructive interference. Two waves of equal amplitude X, 180^\circ out of phase, give a resultant of zero.

In the case of longitudinal waves, destructive interference occurs when the compressions of one wave overlap with the rarefactions of the other.

Destructive interference: When two waves overlap at a point out of phase (180^\circ), the two waves cancel each other.

	Constructive interference	Destructive interference
Phase of the two waves at the point	in phase	out of phase (180^\circ)
What overlaps (transverse)	crest on crest, trough on trough	crest on trough
What overlaps (longitudinal)	compression on compression, rarefaction on rarefaction	compression on rarefaction
Resultant amplitude (waves of amplitude X)	2X (sum)	X-X=0 (difference)

9.4.1Conditions for interferenceOften examined

The following conditions are necessary for the interference of waves:

The waves must be travelling in the same direction.
The waves must arrive at the same place at the same time.
The waves must be in phase for constructive interference and out of phase (180^\circ) for destructive interference.
The path difference must be an integral multiple of the wavelength \lambda for constructive interference, and an odd integral multiple of \tfrac12\lambda for destructive interference.

Path difference

The path difference d is the difference between the distances travelled by the two waves to reach the point. The conditions of part (iv) are:

d=m\lambda\qquad(m=0,\ 1,\ 2,\ 3,\ \dots)

constructive interference

d=\left(m+\dfrac12\right)\lambda\qquad(m=0,\ 1,\ 2,\ 3,\ \dots)

destructive interference

Path difference d	Phase relation at the point	Result
0,\ \lambda,\ 2\lambda,\ 3\lambda,\ \dots	in phase	constructive (loud sound)
\tfrac12\lambda,\ \tfrac32\lambda,\ \tfrac52\lambda,\ \dots	180^\circ out of phase	destructive (silence)

Worked example: loud or quiet at a point

Two loudspeakers A and B are driven by the same oscillator, so they emit in phase. Each produces sound of frequency 400\,\mathrm{Hz} (speed of sound 340\,\mathrm{m\,s^{-1}}). A listener at P is 5.10\,\mathrm{m} from A and 5.95\,\mathrm{m} from B. (a) Is the sound at P loud or quiet? (b) What happens at P if the frequency is changed to 200\,\mathrm{Hz}?

Given

f = 400\,\mathrm{Hz}
r_1 = 5.10\,\mathrm{m}
r_2 = 5.95\,\mathrm{m}
v = 340\,\mathrm{m\,s^{-1}}

To find

(a) loud or quiet at P
(b) the effect of f=200\,\mathrm{Hz}

Two in-phase loudspeakers A and B sending sound to the point P. The longer path r_2 (accent) exceeds r_1 by the path difference d.

Find the wavelength, because the path difference must be compared with it.
\lambda=\dfrac{v}{f}=\dfrac{340}{400}=0.85\,\mathrm{m}
Find the path difference.
d=r_2-r_1=5.95-5.10=0.85\,\mathrm{m}
(a) Divide by the wavelength. A whole number of wavelengths means the waves arrive in step (crest on crest), so the interference is constructive.
\dfrac{d}{\lambda}=\dfrac{0.85}{0.85}=1\ \Rightarrow\ d=1\lambda\ (m=1):\ \text{loud sound}
(b) At 200\,\mathrm{Hz} the wavelength doubles, but the path difference does not change (it depends only on the geometry).
\lambda=\dfrac{340}{200}=1.70\,\mathrm{m},\qquad \dfrac{d}{\lambda}=\dfrac{0.85}{1.70}=0.5
A path difference of half a wavelength, d=\left(0+\tfrac12\right)\lambda, puts the waves in antiphase (crest on trough), so the interference is destructive.
d=\left(m+\tfrac12\right)\lambda\ (m=0):\ \text{quiet}

\text{(a) loud at }400\,\mathrm{Hz}\ (d=\lambda);\qquad \text{(b) quiet at }200\,\mathrm{Hz}\ (d=\tfrac12\lambda)

9.4.2Interference of sound wavesOften examined

One simple device for demonstrating the interference of sound waves is shown in Fig. 9.5 (a). Sound from a loudspeaker S is sent into a tube at point P, where there is a T-shaped junction. Half the sound energy travels in one direction and half travels in the opposite direction, so the sound waves that reach the receiver R can travel along either of the two paths. The lower path length r_1 is fixed, but the upper path length r_2 can be varied by sliding the U-shaped tube.

Figure 9.5 (a): Interference of sound. The path length r_2 is changed by sliding the U-shaped tube, while r_1 is fixed.

If the two paths are equal, both waves arrive at R in phase, so a loud sound is heard due to constructive interference.
If the path length r_2 is adjusted so that the path difference r_2-r_1 is an odd integral multiple of \tfrac12\lambda, both waves arriving at R are out of phase. So no sound is heard, due to destructive interference.
If the sound waves are stopped at Q by pinching the rubber portion of the tube, loud sound is heard again. This proves that the silence was due to destructive interference of the two sound waves.

Noise-cancelling headphones

Noise-cancelling headphones (Fig. 9.5 b) use the principle of superposition to attenuate the environmental noise inside the ear cup of the headphone. The basic idea is to sense the unwanted sound wave coming from the noisy environment and produce an appropriate cancelling sound wave through a loudspeaker. The cancelling wave should match the noise in amplitude and have the opposite phase (that is, be 180^\circ out of phase).

Figure 9.5 (b): Noise-cancelling headphones. The cancelling wave is 180^\circ out of phase with the environmental noise, so the two destructively interfere and the unwanted sound is cancelled out.

In this way the cancelling sound wave destructively interferes with the environmental noise, and the noise component is effectively attenuated.

9.5BeatsOften examined

Beats are the interference effect that results from the superposition of two waves with slightly different frequencies. When two sound waves of slightly different frequencies meet, their amplitudes are added and subtracted alternately over a given period. Hence the sound grows louder and softer through the period, and a fluctuating sound (alternate soft and loud) is heard. The loud sensation of sound is produced by constructive interference, and the soft sound by destructive interference.

Beats: The periodic alternations of sound between maximum and minimum loudness produced when two waves of slightly different frequencies are played simultaneously.

Let us consider two sound waves A and B of slightly different frequencies but similar amplitude, propagating in the same medium. A resultant wave C is obtained by the superposition of A and B, as shown in Fig. 9.6. When the two sound waves meet, a fluctuating sound can be heard. For a certain time the crest of A overlaps the crest of B; this causes constructive interference, so the sound intensity rises for this period. For another interval of time the crest of B overlaps the trough of A, which causes destructive interference.

Figure 9.6: Generation of beats. Waves A and B have slightly different frequencies. Their sum C=A+B has an amplitude that grows to a maximum where the waves are in step and shrinks to a minimum where they are out of step.

Beat frequency

Beat frequency: The number of beats generated per second.

The beat frequency is equal to the absolute value of the difference between the frequencies of the two waves. It is denoted by f_b.

f_b=\left|f_1-f_2\right|\qquad(9.10)

f_1, f_2: frequencies of the two waves; f_b: beats per second.

So if two sound waves of frequencies 32\,\mathrm{Hz} and 30\,\mathrm{Hz} are played simultaneously, a beat frequency of 2\,\mathrm{Hz} is detected. If the number of beats per second is more than 10, the beats cannot be heard as separate.

Applications of beats

Beats are very useful to compare frequencies, to find an unknown frequency, to tune musical instruments, in ultrasonic imaging and in radar speed traps.

Beats in musical instruments: tuning a piano

Musicians use the phenomenon of beats to tune a piano string. They pluck the string and tap a tuning fork at the same time. If the two sound sources, the piano string and the tuning fork, produce detectable beats, then their frequencies are not identical. The musician then adjusts the tension of the piano string and repeats the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency is reduced and approaches 0\,\mathrm{Hz}. When beats are no longer heard, the piano string is tuned to the tuning fork: they play the same frequency. This process allows a musician to match the strings' frequency to the frequency of a standardised set of tuning forks.

Example 9.3: beats from two musical instruments

Two musical instruments are sounded together to produce beats. Evaluate the beat frequency of these two sound waves having frequencies 750\,\mathrm{Hz} and 390\,\mathrm{Hz} respectively. These beats cannot be detected as separate by the human ear. Can you tell why?

Given

f_1 = 750\,\mathrm{Hz}
f_2 = 390\,\mathrm{Hz}

To find

beat frequency f_b

Use the beat frequency formula.
f_b=\left|f_1-f_2\right|
Substitute the values and subtract.
f_b=\left|750-390\right|=360\,\mathrm{Hz}
Why no separate beats? The ear can hear beats as separate only if there are not more than about 10 per second. Here there would be 360 per second, far too fast, so the sound is just a mixture of two notes.

f_b=360\,\mathrm{Hz}

Worked example: reading beats from a graph

The graph shows the resultant when two tones are sounded together. (a) How many beats are heard per second? (b) One tone has a frequency of 5\,\mathrm{Hz}. What can the frequency of the other tone be? (c) How many beats occur in 5\,\mathrm{s}, and what is the time between two successive beats? (Low frequencies are used so that the cycles can be counted; the same reasoning holds for 440\,\mathrm{Hz} and 438\,\mathrm{Hz}.)

Sum of two low tones

The resultant (accent) swells and fades. The dashed envelope shows the loud moments: they occur at t=0,\ 0.5,\ 1.0\,\mathrm{s}.

(a) One beat is one loud swell. The graph shows loud swells at 0, 0.5 and 1.0\,\mathrm{s}, so two swells fit into each second.
f_b=2\,\mathrm{Hz}\ (\text{2 beats per second})
(b) The beat frequency is the difference of the two frequencies, so the other tone is 2\,\mathrm{Hz} above or 2\,\mathrm{Hz} below 5\,\mathrm{Hz}. Beats alone cannot tell which.
f_b=\left|f_1-f_2\right|\ \Rightarrow\ \left|5-f_2\right|=2\ \Rightarrow\ f_2=3\,\mathrm{Hz}\ \text{or}\ 7\,\mathrm{Hz}
(c) The number of beats is the beat frequency times the time, and the time between beats is its reciprocal.
N=f_b\,t=2\times5=10,\qquad T_b=\dfrac{1}{f_b}=\dfrac{1}{2}=0.5\,\mathrm{s}

f_b=2\,\mathrm{Hz};\quad f_2=3\,\mathrm{Hz}\text{ or }7\,\mathrm{Hz};\quad 10\text{ beats in }5\,\mathrm{s},\ \text{one every }0.5\,\mathrm{s}

9.6Stationary waves (or standing waves)Often examined

Sometimes waves do not seem to move; rather, they just vibrate about a mean position. Such standing waves can be seen on a rubber band. If a rubber band is tied at both ends and plucked from the middle, stationary waves are produced by the reflections of waves from the ends of the rubber band.

Stationary wave: The wave obtained when two identical waves having the same speed, amplitude and frequency, travelling in opposite directions, are superposed.

A stationary wave is formed by the superposition of two identical waves moving in opposite directions, each having the same amplitude and frequency, as illustrated in Fig. 9.7. The wave reflected at the boundary interferes constructively or destructively with the incoming wave. The two waves move through each other with their displacements adding as they go by. The displacement of the particles of the waves is shown at seven different positions at t=0,\ \tfrac14T,\ \tfrac12T,\ \tfrac34T and T.

Figure 9.7: A standing wave created by the superposition of two identical waves moving in opposite directions. Rows (a) and (b) are the two travelling waves ((a) moving left, (b) moving right) and row (c) is their sum, the stationary wave, at five instants. Positions 1, 3, 5 and 7 are nodes; 2, 4 and 6 are antinodes.

It can be noted from Fig. 9.7 (c) that the displacement of points 1, 3, 5 and 7 is always zero. These points of the medium do not vibrate at all and are called nodes.

The displacement of points 2, 4 and 6 has maximum amplitude, with these points oscillating rapidly up and down. These points are called antinodes. They oscillate with an amplitude equal to the sum of the amplitudes of the component waves. Antinodes are usually represented by A and nodes by N.

Node (N): A point of a stationary wave where the medium does not vibrate at all: the displacement is always zero.
Antinode (A): A point of a stationary wave where the amplitude of vibration is a maximum, equal to the sum of the amplitudes of the two component waves.

This pattern of nodes and antinodes forms a stationary wave because it appears to be standing, not moving through space like progressive waves. So a standing wave is confined to a given region of space.

Characteristics of stationary waves

No energy is transferred from particle to particle in a stationary wave.
The distance between two successive nodes or two successive antinodes is equal to \lambda/2.
The distance between an adjacent node and antinode is equal to \lambda/4.

9.6.1Stationary waves in a stretched stringOften examined

Stationary waves can be set up in any medium, and they do not transmit energy from one point to another. Standing waves are also found on the strings of musical instruments and are produced by the reflections of waves from the ends of the string.

Consider a string of length L stretched by clamping its two ends so that the tension in the string is T, as shown in Fig. 9.8 (a). To find the frequencies of vibration we have to pluck the string at different points. The clamped ends cannot move, so each end must be a node.

Figure 9.8 (a): A stretched string of length L tied (clamped) at both ends, with tension T.

Derivation Fundamental frequency of a stretched string (one loop)

Aim. Find the frequency f_1 when the string vibrates in one loop.

Figure 9.8 (b): Stationary wave having one loop. The nodes N are at the clamped ends and the antinode A is at the middle: L=\lambda_1/2.

Pluck the string at its middle. A stationary wave of one loop is set up (Fig. 9.8 b). Let f_1 and \lambda_1 be the frequency and wavelength of either of the component transverse waves. One loop is half a wavelength, so the length of the string is half a wavelength.
L=\dfrac{\lambda_1}{2}\ \Longrightarrow\ \lambda_1=2L
If M is the total mass of the string, the speed v of the progressive wave along the string is given by the tension T and the length L and mass M of the string.
v=\sqrt{\dfrac{T\times L}{M}}
Use v=f_1\lambda_1 and substitute \lambda_1=2L.
f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{2L}
Replace v by its value.
f_1=\dfrac{1}{2L}\sqrt{\dfrac{T\times L}{M}}\qquad(9.11\,\mathrm{a})
If m=M/L is the mass per unit length of the string, then TL/M=T/m.
f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}\qquad(9.11\,\mathrm{b})

Resultf_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}

This is called the fundamental frequency, the first harmonic or the first mode of vibration.

Derivation Second harmonic of a stretched string (two loops)

Aim. Find the frequency f_2 when the string vibrates in two loops.

Figure 9.8 (c): Stationary wave having two loops: L=\lambda_2.

Pluck the string at one quarter of its length. A stationary wave having two loops is set up (Fig. 9.8 c). Let f_2 and \lambda_2 be the frequency and wavelength of either component wave. Each loop equals half a wavelength, so the length of the string is two halves of a wavelength.
L=\dfrac{\lambda_2}{2}+\dfrac{\lambda_2}{2}=\lambda_2\ \Longrightarrow\ \lambda_2=L
Use v=f_2\lambda_2.
f_2=\dfrac{v}{\lambda_2}=\dfrac{v}{L}
Write v/L as 2\,(v/2L).
f_2=2\left(\dfrac{v}{2L}\right)
Because v/2L=f_1:
f_2=2f_1\qquad(9.12)

Resultf_2=2f_1

When the string vibrates in two loops its frequency of vibration is doubled compared with the fundamental frequency. This is called the second harmonic or second mode of vibration.

Derivation Third harmonic of a stretched string (three loops)

Aim. Find the frequency f_3 when the string vibrates in three loops.

Figure 9.8 (d): Stationary wave having three loops: L=\tfrac32\lambda_3.

Pluck the string at one sixth of its length. A stationary wave having three loops is set up (Fig. 9.8 d). Let f_3 and \lambda_3 be the frequency and wavelength of either component wave. Each loop equals half a wavelength, so the length of the string is three halves of a wavelength.
L=\dfrac{\lambda_3}{2}+\dfrac{\lambda_3}{2}+\dfrac{\lambda_3}{2}=\dfrac{3\lambda_3}{2}\ \Longrightarrow\ \lambda_3=\dfrac{2L}{3}
Use v=f_3\lambda_3.
f_3=\dfrac{v}{\lambda_3}=\dfrac{3v}{2L}=3\left(\dfrac{v}{2L}\right)
Because v/2L=f_1:
f_3=3f_1\qquad(9.13)

Resultf_3=3f_1

When the string vibrates in three loops its frequency is three times the fundamental frequency. This is called the third harmonic or third mode of vibration. Similarly, frequencies 4f_1,\ 5f_1,\ 6f_1,\ \dots give the 4th, 5th, 6th, ... harmonics, and the string vibrates in 4,\ 5,\ 6,\ \dots loops.

Derivation The nth harmonic of a stretched string (n loops)

Aim. Find the wavelength \lambda_n and frequency f_n when the string vibrates in n loops.

Each loop is half a wavelength, so n loops make a length L:
L=\dfrac{\lambda_n}{2}+\dfrac{\lambda_n}{2}+\dfrac{\lambda_n}{2}+\dots=\dfrac{n\lambda_n}{2}
Make \lambda_n the subject.
\lambda_n=\dfrac{2L}{n}\qquad(9.14)
Use f_n=v/\lambda_n and put in \lambda_n=2L/n. The result is n times the fundamental.
f_n=\dfrac{nv}{2L}=nf_1\qquad\text{or}\qquad f_n=\dfrac{n}{2L}\sqrt{\dfrac{T}{m}}\qquad(9.15)

Resultf_n=nf_1=\dfrac{n}{2L}\sqrt{\dfrac{T}{m}},\qquad\lambda_n=\dfrac{2L}{n}

The lowest frequency, the fundamental frequency, corresponds to the longest wavelength.

Mode	Loops	Length–wavelength relation	Wavelength	Frequency
Fundamental (1st harmonic)	1	L=\lambda_1/2	2L	f_1=\dfrac{v}{2L}
2nd harmonic	2	L=\lambda_2	L	f_2=2f_1
3rd harmonic	3	L=3\lambda_3/2	2L/3	f_3=3f_1
nth harmonic	n	L=n\lambda_n/2	2L/n	f_n=nf_1

Conclusions

The string resonates only if its length is a whole number multiple of half a wavelength, that is L=\dfrac{\lambda}{2},\ 2\dfrac{\lambda}{2},\ 3\dfrac{\lambda}{2},\ \dots
Stationary waves on the string can be set up only with a discrete set of frequencies f_1,\ f_2,\ f_3,\ \dots,\ f_n. This is called quantisation of frequency.
The lowest frequency f_1 is called the fundamental frequency and its higher multiples are called overtones or harmonics.
As the string vibrates in more and more loops its frequency goes on increasing but the wavelength decreases correspondingly, so that the product of frequency and wavelength is always constant and equal to the speed of the wave.

9.7Stationary waves in air columnOften examined

Standing waves can also be generated in air columns such as organ pipes. An organ pipe is a musical instrument. It consists of a long tube which produces sound by means of a vibrating air column. Flute, trombone and clarinet are familiar examples of organ pipes. Organ pipes can be open or closed:

Open organ pipe: An organ pipe whose both ends are open.
Closed organ pipe: An organ pipe with one end closed (and the other end open).

Suppose we have a tube that is closed at one end and open at the other. If we hold a vibrating tuning fork near the open end of the tube, an incident sound wave travels through the tube and reflects off the closed end. The reflected sound has the same frequency and wavelength as the incident sound wave, but travels in the opposite direction.

At the closed end of the tube, the air molecules have very little freedom to oscillate, and a node arises.
At the open end the molecules are free to move, and at the right frequency an antinode occurs and the air column in the tube resonates loudly.
So the standing wave formed in the tube has an antinode at the open end and a node at the closed end. A portion of the sound wave is reflected back into the tube even at an open end.

The distance from a node to an antinode is one-fourth of a wavelength, and this equals the length of the tube. Thus L=\tfrac14\lambda_1.

9.7.1Modes of vibration in open pipeOften examined

Let a vibrating tuning fork be held at one end of an open pipe of length L. There are antinodes at both ends, with a node at the middle (for the fundamental). The modes of vibration of an open organ pipe are as follows.

Derivation Fundamental frequency of an open pipe

Aim. Find the frequency f_1 of the first mode of an open pipe of length L.

Figure 9.9 (a): Fundamental frequency (first harmonic) of an open pipe: antinodes A at both ends and a node N in the middle, so L=\lambda_1/2.

Fig. 9.9 (a): an antinode at each open end and a node in the middle. The distance from an antinode to the nearest node is \lambda_1/4, and there are two such gaps.
L=\dfrac{\lambda_1}{4}+\dfrac{\lambda_1}{4}=\dfrac{\lambda_1}{2}\ \Longrightarrow\ \lambda_1=2L
Use v=f_1\lambda_1.
f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{2L}\qquad(9.16)

Resultf_1=\dfrac{v}{2L}

This is called the fundamental frequency or first harmonic of the open pipe.

Derivation Second harmonic of an open pipe

Aim. Find the frequency f_2 of the second mode of an open pipe.

Figure 9.9 (b): Second harmonic of an open pipe: L=\lambda_2.

Fig. 9.9 (b): the length holds one whole wavelength.
L=\lambda_2\ \Longrightarrow\ \lambda_2=L
Use v=f_2\lambda_2 and write v/L=2(v/2L).
f_2=\dfrac{v}{\lambda_2}=\dfrac{v}{L}=2\left(\dfrac{v}{2L}\right)
Since v/2L=f_1:
f_2=2f_1\qquad(9.17)

Resultf_2=2f_1

This is called the second harmonic of the open pipe.

Derivation Third harmonic of an open pipe

Aim. Find the frequency f_3 of the third mode of an open pipe.

Figure 9.9 (c): Third harmonic of an open pipe: L=\tfrac32\lambda_3.

Fig. 9.9 (c): the length holds three half-wavelengths.
L=3\,\dfrac{\lambda_3}{2}\ \Longrightarrow\ \lambda_3=\dfrac{2L}{3}
Use v=f_3\lambda_3.
f_3=\dfrac{v}{\lambda_3}=\dfrac{3v}{2L}=3\left(\dfrac{v}{2L}\right)
Since v/2L=f_1:
f_3=3f_1\qquad(9.18)

Resultf_3=3f_1

This is the third harmonic of the open pipe.

Similarly, from the above results, the wavelength \lambda_n and frequency f_n of the nth harmonic are given by:

\lambda_n=\dfrac{2L}{n}\qquad(9.19)

f_n=nf_1=n\dfrac{v}{2L}\qquad(9.20)\qquad\text{where }n=1,2,3,\dots

9.7.2Modes of vibration in closed pipeOften examined

Let a vibrating tuning fork be held at one end of a closed pipe of length L. There is an antinode at the open end and a node at the closed end. The modes of vibration of a closed organ pipe are as follows.

Derivation First harmonic (fundamental) of a closed pipe

Aim. Find the frequency f_1 of the first mode of a closed pipe of length L.

Figure 9.10 (a): First harmonic of a closed pipe: an antinode at the open end and a node at the closed end, so L=\lambda_1/4.

Fig. 9.10 (a): one node-to-antinode gap fits in the pipe.
L=\dfrac{\lambda_1}{4}\ \Longrightarrow\ \lambda_1=4L
Use v=f_1\lambda_1.
f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{4L}\qquad(9.21)

Resultf_1=\dfrac{v}{4L}

This is the fundamental frequency or first harmonic of the closed pipe. It is half the fundamental frequency v/2L of an open pipe of the same length.

Derivation Second mode of a closed pipe

Aim. Find the frequency f_2 of the second mode of a closed pipe.

Figure 9.10 (b): Second mode of a closed pipe: L=\tfrac34\lambda_2.

Fig. 9.10 (b): the length holds a half-wavelength and a quarter-wavelength.
L=\dfrac{\lambda_2}{2}+\dfrac{\lambda_2}{4}=\dfrac{3\lambda_2}{4}\ \Longrightarrow\ \lambda_2=\dfrac{4L}{3}
Use v=f_2\lambda_2.
f_2=\dfrac{v}{\lambda_2}=3\,\dfrac{v}{4L}
Since v/4L=f_1:
f_2=3f_1\qquad(9.22)

Resultf_2=3f_1

The next frequency of a closed pipe is three times the fundamental.

Derivation Third mode of a closed pipe

Aim. Find the frequency f_3 of the third mode of a closed pipe.

Figure 9.10 (c): Third mode of a closed pipe: L=\tfrac54\lambda_3.

Fig. 9.10 (c): the length holds one whole wavelength and a quarter-wavelength.
L=\lambda_3+\dfrac{\lambda_3}{4}=\dfrac{5\lambda_3}{4}\ \Longrightarrow\ \lambda_3=\dfrac{4L}{5}
Use v=f_3\lambda_3.
f_3=\dfrac{v}{\lambda_3}=5\,\dfrac{v}{4L}
Since v/4L=f_1:
f_3=5f_1\qquad(9.23)

Resultf_3=5f_1

The third mode of a closed pipe has five times the fundamental frequency.

Similarly, from the above results, the wavelength \lambda_n and frequency f_n for the nth mode are given by:

\lambda_n=\dfrac{4L}{2n-1}\qquad(9.24)

\begin{aligned}f_n&=(2n-1)f_1\\ \text{or}\quad f_n&=(2n-1)\dfrac{v}{4L}&&(9.25)\\ \text{where }n&=1,2,3,\dots\end{aligned}

Thus in an open pipe the fundamental frequency is \dfrac{v}{2L} and all harmonics are present, while in a closed pipe the fundamental frequency is \dfrac{v}{4L} and only odd harmonics are present.

	Open pipe	Closed pipe
Ends	antinode at both ends	antinode at open end, node at closed end
Fundamental frequency	f_1=\dfrac{v}{2L}	f_1=\dfrac{v}{4L}
Wavelength of the nth mode	\lambda_n=\dfrac{2L}{n}	\lambda_n=\dfrac{4L}{2n-1}
Frequencies	f_n=nf_1: f_1,\ 2f_1,\ 3f_1,\ \dots	f_n=(2n-1)f_1: f_1,\ 3f_1,\ 5f_1,\ \dots
Harmonics present	all	odd only

For your information

During earthquakes, often buildings of a certain height are devastated while other, taller buildings remain intact. The building height matches the condition for setting up a standing wave for that particular height, which gives evidence for conditions appropriate for resonance, and constructive and destructive interference. As the earthquake waves travel along the surface of the Earth and reflect off denser rocks, constructive interference occurs at certain points. A building may be vibrated for several seconds with a driving frequency matching that of the natural frequency of vibration of the building, producing a resonance that results in one building collapsing while neighbouring buildings do not. Often areas closer to the epicentre are not damaged while areas farther away are damaged.

Worked example: open pipe against closed pipe of the same length

An open pipe and a closed pipe are each 0.85\,\mathrm{m} long. The speed of sound is 340\,\mathrm{m\,s^{-1}}. Find the first three frequencies each pipe can sound, and say which frequencies one pipe has that the other lacks.

Given

L = 0.85\,\mathrm{m}
v = 340\,\mathrm{m\,s^{-1}}

To find

frequencies of the open pipe
frequencies of the closed pipe

The fundamental mode of an open pipe and a closed pipe of the same length (L=0.85\,\mathrm{m}, v=340\,\mathrm{m\,s^{-1}}). N is a node and A an antinode. The closed pipe has half the frequency.

Open pipe. Antinodes at both ends give L=\lambda/2 for the fundamental, so f_1=v/2L.
f_1=\dfrac{v}{2L}=\dfrac{340}{2(0.85)}=200\,\mathrm{Hz}
All harmonics are present in an open pipe, f_n=nf_1.
f_1,\ f_2,\ f_3=200,\ 400,\ 600\,\mathrm{Hz}
Closed pipe. A node at the closed end and an antinode at the open end give L=\lambda/4, so f_1=v/4L.
f_1=\dfrac{v}{4L}=\dfrac{340}{4(0.85)}=100\,\mathrm{Hz}
Only odd multiples are present in a closed pipe, f_n=(2n-1)f_1 with n=1,2,3.
f=100,\ 300,\ 500\,\mathrm{Hz}
Compare. The closed pipe has half the fundamental frequency of the open pipe of equal length. Its series 100,\,300,\,500\,\mathrm{Hz} has only odd multiples of 100\,\mathrm{Hz}, while the open pipe series 200,\,400,\,600\,\mathrm{Hz} has every multiple of 200\,\mathrm{Hz}.

\text{open: }200,\,400,\,600\,\mathrm{Hz};\qquad \text{closed: }100,\,300,\,500\,\mathrm{Hz}

9.7.3Experiment to demonstrate stationary wavesOften examined

A stationary wave can be set up in an air column by holding a sounded tuning fork over the open end of a glass tube filled with water, as shown in Fig. 9.10 (d). If the water surface in the tube is lowered with the help of a reservoir, then at a certain height of water the air column in the tube resonates loudly. There are several other heights at which the air column in the tube resonates.

Figure 9.10 (d): Resonance tube. The water level (and so the length L of the air column) is changed by raising or lowering the reservoir. The air column is a closed pipe: a node at the water surface and an antinode at the open end. Loud sound is heard at the resonant lengths L=\lambda/4,\ 3\lambda/4,\ \dots

Example 9.4: harmonics of a guitar string

A guitar string with a length of 80.0\,\mathrm{cm} is plucked. The speed of a wave in the string is 400\,\mathrm{m\,s^{-1}}. Calculate the frequency of the first, second and third harmonics.

Given

L = 80.0\,\mathrm{cm}=0.80\,\mathrm{m}
v = 400\,\mathrm{m\,s^{-1}}

To find

first harmonic f_1
second harmonic f_2
third harmonic f_3

A string fixed at both ends has fundamental frequency f_1=v/2L.
f_1=\dfrac{v}{2L}=\dfrac{400}{2(0.80)}=250\,\mathrm{Hz}
The second harmonic is twice the fundamental.
f_2=2f_1=2(250)=500\,\mathrm{Hz}
The third harmonic is three times the fundamental.
f_3=3f_1=3(250)=750\,\mathrm{Hz}

f_1=250\,\mathrm{Hz},\quad f_2=500\,\mathrm{Hz},\quad f_3=750\,\mathrm{Hz}

Assignment 9.3: length of a guitar string

The frequency of the first harmonic of a vibrating guitar string is 587\,\mathrm{Hz}. The speed of the wave is 600\,\mathrm{m\,s^{-1}}. Find the length of the string.

Given

f_1 = 587\,\mathrm{Hz}
v = 600\,\mathrm{m\,s^{-1}}

To find

length L

For the first harmonic f_1=v/2L. Make L the subject.
L=\dfrac{v}{2f_1}
Substitute the values.
L=\dfrac{600}{2(587)}=\dfrac{600}{1174}=0.511\,\mathrm{m}

L=0.511\,\mathrm{m}\ \ (\approx51\,\mathrm{cm})

9.8PolarisationOften examined

Polarisation is a phenomenon associated with transverse waves. A light wave is an electromagnetic wave. An electromagnetic wave is a special type of transverse wave that has both an electric and a magnetic component, as shown in Fig. 9.11 (a).

Figure 9.11 (a): An electromagnetic wave. The electric field E and the magnetic field B vibrate at right angles to each other and to the direction of travel.

Ordinarily, a ray of light consists of a number of waves vibrating in all the directions perpendicular to its line of propagation. Light emitted by the sun, a lamp or a candle flame is created by electric charges that vibrate in various directions, thus creating a variety of electromagnetic waves that vibrate in several planes perpendicular to the line of propagation, as shown in Fig. 9.11 (b). A light wave that is vibrating in more than one plane is referred to as unpolarised light. It is possible to transform unpolarised light into polarised light.

Polarised light: Light waves in which the vibrations occur in a single plane.
Polarisation: The process of transforming unpolarised light into polarised light.

There are a variety of methods of polarising light, discussed below.

9.8.1Polarisation by a polaroid filterOften examined

The most common method of polarisation uses a Polaroid filter. Polaroid filters are made of a special material that is capable of blocking all the planes of vibration of an electromagnetic wave except one, and transmitting only that one plane of vibration, as in Fig. 9.11 (b). So a Polaroid serves as a device that filters out all but one plane of the vibrations upon transmission of light through the filter. When unpolarised light is transmitted through a Polaroid filter, it emerges as polarised light: its intensity is reduced to half and its vibrations are in a single plane.

Figure 9.11 (b): A Polaroid filter blocks all the planes of vibration of an electromagnetic wave except one and transmits only one plane of vibration.

A Polaroid filter is able to polarise light because of the chemical composition of the filter's material. In the same manner, two Polaroid filters oriented with their polarisation axes perpendicular to each other will block all the light. This observation could never be explained by a particle nature of light. When unpolarised light passes through a Polaroid filter its intensity is reduced and only light parallel to the grid axis within the polarising filter is allowed to pass.

The axis of polarisation of the transmitted light is the same as that of the filter that polarised it, and the light is then perfectly plane polarised. When perfectly plane polarised light is incident on an analyser, the intensity I of the light transmitted by the analyser is directly proportional to the square of the cosine of the angle between the transmission axis of the analyser and that of the polariser:

I\propto\cos^{2}\theta

I=I_0\cos^{2}\theta\qquad(9.26)

\theta: angle between the transmission axis of the analyser and the polariser; I_0: intensity of the plane polarised light falling on the analyser; I: intensity transmitted.

Malus's law: The intensity of plane polarised light transmitted by an analyser is directly proportional to the square of the cosine of the angle \theta between the transmission axis of the analyser and the polariser: I=I_0\cos^{2}\theta.

Angle \theta	\cos^{2}\theta	Transmitted intensity
0^\circ (axes parallel)	1	I=I_0 (all passes)
90^\circ (axes perpendicular)	0	I=0 (all blocked)

Worked example: two polaroid filters in a row

Plane-polarised light of intensity 80\,\mathrm{W\,m^{-2}} is polarised in the vertical plane. It passes through filter A, whose axis is 30^\circ from the vertical, and then through filter B, whose axis is 30^\circ further round (that is, 60^\circ from the vertical). Find the intensity after A and after B. A student says "the total angle is 60^\circ, so the final intensity is 80\cos^{2}60^\circ". Is the student right?

Given

I_0 = 80\,\mathrm{W\,m^{-2}}
\theta_A = 30^\circ
\theta_{AB} = 30^\circ

To find

I_A
I_B
test of the student's claim

Plane-polarised light passes through two polaroid filters. The axis of A is 30° from the vertical polarisation of the light and the axis of B is 30° beyond A. The intensity after each filter is shown above the beam.

Apply Malus's law at filter A. The angle is between the vertical polarisation and the axis of A.
I_A=I_0\cos^{2}30^\circ=80\times(0.866)^2=80\times0.75=60.0\,\mathrm{W\,m^{-2}}
The light leaving A is now polarised along the axis of A. So at filter B the angle that matters is between the axes of A and B, which is 30^\circ.
I_B=I_A\cos^{2}30^\circ=60.0\times0.75=45.0\,\mathrm{W\,m^{-2}}
Test the student's claim.
80\cos^{2}60^\circ=80\times0.25=20\,\mathrm{W\,m^{-2}}\neq45.0\,\mathrm{W\,m^{-2}}
The student is wrong. The angles cannot be added up, because each filter re-polarises the light along its own axis, and Malus's law must be applied one filter at a time.

I_A=60.0\,\mathrm{W\,m^{-2}},\qquad I_B=45.0\,\mathrm{W\,m^{-2}}

9.8.2Polarisation by reflection

Unpolarised light can also be polarised by reflection from non-metallic surfaces, as shown in Fig. 9.11 (c). The amount to which polarisation occurs depends upon the angle at which the light is incident on the surface and on the material of the surface. Metallic surfaces reflect light with a variety of vibrational directions, so such reflected light is unpolarised. However, non-metallic surfaces such as water and asphalt roadways reflect light such that there is a large concentration of vibrations in a plane parallel to the reflecting surface.

Figure 9.11 (c): Polarisation by reflection. Unpolarised light falls on a non-metallic surface; the reflected light is polarised, with its vibrations concentrated in a plane parallel to the surface.

A person viewing objects by means of light reflected off non-metallic surfaces will often perceive a glare if the extent of polarisation is large. Fishermen are familiar with this glare, since it prevents them from seeing fish that lie below the water surface. Light reflected off a lake is partially polarised in a direction parallel to the water's surface. Fishermen know that the use of glare-reducing sunglasses with the proper polarisation axis allows for the blocking of this partially polarised light. By blocking the plane-polarised light, the glare is reduced and the fisherman can more easily see fish located under the water.

9.8.3Uses of polaroidsOften examined

Polaroids in sky photography

Polaroid sunglasses are familiar to most of us. They have a special ability to reduce the glare of light reflected from water or from the sky. A camera used to photograph the clouds is fitted with a polaroid before the camera lens. The light coming from the sky is polarised, and the polaroid blocks it. Thus the background becomes sufficiently dark against which a clear photograph is obtained.

Polaroids in stress analysis

In the plastic industry a polaroid is used to perform stress analysis tests on transparent plastics. As light passes through a plastic, each colour of visible light is polarised with its own direction. If such a plastic is placed between two polarising plates, a colourful pattern is observed. As the top plate is turned, the colour pattern changes, as new colours become blocked and the formerly blocked colours are transmitted. Structural stress in plastic is signified at locations where there is a large concentration of coloured bands. This location of stress is usually the location where structural failure will most probably occur.

Polaroids are also used in the entertainment industry to display 3-D movies.

Example 9.5: two polaroid filters

An unpolarised light having intensity I_0=16\,\mathrm{W\,m^{-2}} is incident on a pair of polarisers. The first polaroid filter has its transmission axis at 50^\circ from the vertical. The second polaroid filter has its transmission axis at 20^\circ from the vertical. Calculate the intensity of the light transmitted through both filters.

Given

I_0 = 16\,\mathrm{W\,m^{-2}}\ \text{(unpolarised)}
\text{axis of 1st polariser} = 50^\circ\ \text{from the vertical}
\text{axis of 2nd polariser} = 20^\circ\ \text{from the vertical}

To find

intensity I_2 transmitted through both filters

The light is halved by the first polaroid, then reduced by \cos^{2}30^\circ by the second.

Unpolarised light loses half its intensity in the first polaroid.
I_1=\dfrac{I_0}{2}=\dfrac{16}{2}=8\,\mathrm{W\,m^{-2}}
The light leaving the first filter is linearly polarised at 50^\circ. The angle between this light and the axis of the second filter is the difference of the two axis angles.
\theta=50^\circ-20^\circ=30^\circ
Apply Malus's law to the second filter.
I_2=I_1\cos^{2}30^\circ=8\times(0.866)^{2}=8\times0.75=6\,\mathrm{W\,m^{-2}}

I_2=6\,\mathrm{W\,m^{-2}}

Assignment 9.4: reducing the intensity by 90%

What angle is required between the direction of polarised light and the axis of a polaroid filter to reduce its intensity by 90\%?

Reducing the intensity by 90\% leaves 10\% of it, so I=0.10\,I_0.
\dfrac{I}{I_0}=0.10
Use Malus's law and solve for \cos\theta.
\cos^{2}\theta=0.10\ \Longrightarrow\ \cos\theta=\sqrt{0.10}=0.316
Take the inverse cosine.
\theta=\cos^{-1}(0.316)=71.6^\circ

\theta\approx71.6^\circ

9.9DiffractionOften examined

Diffraction: The bending of waves around the sharp edges or corners of obstacles (or slits) and spreading into the geometrical shadow.

To observe the diffraction effect in water, we generate plane waves of water in a ripple tank and place two obstacles in line in such a way that the separation between them is equal to the wavelength of the water waves, as shown in Fig. 9.12. After passing through the small slit between the two obstacles, the water waves spread around the slit and change into an almost semicircular pattern (Fig. 9.12 a).

Figure 9.12 (a): Diffraction of water waves through a small slit. The slit is comparable with the wavelength, so the waves spread out in semicircles beyond it.

Diffraction of waves can only be observed clearly if the size of the obstacle is comparable with the wavelength of the wave. If the size of the obstacle is larger than the wavelength of the wave, then only a small diffraction occurs near the corners of the obstacle, as shown in Fig. 9.12 (b).

Figure 9.12 (b): Diffraction of water waves through a large slit. Most of the wavefront goes straight through; only the edges bend a little at the corners.

Size of the slit or obstacle compared with \lambda	What is seen
comparable with the wavelength	clear diffraction: the wave spreads into the geometrical shadow
much larger than the wavelength	only a small bending of the waves near the corners

Diffraction of light

Diffraction can also be observed in light waves, when a beam of monochromatic light passes through a narrow slit, or passes over a knife edge. The diffraction of light can be observed only if the size of the opening or obstacle is small enough to be comparable with the wavelength of the light used.

Consider a beam of light passing through a single slit and falling on a photographic film, as shown in Fig. 9.13. The central bright band (also called a fringe) is of high intensity and very much wider than the slit and the other surrounding bands. These bands are the result of an interference effect.

Figure 9.13: Diffraction of light. A beam of monochromatic light passes through a single narrow slit and forms on the screen a wide, intense central bright band with fainter bands on both sides.

9.10Gravitational waves

Albert Einstein predicted the existence of gravitational waves in 1916.

Gravitational wave: A stretch and compression of space, so it can be found by measuring the change in length between two objects. It is a wave of the intensity of gravity generated by the accelerated masses of an orbital binary system, and it propagates outward from its source at the speed of light.

Gravitational waves are actually ripples in spacetime. When objects move, the curvature of spacetime changes, and these changes move away (like ripples on a water surface) as gravitational waves.

Every physical object that accelerates produces gravitational waves. This includes humans, vehicles, aeroplanes and so on. But the masses and accelerations of objects on Earth are too small to make gravitational waves big enough to be detected with our instruments. To find big enough gravitational waves, we have to look far away, outside our solar system. The Universe is filled with extremely massive objects undergoing quick accelerations, which generate gravitational waves that can be detected. Examples of events that could cause a gravitational wave are:

when a star explodes asymmetrically, called a supernova;
when two big stars orbit each other;
when two black holes orbit each other and collide to merge.

Gravitational waves from such far-off celestial events pass through the Earth, but their amplitude is very small. When a gravitational wave passes a body that has mass, the distortion of spacetime can make the body stretch and compress periodically.

9.10.1Interferometer

Interferometers are tools used for investigation in many fields of science and engineering. They are called interferometers because they work on the interference of two or more light waves to create an interference pattern. This pattern can be measured and analysed. The interference patterns generated by interferometers contain information about the object or phenomenon being studied. They are often used to make very small measurements that are not possible by any other way. That is why they are so powerful for detecting gravitational waves. LIGO's interferometers are designed to measure a distance of 1/10\,000th the width of a proton, small enough!

Interferometers are very sensitive detection devices that make use of the interference of laser beams. (Their working and set-up details are not required.) They were used to first detect the existence of gravitational waves.

For your information

The Laser Interferometer Gravitational-Wave Observatory (LIGO) is a large-scale physics experiment and observatory designed to detect gravitational waves and to develop gravitational-wave observations as an astronomical tool. This project is a collaboration between the USA, India, Germany, Australia and the U.K. LIGO consists of two instruments called interferometers, each with two 4\,\mathrm{km} (2.5 mile) long arms arranged in the shape of an "L". The interferometers act as "antennae" to detect gravitational waves. Gravitational waves cause space itself to stretch in one direction and simultaneously compress in a perpendicular direction. In LIGO, this causes one arm of the interferometer to get longer while the other gets shorter, then vice versa, back and forth as long as the wave is passing. Since the arms are simultaneously changing lengths in opposing ways, this effect is measured by laser beam in the interferometer. Hence gravitational waves are detected.

Schematic of an L-shaped interferometer such as LIGO: two arms, each 4\,\mathrm{km} long, carry laser beams. A passing gravitational wave lengthens one arm while shortening the other, and the change is measured by the interference of the laser beams.

Formula sheet

Intensity of a wave

I=\dfrac{P}{A}=\dfrac{E}{A\times t}

P: power through area A; E: energy in time t

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

Intensity and amplitude

I\propto(\text{amplitude})^{2}

for a progressive wave: doubling the amplitude gives four times the intensity

Wave speed

v=f\lambda

v: speed of sound; f: frequency; \lambda: wavelength

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

Doppler: listener moves towards source

f\prime=\dfrac{v+v_L}{v}\,f

v_L: speed of the listener; source at rest

Unit: \mathrm{Hz}

Doppler: listener moves away from source

f\prime=\dfrac{v-v_L}{v}\,f

v_L: speed of the listener; source at rest

Unit: \mathrm{Hz}

Doppler: source moves towards listener

f\prime=\dfrac{v}{v-v_s}\,f

v_s: speed of the source; listener at rest

Unit: \mathrm{Hz}

Doppler: source moves away from listener

f\prime=\dfrac{v}{v+v_s}\,f

v_s: speed of the source; listener at rest

Unit: \mathrm{Hz}

Superposition of waves

y=y_1+y_2+y_3+\dots+y_n

y_i: displacement due to the ith wave at the point

Unit: same as displacement

Path difference for constructive interference

d=m\lambda

m=0,1,2,3,\dots; d: path difference

Unit: \mathrm{m}

Path difference for destructive interference

d=\left(m+\dfrac12\right)\lambda

m=0,1,2,3,\dots; d: path difference

Unit: \mathrm{m}

Beat frequency

f_b=\left|f_1-f_2\right|

f_1, f_2: frequencies of the two waves

Unit: \mathrm{Hz} (beats per second)

Fundamental frequency of a stretched string

f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}

L: length; T: tension; m: mass per unit length

Unit: \mathrm{Hz}

Wavelength and frequency of the nth harmonic of a string

\lambda_n=\dfrac{2L}{n},\qquad f_n=nf_1=\dfrac{n}{2L}\sqrt{\dfrac{T}{m}}

n=1,2,3,\dots (number of loops)

Units: \mathrm{m}; \mathrm{Hz}

Open pipe

\lambda_n=\dfrac{2L}{n},\qquad f_n=nf_1=n\dfrac{v}{2L}

n=1,2,3,\dots; all harmonics present

Units: \mathrm{m}; \mathrm{Hz}

Closed pipe

\lambda_n=\dfrac{4L}{2n-1},\qquad f_n=(2n-1)f_1=(2n-1)\dfrac{v}{4L}

n=1,2,3,\dots; only odd harmonics present

Units: \mathrm{m}; \mathrm{Hz}

Malus's law

I=I_0\cos^{2}\theta

I_0: intensity of the plane polarised light on the analyser; \theta: angle between the transmission axes

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

Key terms

Intensity: The energy carried by a wave per unit area in unit time; I=P/A, measured in \mathrm{W\,m^{-2}}.
Doppler's effect: The apparent change in the frequency of a wave due to the relative motion between the observer and the source.
Superposition principle: When two or more waves pass through the same region at the same time, the total displacement at a point is the vector sum of the individual displacements.
Coherent waves: Waves that have a constant phase difference and the same frequency (and wavelength).
Coherent source: A source that emits waves of the same frequency, wavelength and phase, or with a constant phase difference.
Interference: The effect produced by the superposition of waves from two coherent sources.
Constructive interference: Two waves overlap at a point in the same phase and reinforce each other.
Destructive interference: Two waves overlap at a point out of phase (180^\circ) and cancel each other.
Path difference: The difference in the distances travelled by two waves to reach a point: d=m\lambda for constructive and d=(m+\tfrac12)\lambda for destructive interference.
Beats: The periodic alternations of sound between maximum and minimum loudness produced when two waves of slightly different frequencies are played together.
Beat frequency: The number of beats produced per second, equal to |f_1-f_2|.
Stationary (standing) wave: The wave obtained when two identical waves with the same speed, amplitude and frequency, travelling in opposite directions, are superposed.
Node: A point of a stationary wave where the displacement is always zero.
Antinode: A point of a stationary wave where the amplitude is a maximum.
Fundamental frequency: The lowest frequency at which a stationary wave can be set up in a string or pipe; it corresponds to the longest wavelength (first harmonic).
Harmonics (overtones): The higher frequencies, whole-number multiples of the fundamental, at which a string or pipe can resonate.
Quantisation of frequency: The fact that stationary waves on a string can be set up only at a discrete set of frequencies f_1,\ f_2,\ f_3,\ \dots
Open organ pipe: An organ pipe with both ends open; it has an antinode at each end and contains all harmonics.
Closed organ pipe: An organ pipe with one end closed; it has a node at the closed end and an antinode at the open end, and contains only odd harmonics.
Polarised light: Light waves in which the vibrations occur in a single plane.
Polarisation: The process of transforming unpolarised light into polarised light.
Polaroid filter: A filter made of a special material that blocks all planes of vibration of an electromagnetic wave except one and transmits only that plane.
Malus's law: The intensity of plane polarised light transmitted by an analyser is I=I_0\cos^{2}\theta, where \theta is the angle between the transmission axes of the analyser and the polariser.
Diffraction: The bending of waves around the sharp edges or corners of obstacles (or slits) and their spreading into the geometrical shadow.
Gravitational wave: A ripple in spacetime, a stretch and compression of space, produced by accelerated masses and travelling outward at the speed of light.
Interferometer: A very sensitive instrument that uses the interference of two or more light (laser) beams to make very small measurements, used to detect gravitational waves.

Multiple choice questionsp. 180–181

The book prints no answer key for these multiple choice questions. Each key below is worked out from the unit's own teaching and the reason is given with it.

On doubling amplitude of the wave, intensity is increased:

A\dfrac{1}{2} times
B2 times
C4 timesAnswer
D3 times

Why

Intensity is proportional to the square of the amplitude, I\propto A^{2}. Doubling the amplitude gives I'=(2A)^{2}=4A^{2}, so the intensity is 4 times larger.

\dfrac{I'}{I}=\dfrac{(2A)^{2}}{A^{2}}=4

Option B treats intensity as proportional to amplitude, which is wrong; the square is the reason for the factor 4.

Two waves, each with amplitude of 0.5\,\mathrm{m} are superimposed with constructive interference such that they are in phase. What is the resultant amplitude?

A0.25\,\mathrm{m}
B0.5\,\mathrm{m}
C0\,\mathrm{m}
D1\,\mathrm{m}Answer

Why

In phase waves reinforce, so their amplitudes add: A_R=0.5\,\mathrm{m}+0.5\,\mathrm{m}=1\,\mathrm{m}.

Option C is the result for destructive interference of equal waves, and option B is the amplitude of one wave alone.

What happens when two sound waves of frequencies differing by more than 10\,\mathrm{Hz} reach our ear simultaneously?

Abeats are not produced.
Bthe waves destroy each other's effect.
Cinterference of sound does not take place.
Dbeats are produced but cannot be heard by human ear.Answer

Why

Beats are produced whenever two close frequencies are superposed. The beat frequency is |f_1-f_2|, and when it is more than about 10 per second the loudness changes are too fast to be heard as separate beats.

Option A is wrong because beats are always produced. Option B is wrong because the waves pass through each other unchanged, and option C is wrong because interference does take place.

A car has two horns; one is emitting a frequency of 199\,\mathrm{Hz} and the other is emitting a frequency of 203\,\mathrm{Hz}. What beat frequency do they produce?

A4\,\mathrm{Hz}Answer
B199\,\mathrm{Hz}
C201\,\mathrm{Hz}
D203\,\mathrm{Hz}

Why

The beat frequency is the difference of the two frequencies:

f_b=|f_1-f_2|=|203-199|=4\,\mathrm{Hz}

Option C is the average of the two frequencies, which is not the beat frequency.

Which of the following frequencies are higher harmonics of a string with fundamental frequency of 150\,\mathrm{Hz}?

A200\,\mathrm{Hz}, 300\,\mathrm{Hz}
B300\,\mathrm{Hz}, 600\,\mathrm{Hz}Answer
C250\,\mathrm{Hz}, 450\,\mathrm{Hz}
D250\,\mathrm{Hz}, 500\,\mathrm{Hz}

Why

The harmonics are whole number multiples of the fundamental, f_n=nf_1. Here 300=2\times150 and 600=4\times150, so both are harmonics.

Each other option contains a value that is not a multiple of 150\,\mathrm{Hz} (200, 250), so it is not a harmonic.

What is the wavelength of the third harmonic (n=3) of a standing wave established on a string of length 3\,\mathrm{m} fixed at both ends?

A1\,\mathrm{m}
B1.5\,\mathrm{m}
C2\,\mathrm{m}Answer
D3\,\mathrm{m}

Why

For a string fixed at both ends, L=n\dfrac{\lambda_n}{2}. For n=3:

\lambda_3=\dfrac{2L}{3}=\dfrac{2\times3\,\mathrm{m}}{3}=2\,\mathrm{m}

Option B is the wavelength of the fourth harmonic, 2L/4, and option D is the length of the string, not a wavelength.

A node is a point located along the medium where there is always ___.

Aa double crest.
Bconstructive interference.
Cdestructive interference.Answer
Da double rarefaction.

Why

At a node the two superposed waves are always out of phase, so they cancel and the displacement is zero at all times. This is destructive interference.

A double crest or a double rarefaction describes a point of maximum amplitude, which is an antinode. Constructive interference gives the antinodes, not the nodes.

Which phenomenon is produced when two or more waves passing simultaneously through the same medium meet up with one another?

Arefraction
Bdiffraction
CinterferenceAnswer
Dreflection

Why

When two or more waves meet in the same region, their displacements add. This superposition is interference (the book's definition for coherent sources).

Refraction is the change of direction at a boundary, diffraction is spreading round an obstacle, and reflection is bouncing back from a surface; none of them is the meeting of waves.

A standing wave is formed by waves of frequency 256\,\mathrm{Hz}. The speed of the waves is 128\,\mathrm{m\,s^{-1}}. The distance between the nodes must be:

A2.00\,\mathrm{m}
B1.00\,\mathrm{m}
C0.500\,\mathrm{m}
D0.250\,\mathrm{m}Answer

Why

The wavelength is \lambda=\dfrac{v}{f}=\dfrac{128}{256}=0.500\,\mathrm{m}. Successive nodes are \lambda/2 apart:

d_{\text{node}}=\dfrac{\lambda}{2}=0.250\,\mathrm{m}

Option C is the wavelength, not the node spacing.

An air column that is closed at one end is used to determine the speed of sound. The frequency of the tuning fork used is 329.6\,\mathrm{Hz}. The length of the shortest air column producing the resonance is 25.0\,\mathrm{cm}. The speed of the sound must be:

A380.6\,\mathrm{m\,s^{-1}}
B282.4\,\mathrm{m\,s^{-1}}
C3.30\times10^{2}\,\mathrm{m\,s^{-1}}Answer
D3.50\times10^{2}\,\mathrm{m\,s^{-1}}

Why

For a column closed at one end the shortest resonance has length L=\lambda/4, so \lambda=4L=4\times0.250\,\mathrm{m}=1.00\,\mathrm{m}.

v=f\lambda=329.6\times1.00=3.30\times10^{2}\,\mathrm{m\,s^{-1}}

End corrections are neglected, as the syllabus requires.

The intensity of beam observed through polarizer:

Ais increased
Bis decreasedAnswer
Cis always zero
Dremains unchanged

Why

A polaroid transmits only one plane of vibration, so the beam leaving it has less intensity: half of the unpolarized intensity, and I_0\cos^{2}\theta for polarized light.

Option C is true only when the two axes are crossed, not in general, and option A and D are wrong because a polaroid always removes some light.

Note. The book prints no key for the MCQs. The key here is worked out for the usual case of a beam passed through a polaroid, which is reduced in intensity. The question does not say whether the beam is already polarized or whether the polaroid is turned, so 'always zero' is taken as the wrong option.

Short questionsp. 181–182

12 marks

Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.

Solution

No. In pure constructive interference the two waves are in phase, so their displacements add and the resultant amplitude is 2A, not A.

A_R=A+A=2A

Intensity is proportional to the square of the amplitude, so the resultant intensity is four times that of each wave:

\dfrac{I_R}{I}=\dfrac{(2A)^{2}}{A^{2}}=4

Twice the intensity would be obtained only if the two intensities were simply added, which happens when the waves do not interfere.

I_R=4I

22 marks

Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.

Solution

The rock gives the water a fixed amount of energy. As the wave moves outwards this energy is spread over a circle of increasing circumference, so less energy falls on each unit of area of the surface. The intensity, which is power per unit area, falls with distance from the source:

I=\dfrac{P}{A}

Because intensity is proportional to the square of the amplitude, a smaller intensity means a smaller amplitude. The energy is not destroyed; it is shared over a wider ring of water.

32026 Annual I, Set B, Q.2 (x)3 marks

Differentiate between constructive and destructive interference.

Solution

Both are effects of the superposition of waves from two coherent sources. They differ in the phase of the waves when they meet and in the result.

Feature	Constructive interference	Destructive interference
Phase	Waves in phase: crest on crest, trough on trough	Waves out of phase by 180^\circ: crest on trough
Path difference	d=m\lambda, m=0,1,2,\dots	d=\left(m+\tfrac{1}{2}\right)\lambda, m=0,1,2,\dots
Resultant amplitude	Amplitudes add: X+X=2X	Amplitudes subtract: X-X=0 for equal waves
Result	Loud sound, bright band, greater intensity	Silence, dark band, intensity falls to zero for equal waves

43 marks

The frequency of a stretched string depends on its length. Give two other factors that affect the frequency of a stretched string.

Solution

The fundamental frequency of a stretched string of length L, tension T and mass per unit length m is

f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}

The two other factors are:

**Tension T.** The frequency is proportional to the square root of the tension, so a tighter string has a higher frequency.
**Mass per unit length m.** The frequency is inversely proportional to the square root of m, so a heavier string has a lower frequency.

\text{Tension } T \text{ and mass per unit length } m

53 marks

The pitch of the sound emitted by the siren of a moving fire engine appears to change as it passes a stationary observer. (i) Name this phenomenon. (ii) Will the crew in the fire engine notice this phenomenon? Give a reason for your answer. (iii) Give an application of this phenomenon.

Solution

(i) This is the Doppler effect: the apparent change in frequency of a wave due to relative motion between the source and the observer.

(ii) No. The crew travels with the siren, so there is no relative motion between the source and the crew. They hear the same frequency that the siren emits.

(iii) Radar speed traps: a radar sends out radio waves, and the change in frequency of the wave reflected from a moving car gives the speed of the car. Other applications are sonar, astronomy and Doppler ultrasound for heart diagnosis.

62 marks

How should a sound source move with respect to an observer so that the frequency of its sound does not change?

Solution

The frequency changes only when the source and the observer have relative motion along the line joining them. From the Doppler formula for a moving source,

f'=\dfrac{v}{v\mp v_s}\,f

the frequency does not change when v_s=0 relative to the observer. So the source must be at rest with respect to the observer: it may be stationary, or it may move together with the observer at the same velocity, as the siren moves with the crew in question 5.

72025 Annual I, Set D, Q.2 (xii) OR2 marks

Can sound waves be polarized? Explain.

Solution

No. Polarization is a phenomenon of transverse waves, whose vibrations can lie in a single plane. Sound is a longitudinal wave: the particles of the medium vibrate along the direction in which the wave travels, so there is no transverse plane of vibration for a polaroid to select. A polaroid filter therefore cannot polarize sound.

82 marks

Suppose that light passes through two Polaroid filters whose polarization axes are parallel to each other. What would be the result?

Solution

The first polaroid polarizes the unpolarized light, so the intensity after it is I_0/2. The light then meets the second filter with its axis parallel to the first, so the angle between the polarized light and the second axis is \theta=0^\circ:

I=\dfrac{I_0}{2}\cos^{2}0^\circ=\dfrac{I_0}{2}

The second filter transmits all of the polarized light, so the light is passed through with no further reduction. Crossed axes (at 90^\circ) would block it completely.

92 marks

Give any three examples of a gravitational wave?

Solution

Gravitational waves are produced by accelerating masses. Three examples given in the book are:

A star exploding asymmetrically, which is called a supernova.
Two big stars orbiting each other.
Two black holes orbiting each other and colliding to merge.

103 marks

Consider the three pairs of sunglasses below. Which pair of the glasses is capable of eliminating the glare resulting from sunlight reflecting off the calm waters of a lake? Explain your option. (The polarization axes are shown by the straight lines)

Solution

Pair A. Light reflected from calm water is partly polarized with its vibrations parallel to the water surface, which is horizontal. The glare can be blocked only by a filter whose axis is perpendicular to this plane of vibration, that is, a vertical axis.

Pair A has vertical axes in both lenses, so it blocks the horizontally polarized glare and still transmits the other light.

Pair B has horizontal axes, which pass the glare, so it does not help. Pair C has crossed axes; it does block the glare, but the crossed lenses block all the light, so it is not a useful pair of sunglasses.

Comprehensive questionsp. 182

15 marks

What is superposition principle? Discuss.

Solution

Superposition principle

The superposition principle states that when two or more waves pass through the same region at the same time, the total displacement at a point where they meet is the vector sum of the individual displacements due to each wave at that point.

y=y_1+y_2+y_3+\cdots+y_n

Here y_1,y_2,\dots,y_n are the displacements of the particle due to each of the n waves, and y is the resultant displacement. Displacements carry their signs, because displacement is a vector quantity.

Discussion

Each wave keeps its own shape and speed after it has passed through the other wave. Waves that meet are not destroyed or changed; they only add at the points where they overlap. The book's Fig. 9.3 shows two waves X and Y superposing to give the resultant Z. At t=0 the displacements are -0.9 and -2.1, so the resultant is

y=(-0.9)+(-2.1)=-3

and at a later time, when both displacements are +3, the resultant is +6. Adding the displacements point by point gives the whole curve Z.

The principle is the basis of interference (question 2), stationary waves (question 3) and beats (question 5).

22025 Annual I, Set B, Q.2 (xiv) OR6 marks

What is interference? Discuss its types and corresponding conditions.

Solution

What is interference

Interference is the effect produced by the superposition of waves from two coherent sources. Coherent waves have the same frequency and wavelength and a constant phase difference, so the pattern of reinforcement and cancellation does not change with time.

Conditions for interference

The waves must travel in the same direction through the same medium.
They must arrive at the same place at the same time.
The waves must be in phase for constructive interference, or out of phase by 180^\circ for destructive interference.
The path difference d must be an integral multiple of \lambda for constructive interference, or an odd multiple of \lambda/2 for destructive interference.

Types of interference

Type	Phase	Path difference	Resultant	Effect
Constructive	In phase	d=m\lambda, m=0,1,2,\dots	Amplitudes add: 2X for two waves of amplitude X	Loud sound, bright band, greater intensity
Destructive	Out of phase by 180^\circ	d=\left(m+\tfrac{1}{2}\right)\lambda, m=0,1,2,\dots	Amplitudes subtract: X-X=0	Silence, dark band, intensity falls to zero for equal waves

Constructive interference

Figure 9.4 (a): two waves of amplitude X in phase give a resultant of amplitude 2X.

Destructive interference

Figure 9.4 (b): two waves 180 degrees out of phase cancel each other.

Sound can show both types. Loud and soft points are produced by constructive and destructive interference of the sound from a loudspeaker, as in the tube experiment of question 8.

32025 Annual I, Set D, Q.56 marks

What are stationary waves? How stationary wave is formed? Discuss.

Solution

What are stationary waves

A stationary (standing) wave is formed when two identical waves, with the same speed, amplitude and frequency, travel in opposite directions and superpose. The pattern does not move through the medium, so it appears to stand still, and no energy is transferred along it.

Formation

The wave reflected at a boundary interferes with the incoming wave. The two waves pass through each other, and at each point their displacements add. Some points are always out of phase, so they cancel and never move. Other points are always in phase and reinforce, so they vibrate with the largest amplitude. Fig. 9.7 of the book shows the pattern at t=0, T/4, T/2, 3T/4 and T.

Points that never move are called nodes (N). Points with maximum amplitude are called antinodes (A). The amplitude of an antinode equals the sum of the amplitudes of the two component waves.

Standing wave

A standing wave at t=0 (solid) and at t=T/8 (dashed). Nodes N have zero displacement at all times; antinodes A have the largest amplitude.

Characteristics

No energy is transferred from particle to particle in a stationary wave.
The distance between two successive nodes, or two successive antinodes, is \lambda/2.
The distance between an adjacent node and antinode is \lambda/4.

42026 Annual I, Set B, Q.6 OR2025 Annual I, Set D, Q.58 marks

Discuss the formation of stationary waves in (a) stretched string (b) open pipe (c) closed pipe.

Solution

Stationary waves in a stretched string

A string of length L fixed at both ends has nodes at both ends. A stationary wave forms when the length is a whole number of half wavelengths, L=n\dfrac{\lambda_n}{2}, and the frequencies are

f_n=nf_1,\qquad f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}

Plucking at the middle gives one loop, the first harmonic. The nth harmonic has n loops.

Stationary waves in an open pipe

In an open pipe both ends are antinodes and there is a node in the middle for the first harmonic. The length of the tube is L=n\dfrac{\lambda_n}{2}, so

\lambda_n=\dfrac{2L}{n},\qquad f_n=n\dfrac{v}{2L}

All harmonics are present in an open pipe.

Stationary waves in a closed pipe

In a pipe closed at one end the closed end is a node and the open end is an antinode. The first harmonic has length \lambda_1/4, and only odd harmonics are present:

L=\dfrac{(2n-1)\lambda_n}{4},\qquad f_n=(2n-1)\dfrac{v}{4L}

Comparison

	Stretched string	Open pipe	Closed pipe
Ends	Nodes at both ends	Antinodes at both ends	Node at closed end, antinode at open end
First harmonic	L=\lambda_1/2, f_1=\dfrac{1}{2L}\sqrt{T/m}	L=\lambda_1/2, f_1=\dfrac{v}{2L}	L=\lambda_1/4, f_1=\dfrac{v}{4L}
Harmonics present	All: f_n=nf_1	All: f_n=nf_1	Odd only: f_n=(2n-1)f_1

First harmonic patterns in (a) a stretched string, (b) an open pipe and (c) a closed pipe. N marks nodes and A antinodes.

52026 Annual I, Set D, Q.2 (x)6 marks

What are beats? Explain.

Solution

Definition

When two sound waves of slightly different frequencies are sounded together, their interference makes the sound rise and fall in loudness. The periodic alternation between maximum and minimum loudness is called beats.

Explanation

The two waves are in phase at some moments, so they reinforce (constructive interference and a loud sound). At other moments the crest of one overlaps the trough of the other, so they cancel (destructive interference and a soft sound). Because the frequencies are slightly different, these moments of agreement and disagreement change slowly, and the loudness rises and falls. Fig. 9.6 of the book shows waves A and B and their sum C.

Generation of beats

Figure 9.6: two waves of nearly equal frequency (dashed) and their sum (solid). The sum is loud where the waves are in phase and soft where they are out of phase.

Beat frequency

The beat frequency f_b is the number of beats per second. It equals the absolute difference of the two frequencies:

f_b=|f_1-f_2|

Beats can be heard as separate beats only when there are about 10 or fewer per second. Beats of higher frequency blur into a single sound.

Example 9.3

Two musical instruments are sounded together to produce beats. Evaluate the beat frequency of two sound waves having frequencies 750\,\mathrm{Hz} and 390\,\mathrm{Hz}.

Given

f_1 = 750\,\mathrm{Hz}
f_2 = 390\,\mathrm{Hz}

To find

Use the beat frequency formula.
f_b=|f_1-f_2|
Substitute the values.
f_b=|750-390|\,\mathrm{Hz}=360\,\mathrm{Hz}

f_b=360\,\mathrm{Hz}

These beats cannot be heard as separate beats, since there are far more than 10 per second.

Applications

Comparing two frequencies and finding an unknown frequency.
Tuning musical instruments: a piano string is tuned until the beats with a tuning fork disappear.
Ultrasonic imaging and radar speed traps.

Note. Example 9.3 in the book gives the beat frequency as 380 Hz. This is a misprint: 750 - 390 = 360 Hz. The answer here is 360 Hz.

65 marks

Explain the use of polaroids in (a) sky photography (b) stress analysis of materials.

Solution

(a) Sky photography

Light from the sky is polarized. A polaroid fitted in front of the camera lens transmits only one plane of vibration, so the polarized light from the sky is reduced. The background becomes dark enough that a clear photograph of the clouds is obtained.

(b) Stress analysis of materials

Transparent plastics are used to study stress. When light passes through a stressed plastic, each colour is polarized in its own direction. If the plastic is placed between two polarizing plates, a coloured pattern is seen. As the top plate is turned, the colours change, because different colours are blocked and transmitted in turn.

Stress in the plastic shows as a large concentration of coloured bands. These locations are where structural failure is most likely to occur.

72025 Annual I, Set D, Q.2 (xi) OR8 marks

What is meant by the Doppler's effect? Discuss its cases. Also give some application of the Doppler's effect.

Solution

Doppler's effect

The Doppler effect is the apparent change in the frequency of a wave due to the relative motion between the observer and the source. It is common to all waves, including sound and light.

Case 1: source at rest, listener moving

If the listener moves with speed v_L towards a stationary source, the speed of sound relative to the listener increases to v+v_L while the wavelength stays the same. The apparent frequency is

f'=\dfrac{v+v_L}{v}\,f \quad(\text{towards}), \qquad f'=\dfrac{v-v_L}{v}\,f \quad(\text{away})

The frequency is higher when the listener approaches and lower when the listener recedes.

Case 2: source moving, listener at rest

Let the source move with speed v_s towards a stationary listener. Each new wave is emitted from a position closer to the listener, so the wavelength heard is shorter:

The wavelength measured by the listener is shortened by the distance the source moves in one period, v_s T=v_s/f.
\lambda'=\lambda-\Delta\lambda=\dfrac{v}{f}-\dfrac{v_s}{f}=\dfrac{v-v_s}{f}
The apparent frequency is the wave speed divided by the apparent wavelength.
f'=\dfrac{v}{\lambda'}=\dfrac{v}{v-v_s}\,f
When the source moves away, the wavelength is lengthened and v_s takes the opposite sign in the denominator.
f'=\dfrac{v}{v+v_s}\,f

Figure 9.1 (c): the source moves to the right, so the wavefronts ahead are closer together and the listener ahead hears a higher frequency.

The frequency is higher when the source approaches and lower when it recedes.

Applications

Radar speed traps: the change in frequency of a radio wave reflected from a moving car gives the car's speed.
Sonar: the speed and direction of ships and submarines are found from the change in frequency of reflected sound waves.
Astronomy: a red shift (lower frequency) shows that a star or galaxy is moving away, and a blue shift shows that it is moving towards the Earth.
Satellites: the velocity of Earth's satellites is found from the Doppler shift in the frequency of the radio waves they transmit.
Medicine: Doppler ultrasound (Doppler echocardiography) shows blood flow through the heart and blood vessels.

86 marks

Describe an experiment to demonstrate the interference of sound.

Solution

Apparatus

A loudspeaker S sends sound into a tube at a T-shaped junction P. Half of the sound energy travels along each of two paths, and the two parts meet again at the receiver R. The lower path of length r_1 is fixed. The upper path of length r_2 can be changed by sliding a U-shaped tube.

Figure 9.5 (a): sound from S divides at P and recombines at R. The path difference is r_2-r_1.

Procedure and results

Make the two paths equal, so that r_2=r_1. Both waves arrive at R in phase, and a loud sound is heard because of constructive interference.
Slide the U-tube until the path difference is an odd multiple of \lambda/2. The waves arrive out of phase, and the sound is no longer heard because of destructive interference.
r_2-r_1=\left(m+\tfrac{1}{2}\right)\lambda
Pinch the rubber portion of the tube at Q to stop one of the two waves. The loud sound returns, which proves that the silence was caused by the interference of the two sound waves and not by the absence of sound.

Conclusion

The experiment shows that sound waves from two paths can reinforce each other to give a loud sound, or cancel each other to give silence, depending on the path difference.

92026 Annual I, Set D, Q.32025 Annual I, Set B, Q.2 (i) OR6 marks

What is polarization? How is plane polarized light produced and detected?

Solution

Polarization

Polarization is the process of transforming unpolarized light, whose vibrations occur in all planes perpendicular to the direction of travel, into plane polarized light, whose vibrations occur in a single plane. Polarization is a property of transverse waves, and light is an electromagnetic transverse wave.

Production of plane polarized light

By a polaroid filter. A polaroid transmits only one plane of vibration. When unpolarized light passes through it, its intensity falls to half, and the light leaving the filter is plane polarized.
By reflection. Light reflected from a non-metallic surface, such as water or an asphalt road, is partly polarized, with its vibrations parallel to the surface. This is why glare off a lake is polarized.

Detection

Plane polarized light is detected with a second polaroid, called the analyser. The analyser is rotated, and the intensity of the light it transmits is found. For a plane polarized beam of intensity I_0 the transmitted intensity is given by Malus's law:

I=I_0\cos^{2}\theta

where \theta is the angle between the transmission axes of the analyser and the polarizer. The transmitted intensity is greatest when \theta=0 and zero when the axes are crossed (\theta=90^\circ).

Malus's law

Transmitted intensity through an analyser as the angle \theta between the axes changes.

106 marks

What is meant by the term 'Gravitational Waves'? Explain in detail.

Solution

Definition

A gravitational wave is a stretch and compression of space. Gravitational waves are ripples in spacetime: when masses move, the curvature of spacetime changes, and these changes travel outwards from the source at the speed of light. They can be found by measuring the change in length between two objects.

Sources

Every accelerating object produces gravitational waves, but the masses and accelerations of objects on Earth are far too small to produce waves large enough to detect. The waves must come from extremely massive objects undergoing rapid acceleration, such as:

a star that explodes asymmetrically, which is called a supernova;
two big stars orbiting each other;
two black holes orbiting each other and colliding to merge.

Effect on matter

A gravitational wave passing a body with mass distorts space-time so that the body is stretched and compressed periodically. Gravitational waves reach the Earth from far off celestial events, but their amplitude is very small.

116 marks

Explain the use of interferometer in detecting gravitational waves.

Solution

The interferometer

An interferometer is a device that makes an interference pattern from two or more beams of light. The pattern contains information about the arms of the device, so the interferometer can measure very small changes in length that cannot be measured in any other way.

How it detects gravitational waves

A laser beam is split into two beams, which travel along two perpendicular arms, are reflected back and recombine to form an interference pattern.
Without a gravitational wave the two arms have fixed lengths, and the pattern is steady.
A gravitational wave stretches space in one direction and compresses it in the perpendicular direction. One arm becomes longer while the other becomes shorter, and then the reverse happens.
The change in arm length changes the path difference between the two beams, so the interference pattern shifts. The shift is measured and analysed.

LIGO's interferometers have two 4 km arms in the shape of an L. They are designed to measure a distance of one ten-thousandth of the width of a proton. The interferometer was used to make the first direct detection of gravitational waves.

Numerical problemsp. 182–183

Ultrasound of intensity 1.50\times10^{2}\,\mathrm{W\,m^{-2}} is produced by the rectangular head of a medical imaging device measuring 3.00\,\mathrm{cm} by 5.00\,\mathrm{cm}. What is its power output?

Solution

Given

I = 1.50\times10^{2}\,\mathrm{W\,m^{-2}}
a = 3.00\,\mathrm{cm}
b = 5.00\,\mathrm{cm}

To find

Find the area of the rectangular head, in SI units.
A=ab=(3.00\times10^{-2}\,\mathrm{m})(5.00\times10^{-2}\,\mathrm{m})=1.50\times10^{-3}\,\mathrm{m^{2}}
Intensity is power per unit area, so the power is intensity times area.
I=\dfrac{P}{A}\quad\Rightarrow\quad P=IA
Substitute the values.
P=(1.50\times10^{2})(1.50\times10^{-3})=0.225\,\mathrm{W}

P=0.225\,\mathrm{W}

Book answer. 0.225 W

Two tuning forks of frequencies 440\,\mathrm{Hz} and 437\,\mathrm{Hz} are sounded together. How many beats will be heard over a period of 10 seconds?

Solution

Given

f_1 = 440\,\mathrm{Hz}
f_2 = 437\,\mathrm{Hz}
t = 10\,\mathrm{s}

To find

number of beats N

Beat frequency is the difference of the two frequencies.
f_b=|f_1-f_2|=|440-437|=3\,\mathrm{Hz}
Beats per second times the time gives the number of beats.
N=f_b\,t=3\times10=30

N=30\text{ beats}

Book answer. 30 beats

Suppose that a string is 1.2\,\mathrm{m} long and vibrates in first, second and third harmonic standing wave patterns. Determine the wavelength of the waves for each of the three patterns.

Solution

Given

L = 1.2\,\mathrm{m}

To find

\lambda_1,\lambda_2,\lambda_3

For a string fixed at both ends, L=n\dfrac{\lambda_n}{2}, so \lambda_n=\dfrac{2L}{n}.
\lambda_n=\dfrac{2L}{n}
First harmonic, n=1.
\lambda_1=\dfrac{2\times1.2}{1}=2.4\,\mathrm{m}
Second harmonic, n=2.
\lambda_2=\dfrac{2\times1.2}{2}=1.2\,\mathrm{m}
Third harmonic, n=3.
\lambda_3=\dfrac{2\times1.2}{3}=0.8\,\mathrm{m}

\lambda_1=2.4\,\mathrm{m},\quad \lambda_2=1.2\,\mathrm{m},\quad \lambda_3=0.8\,\mathrm{m}

Book answer. 2.4 m, 1.2 m, 0.8 m

The string is 6.0\,\mathrm{m} long and is vibrating at the third harmonic. The string vibrates up and down with 45 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.

Solution

Given

L = 6.0\,\mathrm{m}
n = 3\ (\text{third harmonic})
N = 45\text{ cycles}
t = 10\,\mathrm{s}

To find

f
T
\lambda_3
v

Frequency is the number of cycles per second.
f=\dfrac{N}{t}=\dfrac{45}{10}=4.5\,\mathrm{Hz}
Period is the reciprocal of the frequency.
T=\dfrac{1}{f}=\dfrac{1}{4.5}=0.22\,\mathrm{s}
For the third harmonic, L=\dfrac{3\lambda_3}{2}, so the wavelength is
\lambda_3=\dfrac{2L}{3}=\dfrac{2\times6.0}{3}=4.0\,\mathrm{m}
Speed is frequency times wavelength.
v=f\lambda_3=4.5\times4.0=18\,\mathrm{m\,s^{-1}}

f=4.5\,\mathrm{Hz},\quad T=0.22\,\mathrm{s},\quad \lambda=4.0\,\mathrm{m},\quad v=18\,\mathrm{m\,s^{-1}}

Book answer. 4.5 Hz, 0.22 s, 4.0 m, 18 m s^-1

An air column closed at one end resonates at the second maximum or the second resonant length. The frequency of the sound wave is 1024\,\mathrm{Hz}. The air temperature is 18.6\,^\circ\mathrm{C}. (a) Draw a diagram of the displacement wave pattern inside the column. (b) Calculate the speed of sound in air. (c) Calculate the wavelength. (d) Calculate the length of the closed air column.

Solution

Given

f = 1024\,\mathrm{Hz}
\theta = 18.6\,^\circ\mathrm{C}
\text{resonance} = \text{second (closed pipe)}

To find

(a) the displacement pattern
v
\lambda
L

(a) Draw a diagram of the displacement wave pattern inside the column.

A closed column has a node at the closed end and an antinode at the open end. The second resonance is the three quarter wavelength pattern: L=\dfrac{3\lambda}{4}. Measured from the closed end, the pattern has a node at the closed end, an antinode at L/3, a node at 2L/3 and an antinode at the open end.

Displacement pattern for the second resonance in a column closed at the bottom: nodes at the closed end and two thirds up, antinodes one third up and at the open end.

(b) Calculate the speed of sound in air.

The speed of sound at 18.6\,^\circ\mathrm{C} is 343\,\mathrm{m\,s^{-1}}, the value the book uses. It corresponds to 332\,\mathrm{m\,s^{-1}} at 0\,^\circ\mathrm{C} plus 0.6\,\mathrm{m\,s^{-1}} for each degree:

v=332+0.6\times18.6=343\,\mathrm{m\,s^{-1}}

v=343\,\mathrm{m\,s^{-1}}

\lambda=\dfrac{v}{f}=\dfrac{343}{1024}=0.335\,\mathrm{m}

\lambda=0.335\,\mathrm{m}

(d) Calculate the length of the closed air column.

For the second resonance of a column closed at one end, the length is three quarters of the wavelength, L=\dfrac{3\lambda}{4}:

L=\dfrac{3\times0.335}{4}=0.251\,\mathrm{m}

L=0.251\,\mathrm{m}

Book answer. 343 m/s, 0.335 m, 0.251 m

Note. The speed of sound at 18.6 degrees C is not given in the unit. The book's answer 343 m/s is used, which corresponds to 332 + 0.6 T. Using 331 m/s at 0 degrees C instead gives about 342 m/s; the wavelength and length then change only in the third figure.

Suppose a train that has a 150\,\mathrm{Hz} horn is moving at 35.0\,\mathrm{m\,s^{-1}} in still air on a day when the speed of sound is 340\,\mathrm{m\,s^{-1}}. What frequencies are observed by a stationary person at the side of the tracks (a) as the train approaches (b) after it passes. (c) What frequency is observed by the train's engineer travelling on the train?

Solution

Given

f = 150\,\mathrm{Hz}
v_s = 35.0\,\mathrm{m\,s^{-1}}
v = 340\,\mathrm{m\,s^{-1}}

To find

f' approaching
f' receding
f' for the engineer

(a) As the train approaches.

The source moves towards a stationary listener, so the apparent frequency is higher.
f'=\dfrac{v}{v-v_s}\,f
Substitute the values.
f'=\dfrac{340}{340-35.0}\times150=\dfrac{340}{305}\times150=167\,\mathrm{Hz}

f'=167\,\mathrm{Hz}

(b) After the train passes.

The source moves away from the stationary listener, so the apparent frequency is lower.
f'=\dfrac{v}{v+v_s}\,f
Substitute the values.
f'=\dfrac{340}{340+35.0}\times150=\dfrac{340}{375}\times150=136\,\mathrm{Hz}

f'=136\,\mathrm{Hz}

The engineer moves with the horn, so there is no relative motion between the source and the engineer. The engineer hears the frequency the horn emits.

f'=150\,\mathrm{Hz}

Book answer. 167 Hz, 136 Hz, same frequency as emitted by the horn (150 Hz)

72026 Annual I, Set D, Q.3

A polarized light of intensity I_0 is passed through another polarizer whose axis makes an angle of 60^\circ with the axis of the polaroid filter. What is the intensity of transmitted polarized light from second polaroid filter?

Solution

Given

I_0 = \text{intensity of the polarized light}
\theta = 60^\circ

To find

The light is already plane polarized, so no half is lost at the second filter. Malus's law gives the intensity transmitted by the second polaroid.
I=I_0\cos^{2}\theta
Substitute \theta=60^\circ, with \cos60^\circ=\tfrac{1}{2}.
I=I_0\left(\tfrac{1}{2}\right)^{2}=\dfrac{I_0}{4}

I=\dfrac{I_0}{4}

Book answer. I_0/4

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 91 mark

Antinode in a wave is point of:

AZero displacement
BMaximum displacementAnswer
CZero velocity
DNo wave

Why

In a stationary wave an antinode is a point of maximum amplitude, so its displacement is the largest (option B). It is where the two component waves are always in phase.

Zero displacement (A) describes a node, where the waves are always out of phase. An antinode is not a point of zero velocity: it vibrates most rapidly.

2025 Annual I, Set B, MCQ 101 mark

Location of a submarine in deep sea can be detected by:

ADoppler effectAnswer
BCompton's effect
CFaraday's Law
DBernoulli principle

Why

Sonar uses the Doppler effect: sound reflected from a moving submarine has a changed frequency, which gives its speed and direction (option A).

Compton's effect (X-ray scattering), Faraday's law (induced emf) and Bernoulli's principle (fluid flow) are not used for detecting submarines.

2025 Annual I, Set B, Q.2 (i) OR3 marks

How can ordinary light be polarized using polarizers?

Solution

Ordinary light is unpolarised: it is made of waves vibrating in all directions perpendicular to the line of propagation. A Polaroid filter blocks all the planes of vibration except one and transmits only that plane.

Figure 9.11 (b): A Polaroid filter blocks all the planes of vibration of an electromagnetic wave except one and transmits only one plane of vibration.

Pass the ordinary (unpolarised) light through the first polaroid (the polariser). It emerges plane polarised, with its vibrations in a single plane, and with its intensity reduced to half.
Place a second polaroid (the analyser) behind it. Turning it changes the transmitted intensity as I=I_0\cos^{2}\theta (Malus's law): maximum when the axes are parallel and zero when they are crossed at 90^\circ.

2025 Annual I, Set B, Q.2 (v)3 marks

The Doppler's effect is used in radar system. Briefly explain the working of radar system.

Solution

Radar (radio detection and ranging) uses the Doppler effect to find the speed and direction of aeroplanes.

The radar sends out radio waves of fixed frequency towards the aeroplane.
The waves are reflected by the moving aeroplane. Because of its motion the reflected waves have a changed frequency.
If the frequency of the reflected waves increases, the aeroplane is moving towards the radar; if it decreases, the aeroplane is moving away.
The size of the change in frequency is used to calculate the speed. The same principle is used in radar speed traps to measure the speed of cars.

2026 Annual I, Set B, Q.2 (x)1.5+1.5 = 3 marks

Compare constructive and destructive interference.

Solution

Both are effects of the superposition of waves from two coherent sources. They differ in the phase of the waves when they meet and in the result.

Feature	Constructive interference	Destructive interference
Phase	Waves in phase: crest on crest, trough on trough	Waves out of phase by 180^\circ: crest on trough
Path difference	d=m\lambda, m=0,1,2,\dots	d=\left(m+\tfrac{1}{2}\right)\lambda, m=0,1,2,\dots
Resultant amplitude	Amplitudes add: X+X=2X	Amplitudes subtract: X-X=0 for equal waves
Result	Loud sound, bright band, greater intensity	Silence, dark band, intensity falls to zero for equal waves

2025 Annual I, Set B, Q.2 (xi) OR3 marks

To increase the intensity of wave by a factor of 10, calculate the required increase in amplitude of wave.

Solution

Given

I_2/I_1 = 10 required increase of intensity

To find

A_2/A_1

Relation (1 mark). Intensity is proportional to the square of the amplitude.
I\propto A^{2}\Rightarrow\dfrac{I_2}{I_1}=\left(\dfrac{A_2}{A_1}\right)^{2}
Substitute (1 mark).
\left(\dfrac{A_2}{A_1}\right)^{2}=10
Answer (1 mark).
\dfrac{A_2}{A_1}=\sqrt{10}=3.16

The amplitude must be increased by a factor of about 3.16 (it becomes 3.16 times the original).

\dfrac{A_2}{A_1}=\sqrt{10}\approx3.16

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

A stationary observer hears the sound of a car horn at a frequency that is 15% higher than the horn's actual frequency. If the speed of sound is 343\,\mathrm{ms^{-1}}, what is the speed of the car as it approaches the observer?

Solution

Given

f\prime = 1.15f apparent frequency, 15 % higher
v = 343\,\mathrm{m\,s^{-1}} speed of sound
v_L = 0 observer at rest

To find

v_s, the speed of the car

Formula (1 mark). The source moves towards a stationary listener, so the apparent frequency is higher.
f\prime=\dfrac{v}{v-v_s}\,f
Substitute (1 mark). Put f\prime=1.15f; the f cancels.
1.15=\dfrac{343}{343-v_s}
Solve for v_s.
343-v_s=\dfrac{343}{1.15}=298.3\Rightarrow v_s=343-298.3
Answer with unit (1 mark).
v_s\approx44.7\,\mathrm{m\,s^{-1}}

v_s\approx44.7\,\mathrm{m\,s^{-1}}

2025 Annual I, Set B, Q.2 (xiv) OR1+2 = 3 marks

What is meant by interference? Give conditions for interference.

Solution

Interference (1 mark). Interference is the effect produced by the superposition of waves from two coherent sources (waves of the same frequency and a constant phase difference). When they meet in phase they reinforce each other (constructive interference); when they meet 180^\circ out of phase they cancel (destructive interference).

Conditions (2 marks).

The waves must travel in the same direction.
They must arrive at the same place at the same time.
They must be in phase for constructive interference and out of phase (180^\circ) for destructive interference.
The path difference must be a whole number of wavelengths, d=m\lambda, for constructive interference, and an odd multiple of half a wavelength, d=\left(m+\tfrac12\right)\lambda, for destructive interference (m=0,1,2,\dots).

2026 Annual I, Set B, Q.6 OR2+2+2 = 6 marks

Derive f_1, f_2 and f_3 for modes of vibration in a closed organ pipe.

Solution

In a closed pipe of length L the closed end is a node (the air cannot move) and the open end is an antinode. Each node-to-antinode gap is \lambda/4. A vibrating tuning fork is held at the open end. Fig. 9.10 shows the three lowest modes.

Derivation First harmonic (fundamental) of a closed pipe

Aim. Find the frequency f_1 of the first mode of a closed pipe of length L.

Figure 9.10 (a): First harmonic of a closed pipe: an antinode at the open end and a node at the closed end, so L=\lambda_1/4.

Fig. 9.10 (a): one node-to-antinode gap fits in the pipe.
L=\dfrac{\lambda_1}{4}\ \Longrightarrow\ \lambda_1=4L
Use v=f_1\lambda_1.
f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{4L}\qquad(9.21)

Resultf_1=\dfrac{v}{4L}

This is the fundamental frequency or first harmonic of the closed pipe. It is half the fundamental frequency v/2L of an open pipe of the same length.

Derivation Second mode of a closed pipe

Aim. Find the frequency f_2 of the second mode of a closed pipe.

Figure 9.10 (b): Second mode of a closed pipe: L=\tfrac34\lambda_2.

Fig. 9.10 (b): the length holds a half-wavelength and a quarter-wavelength.
L=\dfrac{\lambda_2}{2}+\dfrac{\lambda_2}{4}=\dfrac{3\lambda_2}{4}\ \Longrightarrow\ \lambda_2=\dfrac{4L}{3}
Use v=f_2\lambda_2.
f_2=\dfrac{v}{\lambda_2}=3\,\dfrac{v}{4L}
Since v/4L=f_1:
f_2=3f_1\qquad(9.22)

Resultf_2=3f_1

The next frequency of a closed pipe is three times the fundamental.

Derivation Third mode of a closed pipe

Aim. Find the frequency f_3 of the third mode of a closed pipe.

Figure 9.10 (c): Third mode of a closed pipe: L=\tfrac54\lambda_3.

Fig. 9.10 (c): the length holds one whole wavelength and a quarter-wavelength.
L=\lambda_3+\dfrac{\lambda_3}{4}=\dfrac{5\lambda_3}{4}\ \Longrightarrow\ \lambda_3=\dfrac{4L}{5}
Use v=f_3\lambda_3.
f_3=\dfrac{v}{\lambda_3}=5\,\dfrac{v}{4L}
Since v/4L=f_1:
f_3=5f_1\qquad(9.23)

Resultf_3=5f_1

The third mode of a closed pipe has five times the fundamental frequency.

In general the nth mode has \lambda_n=\dfrac{4L}{2n-1} and f_n=(2n-1)\dfrac{v}{4L}. A closed pipe therefore has only odd harmonics, f_1,\,3f_1,\,5f_1.

2026 Annual I, Set D, MCQ 101 mark

Stationary waves are formed by:

ADiffraction
BReflection only
CSuperposition of wavesAnswer
DRefraction

Why

A stationary wave is formed by the superposition of two identical waves of the same speed, amplitude and frequency travelling in opposite directions, for example a wave and its reflection (option C).

Reflection alone (B) only supplies the second wave; the pattern of nodes and antinodes comes from their superposition. Diffraction (A) and refraction (D) are bending of waves.

2025 Annual I, Set D, MCQ 111 mark

Stars moving away from the earth show:

ARed shiftAnswer
BBlue shift
CBlack shift
DYellow shift

Why

Light from a star moving away from the Earth is shifted to a lower frequency, towards the red end. This is the red shift (option A), the Doppler effect for light. Hubble observed that all galaxies appeared red-shifted, which showed that the universe is expanding.

Blue shift (B) is the opposite: stars moving towards the Earth. Black and yellow shifts do not exist.

2026 Annual I, Set D, Q.2 (ix) OR1x3 = 3 marks

Write any three characteristics of stationary waves.

Solution

A stationary wave is formed by two identical waves travelling in opposite directions. Its characteristics (any three):

No energy is transferred from particle to particle (no energy is carried along the wave).
The points of zero displacement are nodes and the points of maximum amplitude are antinodes; the amplitude of an antinode is the sum of the amplitudes of the two component waves.
The distance between two successive nodes or two successive antinodes is \lambda/2.
The distance between an adjacent node and antinode is \lambda/4.
The wave appears to stand, confined to a given region of space, and does not move through it like a progressive wave.

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

What are beats? How are they produced?

Solution

Beats (1 mark). Beats are the periodic alternations of sound between maximum and minimum loudness produced when two waves of slightly different frequencies are played simultaneously. The number of beats per second is the beat frequency f_b=|f_1-f_2|.

Production (2 marks). Two sound waves A and B of slightly different frequencies and similar amplitude travel in the same medium. For some time the crest of A falls on the crest of B: constructive interference gives a loud sound. After a short while the crest of one falls on the trough of the other: destructive interference gives a soft sound. Because the frequencies are slightly different, the waves go in and out of step again and again, so the amplitudes add and subtract alternately and a fluctuating sound (loud, soft, loud) is heard.

Generation of beats

Figure 9.6: two waves of nearly equal frequency (dashed) and their sum (solid). The sum is loud where the waves are in phase and soft where they are out of phase.

2025 Annual I, Set D, Q.2 (xi) OR3 marks

How is Doppler's effect helpful in diagnosis of cardiac diseases? Explain briefly.

Solution

Doppler ultrasound (Doppler echocardiography) uses the Doppler effect to examine the heart and blood vessels.

Very high frequency sound waves are sent into the body and bounced off the heart and the blood vessels. The moving blood changes the frequency of the reflected waves.
The returning waves are picked up and turned into pictures showing the flow of blood through the arteries and the heart.
Doctors can then see clearly the blood flow, see and measure obstructions in the arteries, and measure the degree of narrowing or leakage of the heart valves.

2026 Annual I, Set D, Q.2 (xii) OR3 marks

Explain the use of beats in tuning musical instruments.

Solution

Beats tell how close two frequencies are, because the beat frequency is |f_1-f_2|, and this is used to tune instruments such as a piano.

The musician plucks the piano string and taps a tuning fork of the standard frequency at the same time.
If beats are heard, the two frequencies are not identical.
The tension of the string is adjusted and the test repeated. As the string comes closer to the fork, the beat frequency falls and approaches 0\,\mathrm{Hz}.
When no beats are heard the string is in tune: it plays the same frequency as the fork.

In this way the strings are matched to a standardised set of tuning forks.

2025 Annual I, Set D, Q.2 (xii) OR3 marks

Can sound waves be polarized? Explain briefly.

Solution

2025 Annual I, Set D, Q.2 (xiii)3 marks

To increase intensity of waves by a factor of 25, calculate the required increase in amplitude of wave.

Solution

Given

I_2/I_1 = 25 required increase of intensity

To find

A_2/A_1

Relation (1 mark).
I\propto A^{2}\Rightarrow\dfrac{I_2}{I_1}=\left(\dfrac{A_2}{A_1}\right)^{2}
Substitute (1 mark).
\left(\dfrac{A_2}{A_1}\right)^{2}=25
Answer (1 mark).
\dfrac{A_2}{A_1}=\sqrt{25}=5

The amplitude must be made 5 times as large.

\dfrac{A_2}{A_1}=5

2026 Annual I, Set D, Q.31+2+3 = 6 marks

a. What is meant by polarization. b. Discuss one application of polarization. c. A polarized light of intensity I_o is passed through another polarizer whose axis makes an angle of 60^{\circ} with the axis of the polaroid filter. What is the intensity of transmitted polarized light from second polaroid filter?

Solution

(a) Polarisation (1 mark)

Polarisation is the process of transforming unpolarised light into polarised light. An ordinary ray of light consists of waves vibrating in all directions perpendicular to its line of propagation (unpolarised). In polarised light the vibrations occur in a single plane. Only transverse waves can be polarised.

Figure 9.11 (b): A Polaroid filter blocks all the planes of vibration of an electromagnetic wave except one and transmits only one plane of vibration.

(b) An application (2 marks)

Polaroids in sky photography. The light coming from the sky is polarised. A polaroid is fitted before the camera lens and turned so that its axis blocks this polarised light. The sky then becomes sufficiently dark and the clouds stand out against it, so a clear photograph is obtained. (Another use: polaroid sunglasses block the polarised glare reflected from water or roads.)

(c) Intensity through the second polaroid (3 marks)

Given

I_0 = \text{intensity of the polarised light}
\theta = 60^\circ angle between the axes

To find

I transmitted

Law (1 mark). The light is already plane polarised, so Malus's law applies to the second filter.
I=I_0\cos^{2}\theta
Substitute (1 mark). \cos60^\circ=\tfrac12.
I=I_0\left(\tfrac12\right)^{2}
Answer (1 mark).
I=\dfrac{I_0}{4}

I=\dfrac{I_0}{4}

Note. Printed marks read as 1+2+3 for parts a, b, c.

2025 Annual I, Set D, Q.51+6 = 7 marks

What are stationary waves? Discuss the formation of harmonics in stationary waves in an open organ pipe.

Solution

Stationary waves (1 mark)

A stationary (standing) wave is formed when two identical waves of the same speed, amplitude and frequency travel in opposite directions and are superposed, for example a wave and its reflection. Points that never move are nodes and points of maximum amplitude are antinodes. No energy is carried along the wave.

Harmonics in an open pipe (6 marks)

In an open pipe both ends are open, so there is an antinode at each end. A node lies in the middle for the fundamental. A node-to-antinode gap is \lambda/4.

Derivation Fundamental frequency of an open pipe

Aim. Find the frequency f_1 of the first mode of an open pipe of length L.

Figure 9.9 (a): Fundamental frequency (first harmonic) of an open pipe: antinodes A at both ends and a node N in the middle, so L=\lambda_1/2.

Fig. 9.9 (a): an antinode at each open end and a node in the middle. The distance from an antinode to the nearest node is \lambda_1/4, and there are two such gaps.
L=\dfrac{\lambda_1}{4}+\dfrac{\lambda_1}{4}=\dfrac{\lambda_1}{2}\ \Longrightarrow\ \lambda_1=2L
Use v=f_1\lambda_1.
f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{2L}\qquad(9.16)

Resultf_1=\dfrac{v}{2L}

This is called the fundamental frequency or first harmonic of the open pipe.

Derivation Second harmonic of an open pipe

Aim. Find the frequency f_2 of the second mode of an open pipe.

Figure 9.9 (b): Second harmonic of an open pipe: L=\lambda_2.

Fig. 9.9 (b): the length holds one whole wavelength.
L=\lambda_2\ \Longrightarrow\ \lambda_2=L
Use v=f_2\lambda_2 and write v/L=2(v/2L).
f_2=\dfrac{v}{\lambda_2}=\dfrac{v}{L}=2\left(\dfrac{v}{2L}\right)
Since v/2L=f_1:
f_2=2f_1\qquad(9.17)

Resultf_2=2f_1

This is called the second harmonic of the open pipe.

Derivation Third harmonic of an open pipe

Aim. Find the frequency f_3 of the third mode of an open pipe.

Figure 9.9 (c): Third harmonic of an open pipe: L=\tfrac32\lambda_3.

Fig. 9.9 (c): the length holds three half-wavelengths.
L=3\,\dfrac{\lambda_3}{2}\ \Longrightarrow\ \lambda_3=\dfrac{2L}{3}
Use v=f_3\lambda_3.
f_3=\dfrac{v}{\lambda_3}=\dfrac{3v}{2L}=3\left(\dfrac{v}{2L}\right)
Since v/2L=f_1:
f_3=3f_1\qquad(9.18)

Resultf_3=3f_1

This is the third harmonic of the open pipe.

For the nth harmonic:

\lambda_n=\dfrac{2L}{n},\qquad f_n=nf_1=n\dfrac{v}{2L}\qquad(n=1,2,3,\dots)

An open pipe therefore gives all the harmonics, f_1,\,2f_1,\,3f_1,\dots, with fundamental f_1=\dfrac{v}{2L}.

Exam focus

Stationary waves in pipes: derive the harmonics of an open pipe (f_n=nv/2L) and a closed pipe (f_n=(2n-1)v/4L). (long)
2026 Set B Q.6 OR (closed pipe, 6 marks) and 2025 Set D Q.5 (stationary waves and open pipe, 7 marks). Comprehensive questions 3 and 4.
Go to answer
Doppler's effect: apparent frequency, numericals and applications (radar, sonar, cardiac diagnosis, red shift). (short)
2026 Set B Q.2 (xiii) OR (car at 15 % higher frequency), 2025 Set B MCQ 10 and Q.2 (v), 2025 Set D MCQ 11 and Q.2 (xi) OR.
Go to answer
Beats: definition, production, f_b=|f_1-f_2| and tuning of instruments. (short)
2026 Set D Q.2 (x) (what are beats, how produced) and Q.2 (xii) OR (tuning). Comprehensive question 5.
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Polarisation: definition, production and detection, an application and Malus's law numerical. (long)
2026 Set D Q.3 (1+2+3 marks, with the 60^\circ numerical of book numerical 7), 2025 Set B Q.2 (i) OR and 2025 Set D Q.2 (xii) OR (sound cannot be polarised).
Go to answer
Interference: definition and conditions; constructive against destructive. (short)
2026 Set B Q.2 (x) (comparison, 1.5+1.5) and 2025 Set B Q.2 (xiv) OR (definition and conditions).
Go to answer
Intensity and amplitude: I\propto A^{2}. (numerical)
2025 Set B Q.2 (xi) OR (factor 10 gives 3.16) and 2025 Set D Q.2 (xiii) (factor 25 gives 5); book exercise MCQ 1 and Assignment 9.1.
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Characteristics of stationary waves; nodes and antinodes. (short)
2026 Set D Q.2 (ix) OR, 2026 Set B MCQ 9 (antinode) and 2026 Set D MCQ 10 (formation by superposition).
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