Waves

Class 10 Physics, unit 12, book pages 75–92

A wave carries energy from one place to another without carrying matter. This unit describes how waves are produced and propagate, defines wavelength, amplitude, frequency, time period and wave speed, and sorts waves into mechanical or electromagnetic and transverse or longitudinal. It then uses the ripple tank to show reflection, refraction and diffraction, and ends with how tsunamis form and grow near the coast.

Learning outcomes (11)

Prove that waves transfer energy without transferring matter.
Describe what is meant by wave motion (as illustrated by vibrations in ropes and springs and by experiments using water waves).
Describe the features of a wave (in terms of wavefront, wavelength, frequency, time period, crest (peak), trough, compression, rarefaction, amplitude and wave speed).
Define the terms frequency, wavelength and amplitude.
Recall and apply the equation wave speed = frequency × wavelength ( $v = f λ$ ).
Illustrate that for a transverse wave, the direction of vibration is at right angles to the direction of the energy transfer (including examples such as electromagnetic radiation, waves on the surface of water and seismic S-waves (secondary)).
Illustrate that for a longitudinal wave, the direction of vibration is parallel to the direction of the energy transfer (including examples such as sound waves and seismic P-waves (primary)).
Describe how waves can undergo reflection, refraction and diffraction at an edge.
Describe how wavelength affects diffraction at an edge.
Analyse the phenomenon of tsunamis generated under the surface of water (in terms of underwater earthquakes or volcanic activity generating waves that increase in frequency and amplitude as they encounter increasingly shallow water).
Describe how wavelength and gap size affect diffraction through a gap.

12.1Wave motion

Waves are carriers of energy and information over large distances. Every wave needs some oscillating or vibrating source. In this section the book shows how different waves are produced and how they travel, using the vibrations of objects. Water waves, sound waves and electromagnetic waves are three examples, and a wave on a rope is the clearest model of a transverse wave.

Water waves (12.1.1)

When a stone is thrown into water, it creates ripples that spread outwards. These ripples, or water waves, clearly transfer energy: you can see a boat bobbing up and down as the waves pass. The boat itself does not move far from its place. The energy of the wave makes the boat oscillate up and down, but the boat (matter) stays mostly where it was.

A cork on a water wave only moves up and down. The wave, and the energy it carries, travel sideways.

Sound waves (12.1.2)

When you speak, your vocal cords vibrate and produce sound waves that travel through the air. If you place your hand in front of a speaker, you can feel the vibrations of the sound waves, which shows that energy is being transferred. The air molecules themselves do not travel from the speaker to your hand. They oscillate back and forth around their equilibrium positions. This oscillatory motion carries the sound energy through the medium (air), but the individual air molecules stay relatively in place.

Electromagnetic waves (12.1.3)

Light waves and radio waves are types of electromagnetic waves. When light from the Sun reaches the Earth, it brings energy that warms the planet and powers photosynthesis in plants. There is no medium such as air or water involved in the transfer of electromagnetic waves through the vacuum of space. Instead, electric and magnetic fields oscillate perpendicular to each other and to the direction of wave propagation, carrying energy across vast distances without the need for a material medium.

Wave in a rope (12.1.4)

A wave in a rope is a clear example of transverse wave motion: the oscillations of the medium (the rope) are perpendicular to the direction of wave propagation.

Figure 12.4: Waves in rope. The rope vibrates up and down while the energy moves from A to B.

12.2Wave propagationOften examined

The key to understanding how a wave carries energy without carrying matter lies in how a wave propagates. Waves are disturbances that travel through a medium (or through space, in the case of electromagnetic waves) and make the particles of the medium oscillate around their equilibrium positions. The explanation has three steps.

Oscillation (12.2.1)

In a wave, the particles of the medium experience periodic oscillations or vibrations. These oscillations can be of two kinds:

Longitudinal: the particles move parallel to the direction of the wave, as in sound waves.
Transverse: the particles move perpendicular to the direction of the wave, as in water waves and electromagnetic waves.

Energy transfer (12.2.2)

The energy of the wave is transferred through these oscillations. For example, in a sound wave, the compressions and rarefactions of air molecules pass kinetic and potential energy from one particle to the next.

No net movement (12.2.3)

In spite of the oscillatory motion, the particles of the medium do not undergo a net displacement. They return to their original positions after the wave has passed. Therefore there is no overall transfer of matter, only of energy.

Why a wave transfers energy without transferring matter

The source vibrates and makes the nearby particles of the medium vibrate about their equilibrium positions (12.2.1).
Each vibrating particle hands on its kinetic and potential energy to the next particle, so the energy moves along the medium (12.2.2).
Every particle only goes to and fro about its own place. When the wave has passed, each particle is back where it started, so its net displacement is zero (12.2.3).
Energy has moved from one place to another, but no particle has been carried along. So a wave transfers energy without transferring matter.

Worked example: does the cork travel with the wave?

A cork floats on a pond

2 m

from the spot where a stone is dropped. The ripples reach the cork and then pass beyond it. (a) Describe the motion of the cork. (b) Where is the cork after the ripples have passed? (c) Did the cork receive energy? How can you tell?

The ripples make the water under the cork oscillate. The cork rides on this water, so it bobs up and down about its resting place. It is not carried outwards towards the edge of the pond.
After the last ripple has passed, the water is calm again and each particle of water is back at its starting position. So the cork is back where it was, about $2 m$ from the spot where the stone fell. Its net displacement is zero.
Yes. A cork at rest started to move, so it gained kinetic and potential energy from the wave. This energy came across the pond from the stone, but no water or cork travelled across with it. Energy moved, matter did not.

The cork only oscillates and returns to its place, but it receives energy from the wave.

12.3Characteristic wave parametersOften examined

There are many ways to describe or measure a wave. Some characteristics depend on how the wave is produced, whereas others depend on the medium through which the wave travels. The wave parameters below, with their definitions, are used throughout the unit.

Wavelength $λ$ (12.3.1)

Wavelength ( $λ$ ): The shortest distance between points where the wave pattern repeats itself. Its symbol is the Greek letter lambda ( $λ$ ) and its unit is the metre (m).

A shift of the wave pattern by one wavelength to the right (or the left) reproduces the original pattern. The wavelength may be the distance between two successive crests or two successive troughs, or the distance between two successive compressions (in a longitudinal wave).

Wavefront (12.3.2)

Huygens stated that light is a wave propagating through space like ripples in water or sound in air. Hence light spreads out like a wave in all directions from a source.

Wavefront: The locus of all points that have travelled some distance during a fixed interval of time.

Figure 12.6: Wavefronts. Ripples from a point source form circular wavefronts. Far from the source (or from a straight dipper) the wavefronts are plane.

Wave crest and trough (12.3.3)

Crest: A point on a wave where the displacement of the medium is at its maximum upward value within a cycle.
Trough: The opposite of a crest: the minimum or lowest point in a cycle.

Displacement (12.3.4)

Displacement ( $x$ ): The separation between the mean position and any point, to the right or the left of the mean position. Unit: metre (m).

Amplitude (12.3.5)

Amplitude ( $x_{0}$ ): The maximum displacement from the mean position on either side. As amplitude is also a displacement, it is measured in metres (m).

Wavelength, crest, trough and amplitude

x_{0}

- x_{0}

position

displacement

crest

trough

λ

x_{0}

Figures 12.5, 12.7 and 12.8: displacement against position at one instant. The wavelength

λ

is the distance between two successive crests (or troughs). The amplitude

x_{0}

is the largest displacement from the mean position (the dashed line).

Changing the amplitude

x_{0}

1.6 x_{0}

position

displacement

small amplitude

larger amplitude

Both waves have the same wavelength. The larger wave has the larger amplitude, so its crests are higher and its troughs lower.

Changing the wavelength

x_{0}

- x_{0}

position

displacement

longer wavelength

shorter wavelength

Both waves have the same amplitude. The accent curve repeats twice as often, so its wavelength is half as long.

Time period (12.3.6)

Time period ( $T$ ): The time required for one wave cycle to pass a certain point. It is the time any vibrating element of the wave takes to move through one complete oscillation. Unit: second (s).

Frequency (12.3.7)

Frequency ( $f$ ): The number of cycles completed by an oscillating body in one second. It is measured in cycles per second (cps) or hertz (Hz).

Relation between time period and frequency (12.3.8)

Time period and frequency are reciprocals of each other. The greater the time period, the smaller the frequency, and the other way round.

T = \frac{1}{f}

(12.1)

Time period against frequency

0.2

0.4

0.6

0.8

f (Hz)

T (s)

(5 Hz, 0.2 s)

(2 Hz, 0.5 s)

T = \frac{1}{f}

T = 1/ f

: the larger the frequency, the smaller the time period. The marked point is Example 12.1.

Example 12.1: time period of a pendulum

A student oscillates a pendulum having

5

vibrations per second. Calculate its time period.

Given

$f = 5 Hz$ frequency

To find

time period $T$

Use the relation between time period and frequency.
$T = \frac{1}{f}$
Substitute $f = 5 Hz$ .
$T = \frac{1}{5 Hz}$
Divide.
$T = 0.2 s$

So the pendulum oscillates with a period of $0.2 s$ .

T = 0.2 s

Worked example: reading a wave graph

The graph shows a wave on a rope at one instant. The person holding the end of the rope makes

10

complete up-and-down movements in

4 s

. Find (a) the amplitude, (b) the wavelength, (c) the frequency and (d) the time period of the wave.

A wave on a rope at one instant

100

120

140

160

−8

−6

−4

−2

distance along rope (cm)

y (cm)

Displacement

y

against distance along the rope.

Amplitude is the largest displacement from the mean position (the zero line). The crests reach $+ 6 cm$ and the troughs $- 6 cm$ , so $x_{0} = 6 cm$ . The trough-to-crest height ( $12 cm$ ) is twice the amplitude, so it is not the answer.
$x_{0} = 6 cm$
Wavelength is the distance between two successive crests. The crests are at $20 cm$ and $100 cm$ .
$λ = 100 - 20 = 80 cm = 0.80 m$
Check with the troughs, at $60 cm$ and $140 cm$ : the distance is again $80 cm$ .
Frequency is the number of cycles in one second. Ten cycles take $4 s$ .
$f = \frac{10}{4 s} = 2.5 Hz$
The time period is the reciprocal of the frequency.
$T = \frac{1}{f} = \frac{1}{2.5} = 0.4 s$
Check: ten cycles in $4 s$ means $4 s /10 = 0.4 s$ for one cycle. This agrees.

x_{0} = 6 cm, λ = 0.80 m, f = 2.5 Hz, T = 0.4 s

12.4Waves and their types

Waves are carriers of energy and information over large distances.

Wave: A disturbance that moves outward from its point of origin, transferring energy by means of vibrations, with little or no transport of the medium.

Wave motion is related to oscillation, because the particles of the medium execute simple harmonic motion about their mean positions when a wave passes. For example, take a rope and colour a part of it. Attach one end of the rope to a wall and wiggle the other end continuously. A wave train is produced. The coloured part of the rope oscillates about a certain mean position, as shown in Figure 12.9 (b).

Figure 12.9: Direction of energy and oscillation. In (a) and (b) the pulse moves to the right (the direction of energy transfer). The coloured part of the rope only moves up and down.

12.5Types of waves on the basis of mediumOften examined

On the basis of whether a material medium is needed, waves are divided into two types.

A. Mechanical waves

Mechanical waves: Waves that need a material medium to travel from one point to another. If the medium does not exist, mechanical waves cannot travel.

Examples of mechanical waves are sound waves, waves in water and seismic waves.

B. Electromagnetic waves

Electromagnetic waves: Waves that do not need a material medium to travel from one point to another. The existence of a material medium is not necessary for their propagation. They consist of electric and magnetic fields, as shown in Figure 12.10.

Examples of electromagnetic waves are light waves, X-rays, ultraviolet waves and infrared waves.

Figure 12.10: Travelling of an electromagnetic wave. The electric and magnetic fields oscillate at right angles to each other and to the direction of travel.

Mechanical and electromagnetic waves
	Mechanical waves	Electromagnetic waves
Material medium	Needed	Not needed
Travel through a vacuum	No	Yes
What oscillates	The particles of the medium	Electric and magnetic fields
Examples	Sound, waves in water, seismic waves	Light, X-rays, ultraviolet, infrared

12.6.1Transverse wavesOften examined

On the basis of how they propagate (the direction of vibration compared with the direction of travel), waves are divided into two types: transverse waves and longitudinal waves.

Transverse waves: Waves in which the disturbance occurs perpendicular to the direction of motion of the wave. The vibrations are at right angles to the wave's motion, as shown in Figure 12.11.

Transverse waves are common in nature and can be observed in various forms. Waves produced in a string when one end is fixed to a support and the other end is moved up and down are an example.

Figure 12.11: A transverse wave. The particles vibrate up and down, at right angles to the direction of travel. The wavelength is the distance between two successive crests.

A. Electromagnetic radiation

Electromagnetic waves such as light, X-rays and radio waves are examples of transverse waves. These waves do not require a medium to travel. The electric and magnetic fields vibrate at right angles to each other and to the direction of energy transfer. For example, sunlight travels through space as an electromagnetic wave, bringing energy from the Sun to the Earth.

B. Waves on the surface of water

Waves on the surface of water are another example of transverse waves. The water particles move up and down (vibration), while the wave energy travels horizontally across the surface. For instance, when a stone is thrown into a pond, ripples spread outwards from the point of impact, showing the transverse motion of the water particles.

C. Seismic S-waves (secondary waves)

Seismic S-waves, produced during earthquakes, are transverse waves that move through the Earth. In these waves the ground vibrates side to side or up and down, while the wave energy travels horizontally. Unlike P-waves, S-waves cannot travel through liquids, because they need the shear strength of a solid medium for their vibrations. These waves are important in understanding the Earth's interior and in locating the epicentres of earthquakes.

12.6.2Longitudinal wavesOften examined

On the basis of how they propagate, the second type of waves is the longitudinal wave.

Longitudinal waves: Waves in which the disturbance occurs parallel to the line of travel of the wave. The vibrations and the energy transfer occur along the same line.

A longitudinal wave can be generated with a Slinky, as in Figure 12.12. When one end is pushed forward along its length (longitudinally) and then pulled back to its starting point, a region where the coils are compressed together is sent travelling along the Slinky.

Compression: A part of a longitudinal wave with higher density and pressure.
Rarefaction: A part of a longitudinal wave with smaller density and pressure.

Figure 12.12: A longitudinal wave. The particles move to and fro along the line of travel, forming compressions (close together) and rarefactions (spread apart). The wavelength is the distance between two successive compressions.

A. Sound waves

Sound waves are a classic example of longitudinal waves. When we speak, the air particles near our vocal cords compress and expand, creating regions of compression (where particles are close together) and rarefaction (where particles are spread apart). These compressions and rarefactions move through the air, carrying sound energy to the listener. For instance, when we clap, the sound travels as longitudinal waves through the air to reach someone nearby.

B. Seismic P-waves (primary waves)

Seismic P-waves, generated during earthquakes, are another example of longitudinal waves. These waves make particles in the Earth vibrate back and forth in the same direction as the wave's movement. P-waves are the fastest type of seismic wave and can travel through solids and liquids. This property helps scientists study the Earth's structure and detect earthquakes. For example, the time difference between the arrival of P-waves and the slower S-waves at a seismic station helps to find the location of the earthquake's epicentre.

Transverse and longitudinal waves compared
	Transverse waves	Longitudinal waves
Vibration compared with energy transfer	At right angles (perpendicular)	Along the same line (parallel)
Pattern in the medium	Crests and troughs	Compressions and rarefactions
Examples	Electromagnetic waves, waves on the surface of water, string waves, seismic S-waves	Sound waves, seismic P-waves
Through liquids	S-waves cannot travel through liquids (they need the shear strength of a solid)	P-waves can travel through solids and liquids

Worked example: telling the type of wave from the directions

The figure shows, for two waves X and Y, the direction in which the energy travels and the direction in which the particles of the medium vibrate. (a) Which wave is transverse and which is longitudinal? (b) Name one example of each. (c) A student says that the Slinky wave in Figure 12.12 is transverse because the Slinky also moves a little up and down when it is carried. What is wrong with this?

Two waves X and Y: the arrows show the direction of energy transfer and the direction of vibration.

Compare the two directions for each wave. For wave X the vibration is vertical while the energy travels horizontally, so the two are at right angles. For wave Y the vibration is horizontal, along the same line as the energy.
A wave whose vibration is at right angles to the direction of energy transfer is transverse, so X is transverse. A wave whose vibration is parallel to the direction of energy transfer is longitudinal, so Y is longitudinal.
Examples: X could be a wave on a rope, a ripple on water, light or a seismic S-wave. Y could be a sound wave or a seismic P-wave.
The type of a wave is decided only by the vibration of the particles that carry the wave, compared with the direction in which the energy travels. In Figure 12.12 the coils are pushed and pulled along the length of the Slinky, along the same line as the wave travels, so the wave is longitudinal. Any other movement, such as the whole Slinky sagging or being carried, is not the wave.

X is transverse (for example a rope wave); Y is longitudinal (for example sound).

12.7Wave characteristicsOften examined

Some of the main characteristics of a wave are the wave cycle, the frequency, the time period and the wave speed. They are linked by the universal wave equation.

A. Wave cycle

As a wave passes a given point along its path, that point undergoes cyclic motion. The point is displaced first in one direction and then in the other. Finally it returns to its original equilibrium position, and in this way it completes one cycle. The number of wave cycles is represented by $N$ . It is a pure number, so it has no unit.

B. Frequency

The number of wave cycles passing through some point per unit time is called the frequency of the wave. Mathematically:

f = \frac{N}{t}

(12.2)

Frequency is measured in hertz (Hz).

C. Time period

The time required for one wave cycle to pass a certain point is called the time period. It is represented by $T$ and its unit is the second (s). It is the time any vibrating element of the wave takes to move through one complete oscillation. The time period is the reciprocal of the frequency:

T = \frac{1}{f}

(12.3)

D. Wave speed

The distance travelled by a wave per unit time is called the wave speed. Mathematically:

v = \frac{Δ d}{Δ t}

(1)

Derivation The universal wave equation

Aim. Express the wave speed in terms of the frequency and the wavelength.

Assumptions.

The wave moves with a steady speed.
The wave is a regular train of waves, so one wavelength passes in one time period.

Start from the definition of speed: distance travelled divided by time taken.
$v = \frac{Δ d}{Δ t}$
In one time period $T$ the wave travels a distance of one wavelength $λ$ . So put $Δ d = λ$ and $Δ t = T$ .
$v = \frac{λ}{T} (2)$
The frequency is the reciprocal of the time period, $f = 1/ T$ from (12.3). Replace $1/ T$ by $f$ .
$v = λ \times \frac{1}{T} = λ \times f$

Result

v = f λ (12.4)

Equation (12.4) is the universal wave equation. It can be used for all types of waves. For electromagnetic waves travelling with speed $c$ it is written $c = f λ$ (12.5). Rearranged, $f = v / λ$ and $λ = v / f$ .

c = f λ

(12.5)

for electromagnetic waves

Example 12.2: frequency of waves on a rope

An athlete in the gym moves one end of a rope up and down. The other end is fastened to a wall. The waves travel with a speed of

2.5 m s^{- 1}

and have a wavelength of

0.5 m

. Calculate the frequency of the waves generated in the rope.

Given

$v = 2.5 m s^{- 1}$ wave speed
$λ = 0.5 m$ wavelength

To find

frequency $f$

Use the universal wave equation.
$v = f λ$
Make $f$ the subject.
$f = \frac{v}{λ}$
Substitute the values.
$f = \frac{2.5 m s ^{- 1}}{0.5 m}$
Divide.
$f = 5 Hz$

The rope therefore oscillates $5$ times per second.

f = 5 Hz

Example 12.3: wavelengths of the lowest and highest audible sounds

Find the wavelengths of the lowest and highest frequency sounds that a human can hear, which travel with a speed of

343 m s^{- 1}

in air.

Given

$f_{1} = 20 Hz$ lowest frequency
$f_{2} = 20000 Hz$ highest frequency
$v = 343 m s^{- 1}$ speed of sound

To find

wavelengths $λ_{1}$ and $λ_{2}$

Use the universal wave equation and make $λ$ the subject.
$v = f λ \Rightarrow λ = \frac{v}{f}$
For the lowest frequency.
$λ_{1} = \frac{v}{f _{1}} = \frac{343}{20} = 17.15 m$
For the highest frequency.
$λ_{2} = \frac{v}{f _{2}} = \frac{343}{20 000} = 0.01715 m \approx 0.017 m$

λ_{1} = 17.15 m, λ_{2} \approx 0.017 m

Wavelength against frequency for sound in air

100

150

200

250

300

350

400

f (Hz)

λ (m)

(20 Hz, 17.15 m)

(343 Hz, 1 m)

λ = \frac{343}{f}

λ = v / f

with

v = 343 m s^{- 1}

: the higher the frequency, the shorter the wavelength. The lowest audible frequency,

20 Hz

, has the longest wavelength (Example 12.3).

Worked example: speed of sea waves

A small boat rises and falls

12

times in

30 s

as sea waves pass it. Neighbouring crests are

5 m

apart. (a) Find the frequency and the time period. (b) Find the speed of the waves. (c) What would the wavelength be if waves of the same speed had twice this frequency?

Given

$N = 12$ cycles
$t = 30 s$
$λ = 5 m$ distance between neighbouring crests

To find

$f$ , $T$ , $v$ and the new $λ$

Each rise and fall of the boat is one cycle, so the frequency is the number of cycles per second.
$f = \frac{N}{t} = \frac{12}{30 s} = 0.4 Hz$
The time period is the reciprocal of the frequency.
$T = \frac{1}{f} = \frac{1}{0.4} = 2.5 s$
The wavelength is the distance between two successive crests, here $5 m$ . Use the universal wave equation.
$v = f λ = 0.4 \times 5 = 2.0 m s^{- 1}$
Check with $v = λ / T$ .
$v = \frac{5}{2.5} = 2.0 m s^{- 1}$
For (c) the speed stays $2.0 m s^{- 1}$ and the frequency becomes $2 \times 0.4 = 0.8 Hz$ . Make $λ$ the subject.
$λ = \frac{v}{f} = \frac{2.0}{0.8} = 2.5 m$
The new wavelength ( $2.5 m$ ) is half the old one ( $5 m$ ), as the rule $v = f λ$ predicts.

f = 0.4 Hz, T = 2.5 s, v = 2.0 m s^{- 1}, λ = 2.5 m

12.8Properties of waves and the ripple tank

The basic properties of waves, reflection, refraction and diffraction, can be described with the help of a ripple tank that uses water waves. The ripple tank is also used to calculate the frequency, wavelength and speed of waves on the surface of water. It also helps to analyse characteristics of waves such as transverse and longitudinal behaviour.

Figure 12.13: The ripple tank. A lamp shines through the shallow water tray, and the ripples show up as bright and dark lines on the screen below.

Construction and working

A ripple tank consists of a shallow tray of water with a transparent base, usually lit from above so that light shines through the water. The ripples on the water show up as shadows (bright and dark lines) on the screen underneath the tank. The ripples are created by a small electric motor (a vibrator). The motor has an off-centre weight attached. When the motor is switched on, it makes the strip vibrate and so creates ripples on the surface of the water.

Straight waves are set up by using a straight (plane) dipper.
Circular waves are formed by using a spherical dipper.
To get steady plane waves, the motor assembly is lowered until it just touches the surface of the water, and then the motor is switched on. Straight waves, often called plane waves, are then seen crossing the water surface of the tank.

By lighting the wavefronts as they are created, clear peaks and troughs can be seen on the screen, as in Figure 12.13.

Wavefront (in the ripple tank): The imaginary surface representing corresponding points (crests and troughs) of a wave that vibrate in unison. It may be plane or spherical, depending on the dipper used.

12.8.1Reflection of wavesOften examined

Reflection: The change in direction of a wavefront at an interface between two different media, or at a rigid barrier, so that the wavefront returns into the medium from which it originated.

When a wavefront strikes a straight rigid barrier or an irregular surface, it is reflected back, as shown in Figure 12.14. In the ripple tank, a straight barrier placed in the path of the plane waves sends the waves back into the water.

Figure 12.14: Reflection of waves. The incident rays strike the rigid barrier and the reflected rays return into the same medium.

12.8.2Refraction of wavesOften examined

When waves travel from one medium into another, their speed changes, and in some cases they are diverted from their original path. This phenomenon is called refraction. When a water wave enters a medium in which it moves more slowly, its wavelength decreases as well.

Refraction: The change in direction of a wave passing from one medium to another, caused by its change in speed.

Refraction can be shown in the ripple tank by using a boundary between shallow and deep water. A thick plastic sheet is placed in the tray at some angle to the wavefront. When water waves travel from shallow to deep water, their wavelength and hence speed change, along with the direction of the waves, as shown in Figure 12.15.

Figure 12.15: Refraction of waves. The wavefronts are closer together in shallow water and further apart in deep water, so the wavelength is shorter in shallow water.

Worked example: wavelength after refraction

In a ripple tank the vibrator makes 5 waves every second. The waves travel at 0.40\,\mathrm{m\,s^{-1}} in deep water and at 0.25\,\mathrm{m\,s^{-1}} in shallow water. The figure shows the wavefronts crossing the boundary. (a) Find the wavelength in each region and say which is which in the figure. (b) Explain why the wavelength is shorter in shallow water.

Wavefronts crossing the boundary between two regions of water.

Given

f = 5\,\mathrm{Hz} set by the vibrator
v_{\text{deep}} = 0.40\,\mathrm{m\,s^{-1}}
v_{\text{shallow}} = 0.25\,\mathrm{m\,s^{-1}}

To find

wavelengths \lambda_1 (deep) and \lambda_2 (shallow)

Use the universal wave equation in the form \lambda=v/f. In deep water:
\lambda_1=\dfrac{v_{\text{deep}}}{f}=\dfrac{0.40}{5}=0.080\,\mathrm{m}=8.0\,\mathrm{cm}
In shallow water:
\lambda_2=\dfrac{v_{\text{shallow}}}{f}=\dfrac{0.25}{5}=0.050\,\mathrm{m}=5.0\,\mathrm{cm}
Check the ratio: \lambda_1/\lambda_2=8.0/5.0=1.6 and v_{\text{deep}}/v_{\text{shallow}}=0.40/0.25=1.6. The two ratios agree.
In the figure the wavefronts on the left are further apart, so the left side is deep water (\lambda_1) and the right side, where they are closer together, is shallow water (\lambda_2).
The same vibrator keeps making 5 wavefronts every second, so the frequency does not change at the boundary. In shallow water the waves move more slowly, so in each second they travel a shorter distance, and the 5 waves have to fit into a shorter length. That is why the wavelength decreases when the wave moves more slowly, as the book states.

\lambda_1=8.0\,\mathrm{cm}\ (\text{deep}),\quad \lambda_2=5.0\,\mathrm{cm}\ (\text{shallow})

12.8.3Diffraction of wavesOften examined

Diffraction: The bending of waves around the corners (edges) of an obstacle. When waves pass through an aperture or opening, they spread out because of diffraction.

Diffraction can be observed in a ripple tank by generating straight waves and placing two obstacles so that their separation is comparable to the wavelength of the waves. After passing through the narrow opening, the waves spread out in every direction and turn into circular waves. This effect is greatest when the size of the opening is less than or equal to the wavelength of the generated waves, as in Figure 12.16 (a).

Note that the wavelength (and the speed) of the wave is not changed by diffraction. If the separation between the obstacles is large compared with the wavelength, the central part of the wave is not affected and only the part of the wave at the edges diffracts, as in Figure 12.16 (b).

Figure 12.16: Diffraction. (a) A narrow opening makes the waves spread out as circular waves. (b) A wide opening lets the middle of the wave go straight on and only the edges bend.

A. Wavelength and diffraction

The wavelength is the distance between consecutive crests (or troughs) of a wave.

Shorter wavelengths diffract less. If the wavelength is much smaller than the size of the gap, the waves mainly travel straight through with minimal spreading.
Longer wavelengths diffract more. If the wavelength is large compared with the size of the gap, the waves spread out significantly after passing through the gap.

UV light, which has a shorter wavelength, diffracts less than IR light, which has a longer wavelength, as shown in Figure 12.17.

Figure 12.17: Wavelength and diffraction. UV light (wavelength below 400\,\mathrm{nm}) spreads through the gap by a smaller angle (\theta_1) than IR light (above 800\,\mathrm{nm}), which spreads by a larger angle (\theta_2). So \theta_2>\theta_1.

B. Gap size and diffraction

The size of the opening through which the waves pass also matters.

Narrow gaps: when the gap is comparable to or smaller than the wavelength, significant diffraction occurs and the waves spread out widely on the other side of the gap.
Wide gaps: when the gap is much larger than the wavelength, diffraction is minimal. The waves pass through with only slight spreading, and most of the wave energy remains in the original direction of travel.

For example, sound waves have a relatively large wavelength, which is why they can diffract around obstacles and through openings, so you can hear sounds even when you cannot see the source. Visible light has a much shorter wavelength, about 400\,\mathrm{nm} to 700\,\mathrm{nm}. When light passes through a slit or aperture much larger than its wavelength, there is minimal diffraction and the light beam remains well defined. However, if the slit is of the order of the wavelength of the light (as in a diffraction grating), significant diffraction and interference patterns can be observed.

Wavelength, gap size and diffraction
Condition	Amount of diffraction
Wavelength large compared with the gap (or gap about the same size as the wavelength)	Large: waves spread out widely
Wavelength small compared with the gap (wide gap)	Small: waves go straight through, only the edges bend

Worked example: why we hear round a corner but cannot see round it

Sound of frequency 343\,\mathrm{Hz} (speed 343\,\mathrm{m\,s^{-1}} in air) and green light of wavelength 500\,\mathrm{nm} both pass through a doorway 0.9\,\mathrm{m} wide. (a) Find the wavelength of the sound. (b) Compare each wavelength with the width of the doorway and say which wave diffracts more. (c) What changes if the sound frequency is raised to 3430\,\mathrm{Hz}?

Given

f = 343\,\mathrm{Hz} sound
v = 343\,\mathrm{m\,s^{-1}} speed of sound
\lambda_{\text{light}} = 500\,\mathrm{nm}=5.0\times10^{-7}\,\mathrm{m}
w = 0.9\,\mathrm{m} width of the doorway

To find

\lambda_{\text{sound}}; compare with w

Wavelength of the sound from the universal wave equation.
\lambda_{\text{sound}}=\dfrac{v}{f}=\dfrac{343}{343}=1.0\,\mathrm{m}
Sound: the wavelength (1.0\,\mathrm{m}) is about the same size as the doorway (0.9\,\mathrm{m}). A gap comparable to the wavelength causes significant diffraction, so the sound spreads out widely after the doorway. This is why you can hear someone talking in the next room even when you cannot see them.
Light: compare the doorway with the wavelength.
\dfrac{w}{\lambda_{\text{light}}}=\dfrac{0.9}{5.0\times10^{-7}}=1.8\times10^{6}
The doorway is nearly two million times wider than the wavelength of light. A wide gap gives minimal diffraction, so the light travels straight on and you cannot see round the corner.
Sound diffracts far more than light through the same doorway, because its wavelength is much larger compared with the gap.
If the frequency is raised to 3430\,\mathrm{Hz} the wavelength becomes \lambda=343/3430=0.10\,\mathrm{m}. This is now much smaller than the doorway (0.9\,\mathrm{m}), so the sound spreads less. Shorter wavelengths diffract less.

\begin{aligned}\lambda_{\text{sound}}&=1.0\,\mathrm{m}\ (\text{about the size of the gap: large diffraction})\\ w/\lambda_{\text{light}}&=1.8\times10^{6}\ (\text{minimal diffraction})\end{aligned}

12.9TsunamisOften examined

A tsunami is a series of large ocean waves created by underwater disturbances. These waves travel quickly across the ocean and can reach great heights when they approach the shore, causing massive flooding and damage to coastal areas (Figure 12.18).

Tsunami generation (12.9.1)

Tsunamis are most commonly caused by underwater earthquakes, especially those that happen at tectonic plate boundaries where one plate is pushed beneath another. This sudden movement of the ocean floor displaces a large amount of water, resulting in powerful waves that travel across the ocean, as shown in Figure 12.19. The other causes are:

Underwater volcanic eruptions: the explosion displaces water and creates waves that spread outward.
Landslides under the ocean: the movement of sediment and rock on the sea floor displaces water and leads to the formation of tsunamis.
Large meteorites (asteroids) hitting the ocean: although rare, they can cause massive waves because of the large displacement of water.

Characteristics of tsunamis (12.9.2)

Tsunami waves move quickly across the ocean after the initial disturbance, often travelling at speeds of 500 to 800\,\mathrm{km\,h^{-1}} (310 to 500 miles per hour). In deep water the waves are not very high, which makes them hard to spot. However, as they near the coast, their speed slows down while their height increases significantly, as the wave energy gets compressed.

When tsunamis hit land, they can cause massive destruction with their towering height and powerful energy, flooding coastal regions, demolishing structures and causing substantial loss of life and property.

Figure 12.19: Tsunami generation and approach. A sudden movement of the seabed starts waves that cross the deep sea fast but low. Near the coast the waves slow down and grow taller.

Formula sheet

Time period and frequency

T=\dfrac{1}{f}

T: time period; f: frequency (12.1, 12.3)

Units: \mathrm{s} for T; \mathrm{Hz} (\mathrm{s^{-1}}) for f

Frequency from the number of cycles

f=\dfrac{N}{t}

N: number of wave cycles passing a point; t: time taken (12.2)

Unit: \mathrm{Hz}

Wave speed

v=\dfrac{\Delta d}{\Delta t}

\Delta d: distance travelled by the wave; \Delta t: time taken

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

Wave speed from wavelength and time period

v=\dfrac{\lambda}{T}

\lambda: wavelength; T: time period (the wave travels one wavelength in one time period)

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

Universal wave equation

v=f\lambda

v: wave speed; f: frequency; \lambda: wavelength (12.4); valid for all types of waves

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

Speed of electromagnetic waves

c=f\lambda

c: speed of the electromagnetic wave; f: frequency; \lambda: wavelength (12.5)

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

Key terms

Wave: A disturbance that moves outward from its point of origin, transferring energy by vibrations with little or no transport of the medium.
Mechanical waves: Waves that need a material medium to travel from one point to another, such as sound waves and seismic waves.
Electromagnetic waves: Waves that do not need a material medium. They consist of electric and magnetic fields, for example light, X-rays, ultraviolet and infrared waves.
Transverse waves: Waves in which the particles of the medium vibrate perpendicular to the direction of motion of the wave.
Longitudinal waves: Waves in which the particles of the medium vibrate parallel to the direction of motion of the wave.
Compression: A part of a longitudinal wave with higher density and pressure.
Rarefaction: A part of a longitudinal wave with smaller density and pressure.
Seismic waves: Waves caused by the sudden movement of materials within the Earth, such as slip along a fault during an earthquake.
Seismic S-waves: Secondary seismic waves: transverse waves that cannot travel through liquids.
Seismic P-waves: Primary seismic waves: the fastest seismic waves, longitudinal, travelling through solids and liquids.
Wavelength (\lambda): The shortest distance between points where the wave pattern repeats itself, for example between two successive crests. Unit: metre.
Wavefront: The locus of all points that have travelled some distance in a fixed interval of time; it may be plane or spherical.
Crest: A point of maximum upward displacement of the medium within a cycle.
Trough: The minimum or lowest point in a cycle, the opposite of a crest.
Displacement (x): The separation between the mean position and a point on either side of it. Unit: metre.
Amplitude (x_0): The maximum displacement from the mean position on either side. Unit: metre.
Wave cycle: One complete to-and-fro movement of a point as a wave passes it; the number of cycles N is a pure number.
Time period (T): The time required for one wave cycle to pass a certain point. Unit: second.
Frequency (f): The number of wave cycles passing a point per unit time. Unit: hertz (Hz).
Wave speed (v): The distance travelled by a wave per unit time, v=f\lambda.
Ripple tank: A shallow tray of water with a vibrator, a lamp and a screen, used to show the properties of waves.
Reflection: The throwing back of a wave by a body or surface without absorbing it, so that it returns into the medium it came from.
Refraction: The change in direction of a wave passing from one medium to another, caused by its change in speed.
Diffraction: The bending of waves around the edges of an obstacle, or their spreading on passing through a gap.
Tsunami: A series of enormous ocean waves caused by earthquakes, underwater landslides, volcanic eruptions or asteroids.

Multiple choice questionsp. 90

If a body performing simple harmonic motion completes its one vibration in 2 seconds, its frequency of vibration is:

A2 Hz
B1 Hz
C0.5 HzAnswer
D10 Hz

Why

One vibration takes a time period T = 2\,\mathrm{s}. Frequency is the reciprocal of the time period, so f = \dfrac{1}{T} = \dfrac{1}{2\,\mathrm{s}} = 0.5\,\mathrm{Hz}.

Option A repeats the time period as the frequency, and option D is far too large for a body that takes 2 s per vibration.

360 waves are passing through a point in a river in one hour, time period of the wave is:

A0.1 s
B1 s
C10 sAnswer
D100 s

Why

The frequency is the number of waves per second: f = \dfrac{360}{3600\,\mathrm{s}} = 0.1\,\mathrm{Hz}.

The time period is the reciprocal of the frequency: T = \dfrac{1}{0.1\,\mathrm{Hz}} = 10\,\mathrm{s}. Option A is the frequency value, not the time period.

Time period of electrical vibrator of ripple tank is 0.5 s, water waves in ripple tank has speed of 10 m/s and amplitude of 5 cm. Calculate the frequency of the waves.

A50 Hz
B0.5 Hz
C200 Hz
D2 HzAnswer

Why

The frequency of the waves equals the frequency of the vibrator, which is the reciprocal of its time period: f = \dfrac{1}{T} = \dfrac{1}{0.5\,\mathrm{s}} = 2\,\mathrm{Hz}.

The speed and the amplitude are not needed for the frequency. Option B is the time period given in the question, not the frequency.

Which of the followings is longitudinal wave?

AX-rays
BLight
CString waves
DSoundAnswer

Why

In a longitudinal wave the particles vibrate parallel to the direction of energy transfer. Sound in air compresses and rarefies the air along the direction the wave travels (section 12.6.2).

X-rays and light are electromagnetic, transverse waves, and waves on a string are transverse (Figure 12.4).

When water waves enter from deep water to shallow water:

AFrequency of water waves increases
BTime period of water waves decreases
CSpeed of water waves decreasesAnswer
DWavelength of water waves increases

Why

In shallow water the wave moves more slowly, so its speed decreases. Its frequency is set by the vibrator and does not change, so the time period does not change either.

With f fixed and v smaller, \lambda = \dfrac{v}{f} becomes smaller, so the wavelength decreases, which rules out option D.

Listening to radio transmission in deep valleys of mountain areas is due to:

AReflection of waves
BRefraction of waves
CInterference of waves
DDiffraction of wavesAnswer

Why

Radio waves have long wavelengths. They bend around the edges of the mountains and spread into the valley, which is diffraction (section 12.8.3).

Reflection would send the waves back, refraction only changes their direction at a boundary between media, and interference needs two overlapping waves, so none of A, B or C explains a signal reaching a hidden valley.

Which of the following are transverse waves:

AX-rays
BLight
CString waves
DAll of theseAnswer

Why

X-rays and light are electromagnetic waves, whose electric and magnetic fields vibrate at right angles to the direction of energy transfer. Waves on a string are transverse (Figure 12.4).

All three are transverse, so option D is correct. Options A, B and C are each correct on their own but incomplete.

Tsunamis are most often generated by:

AFlood
BBomb blast
CEarth quakeAnswer
DNone of these

Why

Tsunamis are most often caused by underwater earthquakes, especially at tectonic plate boundaries (section 12.9.1).

A flood is not a generator of ocean waves and a bomb blast is rare, so option D is wrong because an earthquake is listed.

Waves transfer:

AEnergyAnswer
BFrequency
CWavelength
DVelocity

Why

Waves transfer energy without transferring matter (section 12.2). Frequency, wavelength and velocity describe a wave; they are not carried from one place to another.

Which of the following characteristics of a wave is independent of the others?

ASpeed
BFrequency
CAmplitudeAnswer
DWavelength

Why

Speed, frequency and wavelength are linked by v = f\lambda, so each depends on the other two. Amplitude does not appear in this relation, and changing it does not change the speed, frequency or wavelength.

Frequency is also set by the source, but it still enters v = f\lambda, so amplitude is the characteristic that stands apart.

Note. The book prints no key. Frequency is also fixed by the source, so option B is the closest alternative; amplitude is the one that does not appear in the wave equation.

11FBISE SSC-II 2026 Annual I, Set D, MCQ 10

The relation between v, f and λ of a wave is:

Avf = λ
Bfλ = vAnswer
Cvλ = f
Dv = λ/f

Why

The universal wave equation is v = f\lambda (equation 12.4). Option B, f\lambda = v, is the same equation with the terms in a different order.

Options A and C multiply the wrong quantities together. Option D, v = \dfrac{\lambda}{f}, would make the speed fall as the frequency rises, which is not what the equation says.

Constructed response questionsp. 91

Follow the directions to respond to the following questions. (a) In the graph below represent crest, trough and origin. (b) Also in similar graph drawn again show displacement, amplitude and wavelength.

Solution

(a) In the graph below represent crest, trough and origin.

The origin is the point where the wave crosses the horizontal axis at the start of the wave. Here the displacement is zero, at x = 0.

A crest is the point of maximum upward displacement. The first crest is at x = 1 on the graph, where the displacement reaches its largest positive value.

A trough is the point of maximum downward displacement, the lowest point of a cycle. The first trough is at x = 3 on the graph.

Crest, trough and origin on the wave in part (a).

\text{Origin: } (0,0),\quad \text{crest at } x=1,\quad \text{trough at } x=3

(b) Also in similar graph drawn again show displacement, amplitude and wavelength.

Draw the wave again on the same axes. Mark the three quantities as follows.

Displacement y is the distance of a point of the medium from its mean position at a given instant. The vertical line from the axis to the curve at x = 0.5 shows the displacement at that point.
Amplitude A is the maximum displacement from the mean position, so it is the distance from the axis to a crest. It is measured in metres.
Wavelength \lambda is the distance between two successive crests (or troughs), so it is marked from one crest to the next crest, from x = 1 to x = 5.

Displacement y, amplitude A and wavelength \lambda on the wave in part (b).

y \text{ at a point},\quad A \text{ from the axis to a crest},\quad \lambda \text{ from crest to crest}

Short response questionsp. 91

What will happen to frequency of waves in a ripple tank if time period of electrical vibrator is decreased? What will happen to speed of waves?

Solution

The frequency of the waves equals the frequency of the vibrator, and the two are reciprocals: f = \dfrac{1}{T}.

If the time period T is decreased, the frequency f increases, because the vibrator completes more vibrations in each second and sends out more waves each second.

The speed of the waves does not change. The speed depends on the medium (the depth of the water in the tank), which is the same, so v stays the same.

Since v = f\lambda, the wavelength decreases when f increases and v stays the same.

What will happen to wavelength, speed, frequency and time period of the water waves in ripple tank if water waves enter from deep water to shallow water regions?

Solution

The changes are as follows.

Wavelength decreases, because the wave is slower in shallow water and \lambda = \dfrac{v}{f} with f fixed.
Speed decreases, because the wave moves more slowly in shallow water.
Frequency does not change, because it is set by the vibrator and the wave keeps the same number of cycles each second.
Time period does not change, because T = \dfrac{1}{f} and f is unchanged.

Nuclear fusion explosions are taking place at Sun, we get heat on earth but do not hear sound. Why?

Solution

The heat and light from the Sun are electromagnetic waves. Electromagnetic waves do not need a material medium, so they travel through the vacuum of space to the Earth.

Sound is a mechanical wave, which needs a material medium such as air to propagate. Space between the Sun and the Earth is a vacuum, so the sound of the explosions cannot travel to the Earth, and we do not hear it.

Why do water waves refract at the boundary of shallow water and deep water? Relate this behavior to refraction of light.

Solution

Refraction happens when a wave passes from one medium to another and its speed changes. The speed of water waves depends on the depth of the water, so the speed changes at the boundary between shallow and deep water. The wave direction changes as a result (Figure 12.15).

Light is refracted in the same way at the boundary between two media, such as air and water. Light travels more slowly in water, so its speed changes and its direction changes, which is why a pencil in water looks bent.

In both cases the change of speed causes the change of direction. The wavelength also changes, because the frequency stays the same.

Classify different wave types based on whether they require a medium for propagation. Justify why electromagnetic waves do not need a medium.

Solution

Waves are divided into two types on the basis of medium requirement.

Mechanical waves need a material medium to propagate. Examples are sound waves, waves on a water surface, waves on a rope and seismic waves.
Electromagnetic waves do not need a material medium. Examples are light, X-rays, ultraviolet, infrared and radio waves.

Electromagnetic waves do not need a medium because they are made of electric and magnetic fields that vibrate and produce each other. A changing electric field produces a magnetic field and a changing magnetic field produces an electric field, so the wave carries itself forward through vacuum (Figure 12.10).

6FBISE SSC-II 2026 Annual I, Set D, Q.2 (x) OR

Under what conditions maximum diffraction of waves occurs?

Solution

Diffraction is the bending of waves around the edges of an obstacle or through an opening. Diffraction is greatest when the size of the opening is less than or equal to the wavelength of the wave (Figure 12.16 (a)).

Waves of longer wavelength diffract more than waves of shorter wavelength, so maximum diffraction occurs when the gap is about the size of the wavelength or smaller and the wavelength is long.

Distinguish between reflection and refraction by describing how each process affects wave direction and speed. Design an experiment to demonstrate both phenomena using a ripple tank, and evaluate how changes in medium affect each process.

Solution

Distinction

Feature	Reflection	Refraction
Wave direction	The wave returns into the same medium at a changed direction, with the angle of reflection equal to the angle of incidence.	The wave changes direction when it passes into a different medium.
Speed	Does not change, because the wave stays in the same medium.	Changes, because the speed depends on the medium.
Wavelength	Does not change.	Changes, with the frequency fixed.

Ripple tank experiment

Reflection: make straight waves with a plane dipper and place a straight barrier across the tank at an angle to the wavefronts. Observe the reflected waves and measure the angle of incidence and the angle of reflection; they are equal.
Refraction: place a thick plastic sheet in the tray at an angle to the wavefronts so that part of the tank is shallow. Observe the wavefronts bending and the wavelength becoming shorter over the shallow part.

Evaluation

Reflection needs no change of medium, so changing the depth of the water does not change how the waves reflect, although the reflected waves keep the same speed. Refraction depends on the medium: making the water shallower slows the waves and bends them, so the change in medium decides how much a wave is refracted.

Long response questionsp. 91–92

How does a wave transfer energy without transferring matter? How does this principle apply to different media like ropes, water, and air?

Solution

Energy transfer without transfer of matter

A wave is a disturbance that moves through a medium, or through space in the case of electromagnetic waves. The particles of the medium only oscillate about their mean positions and return to where they started, so the matter stays in place while the energy moves forward.

A source starts vibrating and sets the nearest particles of the medium in motion.
Each particle passes its energy to the next particle through the forces between them.
Each particle oscillates about its mean position and then returns to it, so there is no net displacement of matter.
The disturbance moves on, and the energy moves with it.

Application to ropes, water and air

Rope: when one end is moved up and down, the rope's particles move up and down (transverse), while the energy travels along the rope to the wall (Figure 12.9).
Water: a stone thrown into a pond makes ripples. A cork or a boat bobs up and down as the waves pass, but it stays near its place, so the energy moves outwards while the water does not travel with it.
Air: the vocal cords make the air particles compress and rarefy, and this sound energy travels to the listener. The air molecules only oscillate about their positions (Figure 12.12).

The principle is the same in all three media: energy is passed on from particle to particle while each particle stays near its equilibrium position. Only the direction of vibration differs. In ropes and on water surfaces the particles move at right angles to the energy transfer, and in air they move along it.

Describe the key characteristics of a wave (wavelength, frequency, amplitude) and their relationships to each other. How do these characteristics affect the wave's energy and behavior?

Solution

Characteristics

Wavelength \lambda is the shortest distance between points where the wave pattern repeats, such as two successive crests. Its unit is the metre.
Frequency f is the number of cycles passing a point per second, f = \dfrac{N}{t}, measured in hertz (Hz).
Amplitude A is the maximum displacement of a particle from its mean position. It is measured in metres.
Time period T is the time for one complete cycle, T = \dfrac{1}{f}, measured in seconds.

Relationships

Time period and frequency are reciprocals of each other. The speed of a wave is related to its frequency and wavelength by v = f\lambda. If the speed is fixed, a higher frequency means a shorter wavelength, and a longer wavelength means a lower frequency.

v = f\lambda

Effect on energy and behaviour

A wave with a larger amplitude carries more energy: a larger displacement means more energy is passed on through the medium. For waves of the same amplitude, a high frequency wave carries more energy than a low frequency wave, because it needs more energy per second to generate (information box, section 12.2.2). The wavelength decides how much a wave diffracts through a gap (section 12.8.3), and the wavelength changes when the wave enters a medium of a different speed.

Explain the relationship between wave speed, frequency, and wavelength using the equation v = fλ. Analyze a scenario where one of these variables changes and discuss the impact on the others.

Solution

The relationship

The universal wave equation v = f\lambda says that the speed of a wave equals the product of its frequency and wavelength. It holds for all types of waves (equation 12.4).

Scenario: frequency changes, medium stays the same

Take a rope with the same tension, so the speed of the wave on it is fixed at v = 0.5\,\mathrm{m\,s^{-1}}. A wave of frequency f = 2\,\mathrm{Hz} travels along it.
\lambda = \dfrac{v}{f} = \dfrac{0.5\,\mathrm{m\,s^{-1}}}{2\,\mathrm{Hz}} = 0.25\,\mathrm{m}
Double the frequency to f = 4\,\mathrm{Hz} by moving the end of the rope faster. The speed is still set by the rope, so it stays 0.5\,\mathrm{m\,s^{-1}}.
\lambda = \dfrac{0.5\,\mathrm{m\,s^{-1}}}{4\,\mathrm{Hz}} = 0.125\,\mathrm{m}

Doubling the frequency halves the wavelength. The speed does not change because the medium has not changed.

Scenario: the medium changes

When a water wave passes from deep to shallow water, the frequency stays the same, the speed decreases and so the wavelength decreases. The equation v = f\lambda shows that a fall in speed at fixed frequency must reduce the wavelength.

In general, changing one of the three variables changes at least one of the others, and the equation tells us which. The source fixes the frequency; the medium fixes the speed; the wavelength then follows from the equation.

Compare and contrast transverse and longitudinal waves, providing examples of each. Why is this distinction important in understanding wave behavior?

Solution

Comparison

Feature	Transverse waves	Longitudinal waves
Direction of vibration	At right angles to the direction of energy transfer	Parallel to the direction of energy transfer
Features	Crests and troughs	Compressions and rarefactions
Examples	Electromagnetic waves, waves on a rope, water surface waves, S-waves	Sound waves in air, P-waves
Speed in solids	Slower; cannot travel through liquids	Faster; travel through solids and liquids

Why the distinction matters

The direction of vibration decides what a wave can do. Transverse waves need a rigid medium to keep the vibration at right angles, which is why S-waves do not pass through liquids. Longitudinal waves need a medium that can be compressed, which is why sound needs air or another material. Knowing the type therefore explains which media a wave can pass through and how fast it moves through them.

Describe how waves undergo reflection, refraction, and diffraction. Analyze a real-world scenario to illustrate these phenomena.

Solution

Reflection

Reflection is the change in direction of a wave at an interface between two media or at a rigid barrier, so that the wave returns into the medium it came from (section 12.8.1). The angle of incidence equals the angle of reflection.

Refraction

Refraction is the change in direction of a wave when it passes from one medium into another, caused by the change in its speed. When a wave enters a medium in which it moves more slowly, it bends towards the normal. When it enters a medium in which it moves faster, it bends away from the normal.

Figure 12.15: Refraction at a boundary. The ray bends away from the normal when the wave enters deep water, where it moves faster and its wavelength increases.

Diffraction

Diffraction is the bending of waves around the edges of an obstacle or through an opening (section 12.8.3). It is greatest when the gap is about the size of the wavelength or smaller.

Real-world scenario: radio in a mountain valley

A radio transmitter sends out waves that reach a listener in a deep valley behind a mountain. The waves are reflected from the rock faces and bent around the edges of the mountain, which is why the signal is heard in the valley. The bending around the mountain edges is diffraction, and the long wavelength of radio waves makes the diffraction large enough to reach the valley (MCQ 6).

Reflection and refraction also play their part: the sea waves approaching a shore are refracted as the water becomes shallower, and they are reflected from a rigid seawall.

6FBISE SSC-II 2026 Annual I, Set B, Q.2 (x) OR

How does the wavelength of a wave affect its diffraction pattern? What factors determine the extent of diffraction, and how can this principle be applied in technology?

Solution

Effect of wavelength

Waves of longer wavelength diffract more; waves of shorter wavelength diffract less. For a short wavelength, such as ultraviolet light, the wave travels almost straight through a gap, while a long wavelength, such as infrared light, spreads out after passing through the same gap (Figure 12.17).

Factors that decide the extent of diffraction

The wavelength of the wave: a longer wavelength diffracts more.
The size of the gap or obstacle compared with the wavelength: diffraction is greatest when the gap is less than or equal to the wavelength, and it is minimal when the gap is much larger than the wavelength.

Application in technology

Diffraction is used in the diffraction grating, a plate with many closely spaced slits. When light passes through the slits, the diffraction and interference of the waves spread the colours of white light. Long radio waves are used in broadcasting because they diffract around hills and into valleys (MCQ 6).

Explain the formation of tsunamis due to underwater earthquakes. Discuss how these waves change as they move from deep to shallow water and their potential impacts.

Solution

Formation of a tsunami

A tsunami is a series of large ocean waves created by an underwater disturbance (section 12.9). Underwater earthquakes are the most common cause. Along tectonic plate boundaries, one plate is pushed beneath another, and the sudden movement of the ocean floor displaces a large amount of water. This displacement produces powerful waves that travel across the ocean (Figure 12.19).

Change in deep and shallow water

In deep water the waves travel quickly, at about 500 to 800 km/h, and are not very high, so they are hard to spot.
As they near the coast the speed of the wave decreases, its wavelength decreases, and its height increases significantly, because the wave energy is compressed into a smaller depth.

Potential impacts

When a tsunami reaches land it causes flooding of coastal regions, destruction of buildings, and substantial loss of life and property because of its great height and energy (section 12.9.2).

Note. Learning outcome P-10-D-10 says the tsunami waves increase in frequency as they approach the shore. The frequency of a wave is set by its source and does not change on entering shallower water; what changes is the speed and wavelength, and the height increases. This answer follows the physics.

How does the amplitude of a wave relate to the amount of energy it carries? Use mathematical reasoning and real-world examples to support your answer.

Solution

Amplitude and energy

The energy of a wave is passed on by the oscillations of the particles of the medium. A wave of larger amplitude has particles that are displaced more from their mean positions, so each particle carries more energy as the wave passes. Therefore a wave of greater amplitude carries more energy.

Mathematical reasoning

For a wave of fixed frequency, the energy it carries increases as its amplitude increases.
For waves of the same amplitude, the wave with the higher frequency carries more energy, because it generates more oscillations per second (information box, section 12.2.2).

The exact formula connecting energy with amplitude and frequency is beyond this unit, so the relation here is stated qualitatively, as in the book.

Real-world examples

Sound: a loud sound has a larger amplitude than a soft one, and carries more energy into the ear.
Earthquake: a large amplitude seismic wave carries more energy and causes more damage than a small one.

Differentiate between primary (P) and secondary (S) seismic waves. How do these waves provide insights into the Earth's internal structure?

Solution

Differences

Feature	Primary (P) waves	Secondary (S) waves
Type	Longitudinal	Transverse
Vibration	Particles vibrate parallel to the direction of travel	Ground vibrates side to side or up and down, at right angles to the direction of travel
Speed	Fastest type of seismic wave	Slower than P-waves
Medium	Travel through solids and liquids	Cannot travel through liquids, only through solids

Insights into the Earth's interior

The time difference between the arrival of P-waves and the slower S-waves at a seismic station helps determine the distance to the earthquake's epicentre. Because the two waves travel differently through solids and liquids, the way they arrive at different places tells geophysicists about the structure of the Earth's interior (section 12.6.2).

Discuss practical applications of wave reflection, refraction, and diffraction in technology. Provide examples and analyze their significance.

Solution

Reflection

Reflection of waves is used wherever a wave must be sent back or directed. The ripple tank itself uses reflection of water waves from a barrier, and reflecting surfaces are used to direct waves. The significance is that the reflected wave keeps the same speed and frequency, so the path it follows can be predicted from the angle of incidence.

Refraction

Refraction explains why a pencil in water looks bent, and why water waves change direction at the boundary of shallow and deep water. Its significance is that the change in speed, not the frequency, bends the wave. Knowing how waves refract is essential to predicting where a tsunami's energy arrives on the coast (section 12.9).

Diffraction

Diffraction lets us hear sounds around corners, because sound waves have long wavelengths that spread around obstacles. Diffraction of light through a diffraction grating separates colours. Its significance is that the amount of spreading depends on the wavelength compared with the gap size, so the design of a gap or grating controls the result.

Numerical response questionsp. 92

If 10 waves are produced on a long-stretched string of length 20 m. What is wavelength of the waves? What will be frequency of the wave if speed of waves is 12 m/s?

Solution

Given

L = 20\,\mathrm{m} length of the string
N = 10 number of waves
v = 12\,\mathrm{m\,s^{-1}}

To find

\lambda (wavelength)
f (frequency)

The 10 waves fill the whole 20 m string, so 20 m is 10 wavelengths.
\lambda = \dfrac{L}{N} = \dfrac{20\,\mathrm{m}}{10} = 2\,\mathrm{m}
Use the universal wave equation v = f\lambda and make f the subject.
f = \dfrac{v}{\lambda}
Substitute the values.
f = \dfrac{12\,\mathrm{m\,s^{-1}}}{2\,\mathrm{m}} = 6\,\mathrm{Hz}

\lambda = 2\,\mathrm{m},\qquad f = 6\,\mathrm{Hz}

Book answer. 2 m, 6 Hz

Note. Reads the 10 waves as filling the whole 20 m, so that 20 m equals 10 wavelengths; this is the reading that gives the book's answer.

In a ripple tank of length 1.5 m, waves cover this length in 2 seconds. If distance between two successive crest and trough is 10 cm then find the number of waves, frequency of waves and their time period?

Solution

Given

L = 1.5\,\mathrm{m} length of the ripple tank
t = 2\,\mathrm{s} time taken to cover the length
d = 10\,\mathrm{cm}=0.10\,\mathrm{m} distance from a crest to the next trough

To find

N (number of waves)
f (frequency)
T (time period)

The speed of the waves is the distance covered per unit time.
v = \dfrac{L}{t} = \dfrac{1.5\,\mathrm{m}}{2\,\mathrm{s}} = 0.75\,\mathrm{m\,s^{-1}}
A crest and the next trough are half a wavelength apart, so the wavelength is twice this distance.
\lambda = 2d = 2 \times 0.10\,\mathrm{m} = 0.20\,\mathrm{m}
The number of waves in the length is the length divided by the wavelength.
N = \dfrac{L}{\lambda} = \dfrac{1.5\,\mathrm{m}}{0.20\,\mathrm{m}} = 7.5
The frequency is the speed divided by the wavelength.
f = \dfrac{v}{\lambda} = \dfrac{0.75\,\mathrm{m\,s^{-1}}}{0.20\,\mathrm{m}} = 3.75\,\mathrm{Hz}
Check with the definition f = N/t: f = \dfrac{7.5}{2\,\mathrm{s}} = 3.75\,\mathrm{Hz}, which agrees.
The time period is the reciprocal of the frequency.
T = \dfrac{1}{f} = \dfrac{1}{3.75\,\mathrm{Hz}} = 0.267\,\mathrm{s}

N = 7.5\text{ waves},\qquad f = 3.75\,\mathrm{Hz},\qquad T = 0.267\,\mathrm{s}

Book answer. 7.5 waves, 3.75 Hz, 0.266 s

Note. The book gives 0.266 s. Since 1/3.75 = 0.2667\ldots the book has truncated the value; 0.267 s is the rounded result.

Calculate the frequency of red light of wavelength 700 nm in vacuum?

Solution

Given

\lambda = 700\,\mathrm{nm}=7.0\times10^{-7}\,\mathrm{m} wavelength of red light
c = 3.0\times10^{8}\,\mathrm{m\,s^{-1}} speed of light in vacuum

To find

f (frequency)

Light is an electromagnetic wave, so c = f\lambda holds with c as the speed in vacuum. Make f the subject.
f = \dfrac{c}{\lambda}
Substitute the values.
f = \dfrac{3.0\times10^{8}\,\mathrm{m\,s^{-1}}}{7.0\times10^{-7}\,\mathrm{m}}
Evaluate.
f = 4.29\times10^{14}\,\mathrm{Hz}

f = 4.29\times10^{14}\,\mathrm{Hz}

Book answer. 4.29 × 10^14 Hz

A wooden bar vibrating into the water surface has a frequency of 12 Hz. The resulting wave has a wavelength of 3 cm. What is the speed of the wave?

Solution

Given

f = 12\,\mathrm{Hz} frequency of the wave
\lambda = 3\,\mathrm{cm}=0.03\,\mathrm{m} wavelength

To find

v (speed)

Use the universal wave equation.
v = f\lambda
Substitute the values.
v = 12\,\mathrm{Hz} \times 0.03\,\mathrm{m} = 0.36\,\mathrm{m\,s^{-1}}

v = 0.36\,\mathrm{m\,s^{-1}}

Book answer. 0.36 m s^-1

What is the wavelength of the radio-waves transmitted by an FM station at 90 MHz?

Solution

Given

f = 90\,\mathrm{MHz}=9.0\times10^{7}\,\mathrm{Hz} frequency of the FM station
c = 3.0\times10^{8}\,\mathrm{m\,s^{-1}} speed of radio waves (speed of light)

To find

\lambda (wavelength)

Radio waves are electromagnetic waves, so c = f\lambda. Make \lambda the subject.
\lambda = \dfrac{c}{f}
Substitute the values.
\lambda = \dfrac{3.0\times10^{8}\,\mathrm{m\,s^{-1}}}{9.0\times10^{7}\,\mathrm{Hz}}
Evaluate.
\lambda = 3.33\,\mathrm{m}

\lambda = 3.33\,\mathrm{m}

Book answer. 3.33 m

FBISE 2026 questions

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

FBISE SSC-II 2026 Annual I, Set B, MCQ 101 mark

Compression and rarefaction are features of ________ waves.

ALongitudinalAnswer
BTransverse
CSurface
DElectromagnetic

Why

A longitudinal wave is made of compressions (particles close together, higher pressure) and rarefactions (particles spread apart, lower pressure). Option B is wrong because a transverse wave has crests and troughs, not compressions and rarefactions. Option D is also wrong: electromagnetic waves are transverse and need no medium to be compressed.

Note. The paper tags this item to Unit 13 (Sound), since sound is the usual example; the topic itself is taught in this unit (longitudinal waves).

FBISE SSC-II 2026 Annual I, Set B, Q.2 (iii)1+2 = 3 marks

Make wave form of transverse waves. How is energy transferred in a transverse wave? Write briefly.

Solution

Wave form (1 mark). A transverse wave has crests and troughs. Its particles vibrate up and down about the rest position while the wave moves along.

Transverse wave. Each particle vibrates up and down, at right angles to the direction of travel of the wave, which is the direction of energy transfer.

Energy transfer (2 marks). The source (for example a hand moving the end of a rope up and down) makes the first particle vibrate. Each particle pulls its neighbour with it, so its neighbour starts vibrating a little later, and the disturbance passes from particle to particle along the wave. The particles themselves only vibrate about their mean positions at right angles to the wave and do not travel with it. Only the energy is carried forward, in the direction of travel of the wave.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (x) OR1.5+1.5 = 3 marks

How does wavelength affect the diffraction of a wave through a gap?

Solution

Effect (1.5 marks). Diffraction is the spreading of waves as they pass through a gap or around an obstacle. For a gap of a given size, a longer wavelength diffracts more and a shorter wavelength diffracts less.

A wave whose wavelength is large compared with the gap spreads out widely on the far side. A wave whose wavelength is much smaller than the gap passes almost straight through with very little spreading. Diffraction is greatest when the gap is less than or equal to the wavelength. The wavelength and speed of the wave do not change in diffraction.

Wavelength and diffraction through the same gap. UV light (shorter wavelength) spreads through a small angle, IR light (longer wavelength) spreads through a larger angle.

Example (1.5 marks). Infrared light, with the longer wavelength, diffracts more than ultraviolet light, which has the shorter wavelength, through the same gap (Figure 12.17 of the book). Long radio waves bend round hills and reach valleys for the same reason.

FBISE SSC-II 2026 Annual I, Set D, MCQ 101 mark

Wave speed is given by the relation:

Av=f+\lambda
Bv=f.\lambdaAnswer
Cv=\lambda/f
Dv=f/\lambda

Why

Wave speed is the distance moved by a wave in one second. In one time period T the wave moves one wavelength \lambda, so v=\dfrac{\lambda}{T}=f\lambda. Option C (\lambda/f) is wrong: it has the units \mathrm{m\,s}, not \mathrm{m\,s^{-1}}, because it divides by f where it should divide by T (or multiply by f).

FBISE SSC-II 2026 Annual I, Set D, Q.2 (iii)1.5+1.5 = 3 marks

Draw wave forms to compare the vibration directions of transverse and longitudinal waves.

Solution

Transverse wave (1.5 marks). The particles vibrate at right angles to the direction in which the wave (energy) travels. The wave form has crests and troughs.

Transverse wave. Each particle vibrates up and down, at right angles to the direction of travel of the wave, which is the direction of energy transfer.

Longitudinal wave (1.5 marks). The particles vibrate parallel to the direction in which the wave (energy) travels, along the same line. The wave form has compressions and rarefactions.

Longitudinal wave. Each particle moves to and fro along the same line as the energy travels, forming compressions and rarefactions.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (x) OR1.5+1.5 = 3 marks

How does gap size affect the diffraction of a wave?

Solution

Narrow gap (1.5 marks). When the gap is less than or equal to the wavelength, diffraction is greatest. The waves spread out in all directions beyond the gap and become nearly circular waves (Figure 12.16 (a) of the book).

Wide gap (1.5 marks). When the gap is much larger than the wavelength, diffraction is minimal. The central part of the wave passes straight through and only the part at the edges bends, so most of the energy keeps its original direction (Figure 12.16 (b)). The wavelength and speed of the wave stay the same in both cases.

Gap size and diffraction. (a) A narrow opening makes the waves spread out as circular waves. (b) A wide opening lets the central part pass straight, with bending only at the edges.

FBISE SSC-II 2026 Annual I, Set M, MCQ 91 mark

Maximum displacement of particles refers to:

ASpeed
BAmplitudeAnswer
CFrequency
DWave length

Why

Amplitude is the maximum displacement of a vibrating particle from its mean (rest) position, that is, the height of a crest or the depth of a trough. Option D is wrong: wavelength is the distance between two successive crests (or troughs), a distance along the wave and not a displacement of a particle.

FBISE SSC-II 2026 Annual I, Set M, MCQ 101 mark

Speed of wave is 7 m/s, if amplitude is doubled, the speed will:

ANot changeAnswer
BBe doubled
CBecome zero
DBecome half

Why

The speed of a wave is v=f\lambda and depends on the medium (and the source frequency), not on the amplitude. Doubling the amplitude makes the wave carry more energy, but it is still 7\,\mathrm{m\,s^{-1}}. Option B is the tempting wrong choice: amplitude is not in the relation v=f\lambda.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (iii)3 marks

Explain with the help of diagram how the vibration direction in a transverse wave is perpendicular to the direction of energy transfer.

Solution

Diagram. The figure shows a transverse wave moving to the right (the direction of travel is the direction of energy transfer). The particles are marked with dots; the double arrow shows the direction of their vibration.

Transverse wave. Each particle vibrates up and down, at right angles to the direction of travel of the wave, which is the direction of energy transfer.

Explanation. When one end of a rope is moved up and down, each particle of the rope moves up and down about its rest position, along the vertical line shown by the double arrow. The wave itself, and with it the energy, moves horizontally along the rope. The vertical line of vibration and the horizontal line of travel are at 90^{\circ} to each other, so the vibration direction is perpendicular to the direction of energy transfer. The particles do not move along with the wave: they only pass the energy on to the next particle.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (x) OR3 marks

If the frequency of a wave is 50 Hz and its wavelength is 2 mm, calculate its speed in SI units.

Solution

Given

f = 50\,\mathrm{Hz}
\lambda = 2\,\mathrm{mm}=2\times10^{-3}\,\mathrm{m} converted to metres (1 mm = 10^{-3} m)

To find

Speed v in \mathrm{m\,s^{-1}}

Write the wave equation.
v=f\lambda
Substitute the values in SI units.
v=50\times2\times10^{-3}
Multiply.
v=0.1\,\mathrm{m\,s^{-1}}

v=0.1\,\mathrm{m\,s^{-1}}

Exam focus

Compare transverse and longitudinal waves: draw both wave forms, label crests, troughs, compressions and rarefactions, and state the vibration direction against the direction of energy transfer. (short)
Q.2 (iii) in all three 2026 sets: Set B (transverse wave form and energy transfer), Set D (compare both wave forms) and Set M (why transverse vibration is perpendicular to energy transfer). Also comprehensive question 4 and MCQ 4 of the book.
Go to answer
Explain how wavelength and gap size affect diffraction, with the narrow-gap and wide-gap diagrams. (short)
Q.2 (x) OR in Set B (effect of wavelength) and Set D (effect of gap size); the book asks it as short question 6 and comprehensive question 6. Diffraction is greatest when the gap is less than or equal to the wavelength.
Go to answer
Use v=f\lambda with unit conversion (mm, cm, nm, MHz to SI) to find speed, frequency or wavelength. (numerical)
Q.2 (x) OR in Set M (50 Hz, 2 mm) and MCQ 10 of Set D; the book's numerical problems 1 to 5 all use it. The paper prints v=f\lambda and 1\,\mathrm{mm}=10^{-3}\,\mathrm{m} in its formula list, so the marks are in the conversion and the substitution.
Go to answer
Define the wave parameters (amplitude, wavelength, frequency, time period) and know what each depends on. (mcq)
MCQ 9 and MCQ 10 of Set M (amplitude is the maximum displacement; speed does not change when amplitude is doubled) and the book's MCQ 10 on the independent characteristic.
Go to answer
Name the features of longitudinal waves (compression, rarefaction) and give examples of each wave type. (mcq)
MCQ 10 of Set B (compression and rarefaction belong to longitudinal waves); book exercise MCQ 4 and MCQ 7 ask which waves are longitudinal or transverse.
Go to answer
Describe reflection, refraction and diffraction of waves in the ripple tank and how speed, wavelength and frequency change in each. (long)
Not asked in the 2026 papers of this unit, but comprehensive question 5, short questions 2, 4 and 7 and MCQ 5 of the book cover it, and the three effects are the core of the unit.
Go to answer
Explain how a wave transfers energy without transferring matter, and how amplitude relates to energy. (long)
Set B Q.2 (iii) asks how energy is transferred in a transverse wave; the book's comprehensive questions 1 and 8 ask the same idea in full.
Go to answer

12.1Wave motion

Water waves (12.1.1)

Sound waves (12.1.2)

Electromagnetic waves (12.1.3)

Wave in a rope (12.1.4)

12.2Wave propagationOften examined

Oscillation (12.2.1)

Energy transfer (12.2.2)

No net movement (12.2.3)

12.3Characteristic wave parametersOften examined

Wavelength λ (12.3.1)

Wavefront (12.3.2)

Wave crest and trough (12.3.3)

Displacement (12.3.4)

Amplitude (12.3.5)

Time period (12.3.6)

Frequency (12.3.7)

Relation between time period and frequency (12.3.8)

12.4Waves and their types

12.5Types of waves on the basis of mediumOften examined

A. Mechanical waves

B. Electromagnetic waves

12.6.1Transverse wavesOften examined

A. Electromagnetic radiation

B. Waves on the surface of water

C. Seismic S-waves (secondary waves)

12.6.2Longitudinal wavesOften examined

A. Sound waves

B. Seismic P-waves (primary waves)

12.7Wave characteristicsOften examined

A. Wave cycle

B. Frequency

C. Time period

D. Wave speed

Derivation The universal wave equation

12.8Properties of waves and the ripple tank

Construction and working

12.8.1Reflection of wavesOften examined

12.8.2Refraction of wavesOften examined

12.8.3Diffraction of wavesOften examined

A. Wavelength and diffraction

B. Gap size and diffraction

12.9TsunamisOften examined

Tsunami generation (12.9.1)

Characteristics of tsunamis (12.9.2)

Formula sheet

Key terms

Multiple choice questionsp. 90

Constructed response questionsp. 91

Short response questionsp. 91

Distinction

Ripple tank experiment

Evaluation

Long response questionsp. 91–92

Energy transfer without transfer of matter

Application to ropes, water and air

Characteristics

Relationships

Effect on energy and behaviour

The relationship

Scenario: frequency changes, medium stays the same

Scenario: the medium changes

Comparison

Why the distinction matters

Reflection

Refraction

Diffraction

Real-world scenario: radio in a mountain valley

Effect of wavelength

Factors that decide the extent of diffraction

Application in technology

Formation of a tsunami

Change in deep and shallow water

Potential impacts

Amplitude and energy

Mathematical reasoning

Real-world examples

Differences

Insights into the Earth's interior

Reflection

Wavelength $λ$ (12.3.1)