DSE Physics Recap – Wave Propagation and Wave Phenomena | Ronnie (2026)

Wave Propagation and Wave Phenomena

Subject: Physics

Ronnie (2026)

For use on Tuesday, 15 September 2026

True or False

The distance from the first crest to the sixth crest of a water wave is equal to five wavelengths.
The direction of propagation of a water wave is always parallel to its wavefronts.
When water waves travel across a ripple tank, a small floating object is carried continuously from one side of the tank to the other.
Two water waves travelling at the same speed must have the same frequency.
If the frequency of water waves is doubled without changing the water depth, the distance between two adjacent crests is halved.
When a water wave is reflected by a straight barrier, its speed remains unchanged but its velocity changes.
The angle between an incident wavefront and a straight barrier is equal to the angle of incidence.
After reflection by a straight barrier, the reflected wavefronts are always parallel to the incident wavefronts.
When water waves travel normally from deep water into shallow water, their speed and wavelength decrease, but their direction remains unchanged.
When water waves travel at an angle from shallow water into deep water, they bend away from the normal while their frequency remains unchanged.

10 marks

Wave Propagation

Crest Spacing and Wave Timing

A photograph shows straight water waves travelling across a ripple tank. The distance measured from the first crest to the ninth crest is 28.0 cm.

At a fixed point, the time from the arrival of the first crest to the arrival of the eleventh crest is 4.0 s.

Determine the number of complete wavelengths between the first and ninth crests.
Calculate the wavelength of the water waves in metres.
Determine the number of complete periods between the arrival of the first and eleventh crests.
Calculate:
1. the period of the waves;
2. the frequency of the waves;
3. the speed of the waves.
A student calculates the frequency by dividing 11 by 4.0 and obtains 2.75 Hz. Explain why this answer is incorrect.
A small piece of cork is placed at the fixed point. Describe its motion during the time in which several crests pass the point.
The crests continue to travel across the ripple tank, but the cork remains near the same position. Explain what is being transferred across the ripple tank.

15 marks

Wave Propagation

Comparing Two Wave Patterns

Water waves P and Q travel separately through the same region of a ripple tank. The depth of the water is unchanged.

Water wave	Number of complete waves passing a point	Time taken / s	Wavelength / cm
P	12	3.0	2.5
Q	8	4.0	5.0

Calculate the frequency of:
1. wave P;
2. wave Q.
Calculate the speed of:
1. wave P;
2. wave Q.
The results suggest that P and Q travel at the same speed. Explain why this is expected.
Compare the frequency and wavelength of wave P with those of wave Q.
The frequency of the source producing wave Q is gradually increased until it is equal to the frequency of wave P. Determine the new wavelength of wave Q.
A student claims that the wavefronts of P move faster because more wavefronts pass a fixed point each second. Explain the error in the student’s reasoning.
On a diagram, five parallel lines are used to represent five successive straight wavefronts. The total distance from the first line to the fifth line is 12 cm. Determine the wavelength represented by the diagram.

15 marks

Reflection of Water Waves

Wavefronts, Rays and Reflection

Straight water waves travel towards a straight barrier. The incident wavefronts make an angle of 38° with the barrier.

Determine the angle between:
1. the incident ray and the incident wavefronts;
2. the incident ray and the normal;
3. the reflected ray and the normal;
4. the reflected ray and the barrier.
Calculate the smaller angle between the incident ray and the reflected ray.
The wavelength of the incident waves is 3.2 cm and their frequency is 5.0 Hz. Calculate the speed of the incident waves.
Determine the following quantities for the reflected waves:
1. frequency;
2. wavelength;
3. speed.
The barrier changes the direction of propagation of the waves without changing their speed. Explain why the velocity of the waves has nevertheless changed.
The barrier is rotated through 10°, while the direction of the incident waves remains unchanged. Determine the two possible new values of the angle of incidence.
The waves are now directed along the normal towards the barrier. Determine:
1. the angle of incidence;
2. the angle of reflection;
3. the angle through which the direction of motion changes during reflection.

17 marks

Basic Refraction of Water Waves

Waves Crossing a Change in Water Depth

Straight water waves travel from a deep region into a shallow region of a ripple tank. The frequency of the source is 6.0 Hz.

In the deep region, the distance from the first wavefront to the seventh wavefront is 24 cm. The speed of the waves in the shallow region is 0.15 m s⁻¹.

Determine the number of wavelengths between the first and seventh wavefronts.
Calculate the wavelength of the waves in the deep region.
Calculate the speed of the waves in the deep region.
Determine the frequency of the waves in the shallow region.
Calculate the wavelength of the waves in the shallow region.
Compare the spacing of the wavefronts in the deep and shallow regions.
The waves enter the shallow region at an angle to the normal. State how their direction of propagation changes.
One end of each wavefront enters the shallow region before the other end. Explain how the difference in wave speed causes the wavefront to change direction.
The boundary is turned so that the waves travel along the normal when they reach it. State which of the following change as the waves enter the shallow region:
1. direction;
2. speed;
3. frequency;
4. wavelength.
A student states, “The wavelength decreases in shallow water because the source produces fewer waves per second.” Explain why the student’s statement is incorrect.

18 marks

Question 1

True. Six crests form five spaces between adjacent crests, so the distance from the first crest to the sixth crest is five wavelengths.
False. The direction of wave propagation is perpendicular to the wavefronts.
False. The floating object oscillates around a nearly fixed position. It is not carried continuously across the ripple tank with the wave.
False. Waves travelling at the same speed can have different frequencies if they have different wavelengths.
v = fλ
True. At unchanged water depth, the wave speed is unchanged. Since:
λ = v/f
doubling the frequency halves the wavelength.
True. The speed remains unchanged because the water depth is unchanged, but the direction changes. Velocity changes when direction changes.
True. The incident ray is perpendicular to the incident wavefront, while the normal is perpendicular to the barrier. Therefore, the angle between the wavefront and the barrier equals the angle between the ray and the normal.
False. In general, the reflected wavefronts are not parallel to the incident wavefronts. Their orientation changes during reflection.
True. The waves slow down in shallow water. Their frequency remains unchanged, so their wavelength decreases. At normal incidence, their direction does not change.
True. Waves speed up when travelling from shallow water into deep water. If they enter at an angle, they bend away from the normal. Their frequency remains equal to the source frequency.

Question 2

Nine crests form eight spaces between adjacent crests. There are eight complete wavelengths between the first and ninth crests.
8λ = 28.0 cm
λ = 28.0/8
λ = 3.50 cm
Converting to metres:
λ = 3.50 × 10⁻² m
λ = 0.0350 m
Wavelength = 0.0350 m.
The time between the first and second crests is one period. Therefore, the time from the first crest to the eleventh crest contains ten periods. There are ten complete periods.
1. T = 4.0/10
  T = 0.40 s
  Period = 0.40 s.
2. f = 1/T
  f = 1/0.40
  f = 2.5 Hz
  Frequency = 2.5 Hz.
3. v = fλ
  v = 2.5 × 0.0350
  v = 0.0875 m s⁻¹
  Wave speed = 0.0875 m s⁻¹.
The arrival of eleven crests does not mean that eleven complete waves pass the point. There are only ten time intervals between the arrival of the first crest and the arrival of the eleventh crest.
f = 10/4.0 = 2.5 Hz
The student has counted crests instead of the intervals between the crests.
The cork moves up and down, with possibly a small back-and-forth motion, as the water particles oscillate. It remains close to its original horizontal position and is not transported across the tank with the crests.
The disturbance travels across the water and can cause particles at later positions to oscillate. The wave transfers energy across the ripple tank without a continuous transfer of water in the direction of propagation.

Exam warning: When a distance or time is measured from the first crest to the nth crest, count the intervals between the crests, not the total number of crests.

Question 3

1. For wave P:
  f_P = 12/3.0
  f_P = 4.0 Hz
2. For wave Q:
  f_Q = 8/4.0
  f_Q = 2.0 Hz
1. For wave P:
  λ_P = 2.5 cm = 0.025 m
  
  v_P = fλ
  v_P = 4.0 × 0.025
  v_P = 0.10 m s⁻¹
2. For wave Q:
  λ_Q = 5.0 cm = 0.050 m
  
  v_Q = fλ
  v_Q = 2.0 × 0.050
  v_Q = 0.10 m s⁻¹
Both waves travel at 0.10 m s⁻¹.
The speed of a water wave depends on the conditions of the water, including its depth. P and Q travel through the same region, and the water depth is unchanged. They are therefore expected to travel at the same speed.
Wave P has twice the frequency of wave Q:
f_P = 2f_Q
Wave P has half the wavelength of wave Q:
λ_P = ½λ_Q
This allows both waves to have the same speed.
The new frequency of Q is 4.0 Hz. Its speed remains 0.10 m s⁻¹ because the water depth is unchanged.
λ = v/f
λ = 0.10/4.0
λ = 0.025 m
λ = 2.5 cm
New wavelength of Q = 2.5 cm.
The number of wavefronts passing a point each second gives the frequency, not the wave speed. Wave P has a higher frequency, but it also has a proportionally shorter wavelength.
v_P = 4.0 × 0.025 = 0.10 m s⁻¹
v_Q = 2.0 × 0.050 = 0.10 m s⁻¹
More wavefronts pass each second because P has a higher frequency, not because P travels faster.
Five successive wavefronts form four wavelength intervals.
4λ = 12 cm
λ = 3.0 cm
Wavelength = 3.0 cm.

Question 4

1. A ray is perpendicular to its wavefront.
  Angle between incident ray and incident wavefront = 90°
2. The normal is perpendicular to the barrier. The angle between the wavefront and barrier is therefore equal to the angle between the ray and normal.
  Angle of incidence = 38°
3. By the law of reflection:
  Angle of reflection = angle of incidence
  Angle between reflected ray and normal = 38°
4. The normal is at 90° to the barrier.
  Angle between reflected ray and barrier
  = 90° − 38°
  = 52°
The incident ray and reflected ray are on opposite sides of the normal.
Smaller angle = 38° + 38°
Smaller angle = 76°
The smaller angle between the two rays is 76°.
λ = 3.2 cm = 0.032 m

v = fλ
v = 5.0 × 0.032
v = 0.16 m s⁻¹
Speed of the incident waves = 0.16 m s⁻¹.
Reflection occurs in the same water and the source continues to produce waves at the same rate.
1. Frequency = 5.0 Hz.
2. Wavelength = 3.2 cm.
3. Speed = 0.16 m s⁻¹.
Velocity is a vector quantity. It depends on both speed and direction. Reflection changes the direction of propagation even though the magnitude of the velocity remains unchanged. The velocity changes because the direction changes.
Rotating the barrier through 10° also rotates its normal through 10°. Depending on the direction of rotation, the angle between the incident ray and the normal either decreases or increases by 10°.
New angle of incidence = 38° − 10° = 28°
or
New angle of incidence = 38° + 10° = 48°
The possible new angles of incidence are 28° and 48°.
1. The incident ray travels along the normal.
  Angle of incidence = 0°
2. By the law of reflection:
  Angle of reflection = 0°
3. The reflected wave travels back along the same path in the opposite direction.
  Change in direction = 180°

Exam warning: The angles of incidence and reflection are measured from the normal, not from the barrier.

Question 5

Seven wavefronts form six spaces between successive wavefronts. There are six wavelengths between the first and seventh wavefronts.
6λ = 24 cm
λ = 24/6
λ = 4.0 cm
In metres:
λ = 0.040 m
Wavelength in deep water = 4.0 cm or 0.040 m.
v = fλ
v = 6.0 × 0.040
v = 0.24 m s⁻¹
Speed in deep water = 0.24 m s⁻¹.
The frequency is determined by the source. Crossing the boundary does not change the rate at which complete waves are produced. Frequency in shallow water = 6.0 Hz.
λ = v/f
λ = 0.15/6.0
λ = 0.025 m

λ = 2.5 cm
Wavelength in shallow water = 0.025 m or 2.5 cm.
The wavefront spacing is equal to the wavelength.
Deep water spacing = 4.0 cm
Shallow water spacing = 2.5 cm
The wavefronts are closer together in the shallow region.
The waves slow down when they enter shallow water. When they enter at an angle, they bend towards the normal.
The end of the wavefront that enters the shallow region first slows down first. The other end remains in deep water briefly and continues moving at the higher speed. As a result, one end advances farther than the other, causing the wavefront to rotate. The direction of propagation is perpendicular to the wavefront, so the propagation direction also changes. The wave bends towards the normal because its speed decreases on entering shallow water.
1. Direction: unchanged. At normal incidence, the whole wavefront enters the shallow region at the same time.
2. Speed: decreases.
3. Frequency: unchanged.
4. Wavelength: decreases.
The source continues to produce 6.0 complete waves each second, so the frequency does not decrease. The wavelength decreases because the wave speed decreases while the frequency remains constant.
λ = v/f
The student has incorrectly attributed the shorter wavelength to a decrease in frequency. It is caused by the lower wave speed in shallow water.

Key idea: During refraction, the frequency is unchanged because it is fixed by the source. The speed and wavelength change when the water depth changes.