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A Level H2 Physics Waves Sound Light Quiz

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Questions

A-Level Physics H2 Quiz - Waves Sound Light

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ________ / 60

Duration: 75 minutes
Total Marks: 60

Instructions

Answer ALL questions.
Write your answers in the spaces provided.
The number of marks for each question is shown in brackets [ ].
Show all working clearly for calculation questions. Marks may be awarded for correct method even if the final answer is incorrect.
Non-programmable scientific calculators may be used.
Assume $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Multiple Choice [10 marks]

Each question in this section is worth 1 mark. Choose the ONE best answer.

1. A transverse wave travels along a string. Which of the following correctly describes the relationship between the direction of oscillation of particles in the string and the direction of energy transfer?

A. The particles oscillate perpendicular to the direction of energy transfer.
B. The particles oscillate parallel to the direction of energy transfer.
C. The particles do not oscillate; only the wave moves forward.
D. The particles oscillate in random directions.

Answer: ________ [1]

2. A progressive wave is described by the equation $y = 0.04 \sin(8\pi t - 0.4\pi x)$ , where $y$ is in metres, $t$ is in seconds, and $x$ is in metres. What is the frequency of the wave?

A. 2 Hz
B. 4 Hz
C. 8 Hz
D. 16 Hz

Answer: ________ [1]

3. Two coherent sources of light produce an interference pattern on a screen. At a point where the path difference from the two sources is $1.5\lambda$ , what is observed?

A. A bright fringe (constructive interference)
B. A dark fringe (destructive interference)
C. Neither bright nor dark — an intermediate intensity
D. No light at all

Answer: ________ [1]

4. A diffraction grating has 500 lines per mm. Light of wavelength 600 nm is incident normally on the grating. What is the angle of the first-order maximum?

A. $17.5°$
B. $20.5°$
C. $25.8°$
D. $30.0°$

Answer: ________ [1]

5. Which of the following is a necessary condition for two waves to produce a stable interference pattern?

A. The waves must have the same amplitude.
B. The waves must have the same frequency and a constant phase difference.
C. The waves must travel in opposite directions.
D. The waves must be longitudinal.

Answer: ________ [1]

6. A pipe open at both ends has a fundamental frequency of 256 Hz. The speed of sound in air is 340 m s $^{-1}$ . What is the length of the pipe?

A. 0.33 m
B. 0.66 m
C. 1.33 m
D. 2.66 m

Answer: ________ [1]

7. A wave has a speed of $3.0 \times 10^8$ m s $^{-1}$ and a frequency of $5.0 \times 10^{14}$ Hz. What is its wavelength?

A. 150 nm
B. 400 nm
C. 600 nm
D. 800 nm

Answer: ________ [1]

8. In a standing wave on a string fixed at both ends, the distance between two adjacent nodes is 0.40 m. What is the wavelength of the wave?

A. 0.20 m
B. 0.40 m
C. 0.80 m
D. 1.60 m

Answer: ________ [1]

9. A sound wave travels from air into water. Which of the following quantities remains unchanged?

A. Speed
B. Wavelength
C. Frequency
D. Amplitude

Answer: ________ [1]

10. Polarisation is a phenomenon that can be exhibited by:

A. Sound waves only
B. Longitudinal waves only
C. Transverse waves only
D. Both longitudinal and transverse waves

Answer: ________ [1]

Section B: Structured Questions [30 marks]

11. (a) Define the term displacement of a wave. [1]

(b) Define the term amplitude of a wave. [1]

(d) A water wave has a frequency of 2.0 Hz and a wavelength of 0.75 m. Calculate the speed of the wave. [2]

12. (a) State the principle of superposition of waves. [2]

(b) Two coherent sound waves of equal amplitude meet at a point. When the path difference between them is $2.5\lambda$ , determine whether constructive or destructive interference occurs. Explain your reasoning. [2]

13. A student sets up a demonstration of Young's double-slit experiment using monochromatic green light of wavelength 530 nm. The slit separation is 0.25 mm and the screen is placed 1.5 m from the slits.

(a) Calculate the fringe spacing $\Delta y$ on the screen. [3]

(b) The student replaces the green light with red light of wavelength 650 nm. State and explain what happens to the fringe spacing. [2]

14. The figure below shows a standing wave on a string of length 1.2 m fixed at both ends.

<image_placeholder> id: Q14-fig1 type: diagram linked_question: Q14 description: Standing wave on a string fixed at both ends, length 1.2 m, showing 3 loops (3 antinodes, 4 nodes including endpoints). The string has nodes at each end and two intermediate nodes, with antinodes at the midpoint of each segment. labels: Length L = 1.2 m, 3 loops visible, nodes marked at 0, 0.4, 0.8, 1.2 m, antinodes at 0.2, 0.6, 1.0 m values: L = 1.2 m, number of loops = 3, harmonic number n = 3 must_show: String fixed at both ends, 3 complete loops, nodes and antinodes clearly labelled, length dimension marked </image_placeholder>

(a) State the harmonic number shown in the diagram. [1]

(b) Calculate the wavelength of the standing wave. [2]

15. (a) Explain what is meant by diffraction of a wave. [2]

(b) A single slit of width 0.12 mm is illuminated with light of wavelength 580 nm. The diffraction pattern is observed on a screen 2.0 m away.

(i) Calculate the angle $\theta$ of the first minimum from the central maximum. [2]

(ii) Calculate the linear distance from the centre of the central maximum to the first minimum on the screen. [2]

16. (a) State two differences between a transverse wave and a longitudinal wave. [2]

(b) Explain how the phenomenon of polarisation provides evidence that light is a transverse wave. [2]

17. A source of sound emits waves of frequency 850 Hz. The speed of sound in air is 340 m s $^{-1}$ .

(a) Calculate the wavelength of the sound wave. [2]

(b) Two speakers connected to the same signal generator are placed 1.60 m apart and emit sound in phase. A detector is moved along a line parallel to the line joining the speakers, at a perpendicular distance of 3.0 m from the midpoint between the speakers. The first loud sound (constructive interference) is detected at a point 0.30 m from the central axis.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: Plan view of two-speaker interference setup. Two speakers S1 and S2 are 1.60 m apart on a horizontal line. A detection line is drawn parallel to the speaker line at a perpendicular distance of 3.0 m. A point P is marked on the detection line at 0.30 m from the central axis (perpendicular bisector). Distances from S1 to P and S2 to P are shown as r1 and r2 respectively. labels: S1 and S2 (speakers, 1.60 m apart), detection line at 3.0 m from midpoint, point P at 0.30 m from central axis, r1 = distance S1 to P, r2 = distance S2 to P values: d = 1.60 m (speaker separation), D = 3.0 m (perpendicular distance), y = 0.30 m (offset from central axis) must_show: Two speakers with separation, detection line, point P offset from centre, path difference r1 and r2 indicated </image_placeholder>

(i) Calculate the path difference between the waves arriving at point P from the two speakers. [3]

(ii) Determine the order of the maximum detected at point P. [1]

Section C: Data and Application Questions [20 marks]

18. A student investigates the relationship between the length of a closed pipe and its resonant frequencies. The student uses a tube closed at one end and a tuning fork of known frequency 512 Hz. The speed of sound is 340 m s $^{-1}$ .

(a) Explain why only odd harmonics are present in a pipe closed at one end. [3]

(b) Calculate the shortest length of the closed pipe that will resonate with the 512 Hz tuning fork. [2]

(c) The student also finds that the next resonant length is 0.50 m. Show that this is consistent with the theory for a closed pipe. [3]

19. The intensity $I$ of a wave is proportional to the square of its amplitude $A$ . A sound wave has an amplitude of $2.5 \times 10^{-6}$ m and an intensity of $1.2 \times 10^{-4}$ W m $^{-2}$ .

(a) State the relationship between intensity and amplitude, including the proportionality. [1]

(b) If the amplitude of the wave is doubled, calculate the new intensity. [2]

(c) A second sound wave of the same frequency but with a phase difference of $\pi$ radians is superimposed on the first wave. Both waves have the same amplitude. Describe the resulting wave and state the amplitude of the resultant. [3]

20. A microwave source emits waves of frequency $1.2 \times 10^{10}$ Hz. The waves pass through a diffraction grating with slit separation 2.5 cm.

(a) Calculate the wavelength of the microwaves. [2]

(b) Determine the angle at which the first-order diffraction maximum occurs. [3]

(c) A student claims that if the slit separation is reduced to 1.0 cm, no diffraction maxima will be observed. Evaluate this claim. [3]

END OF QUIZ

Answers

A-Level Physics H2 Quiz - Waves Sound Light

Answer Key and Marking Scheme

Section A: Multiple Choice [10 marks]

1. A [1]
Explanation: In a transverse wave, the particles oscillate perpendicular to the direction of energy transfer. This is the defining characteristic of a transverse wave. Sound waves in air are longitudinal (particles oscillate parallel to energy transfer), but the question refers to a wave on a string, which is transverse.

2. B [1]
Explanation: The general wave equation is $y = A \sin(\omega t - kx)$ . Comparing with $y = 0.04 \sin(8\pi t - 0.4\pi x)$ , we have $\omega = 8\pi$ rad s $^{-1}$ . Since $\omega = 2\pi f$ , we get $f = \frac{8\pi}{2\pi} = 4$ Hz.
Common mistake: Students may confuse $\omega$ with $f$ and select 8 Hz.

3. B [1]
Explanation: Destructive interference occurs when the path difference is a half-integer multiple of the wavelength: $(n + \frac{1}{2})\lambda$ where $n = 0, 1, 2, \ldots$ . Here, $1.5\lambda = (1 + \frac{1}{2})\lambda$ , so $n = 1$ , which gives a dark fringe.

4. A [1]
Explanation: Grating spacing $d = \frac{1}{500}$ mm $= 2.0 \times 10^{-6}$ m. Using $d \sin\theta = n\lambda$ for $n = 1$ :
$\sin\theta = \frac{\lambda}{d} = \frac{600 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.30$
$\theta = \sin^{-1}(0.30) = 17.5°$

5. B [1]
Explanation: For a stable (stationary) interference pattern, the two sources must be coherent — they must have the same frequency and maintain a constant phase difference. Equal amplitude is not strictly necessary (it only affects the contrast of the pattern), and the waves do not need to travel in opposite directions.

6. B [1]
Explanation: For a pipe open at both ends, the fundamental has $\lambda = 2L$ .
$v = f\lambda = f(2L)$ , so $L = \frac{v}{2f} = \frac{340}{2 \times 256} = \frac{340}{512} = 0.664$ m $\approx 0.66$ m.
Common mistake: Using $\lambda = 4L$ (which applies to a closed pipe) gives 0.33 m (option A).

7. C [1]
Explanation: $v = f\lambda$ , so $\lambda = \frac{v}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7}$ m $= 600$ nm. This is in the visible spectrum (orange-red light).

8. C [1]
Explanation: In a standing wave, the distance between two adjacent nodes is $\frac{\lambda}{2}$ . Given this distance is 0.40 m:
$\frac{\lambda}{2} = 0.40$ m, so $\lambda = 0.80$ m.

9. C [1]
Explanation: When a wave crosses a boundary between two media, the frequency is determined by the source and remains unchanged. The speed and wavelength both change (since $v = f\lambda$ ). Amplitude may also change due to partial reflection/transmission.

10. C [1]
Explanation: Polarisation occurs when the oscillation of a wave is restricted to a single plane. This is only possible for transverse waves, where the oscillation direction can be selected. Longitudinal waves oscillate along the direction of propagation and cannot be polarised.

Section B: Structured Questions [30 marks]

11. (a) Displacement of a wave is the distance and direction of a particle from its equilibrium (mean) position at a given instant. [1]
Note: Direction must be mentioned for full credit.

(b) Amplitude of a wave is the maximum displacement of a particle from its equilibrium position. [1]
Note: "Maximum displacement" is the key phrase.

(d) Using $v = f\lambda$ : [2]
$v = 2.0 \times 0.75$
$v = 1.5$ m s $^{-1}$
[1] for correct formula, [1] for correct answer with unit.

12. (a) The principle of superposition states that when two or more waves meet at a point, the resultant displacement at that point is the vector sum of the individual displacements due to each wave. [2]
Marking: [1] for "when two/more waves overlap/meet", [1] for "resultant displacement is the sum of individual displacements".

(b) The path difference is $2.5\lambda = (2 + \frac{1}{2})\lambda$ . [1]
Since the path difference is a half-integer multiple of $\lambda$ , destructive interference occurs. [1]
Note: The waves arrive out of phase (phase difference = $5\pi$ radians), so they cancel.

13. (a) Using the double-slit formula $\Delta y = \frac{\lambda D}{d}$ : [3]
$\Delta y = \frac{530 \times 10^{-9} \times 1.5}{0.25 \times 10^{-3}}$
$\Delta y = \frac{7.95 \times 10^{-7}}{2.5 \times 10^{-4}}$
$\Delta y = 3.18 \times 10^{-3}$ m $= 3.18$ mm
[1] for correct formula, [1] for correct substitution, [1] for correct answer with unit.

(b) The fringe spacing increases. [1]
Since $\Delta y = \frac{\lambda D}{d}$ and $\lambda_{\text{red}} > \lambda_{\text{green}}$ , the fringe spacing is proportional to wavelength, so it increases when red light is used. [1]
Note: Red light has a longer wavelength than green light.

14. (a) The diagram shows 3 loops, which corresponds to the third harmonic ( $n = 3$ ). [1]

(b) For a string fixed at both ends, $L = n \cdot \frac{\lambda}{2}$ where $n = 3$ : [2]
$1.2 = 3 \times \frac{\lambda}{2}$
$\lambda = \frac{2 \times 1.2}{3} = 0.80$ m
[1] for correct relationship, [1] for correct answer.

(c) Using $v = f\lambda$ : [2]
$v = 450 \times 0.80$
$v = 360$ m s $^{-1}$
[1] for correct formula/substitution, [1] for correct answer with unit.

15. (a) Diffraction is the spreading out of waves when they pass through a gap or around an obstacle. [1]
The effect is most noticeable when the size of the gap or obstacle is comparable to the wavelength of the wave. [1]
Note: Both the definition and the condition for significant diffraction are required.

(b) (i) For the first minimum in single-slit diffraction: $a \sin\theta = \lambda$ [2]
$\sin\theta = \frac{\lambda}{a} = \frac{580 \times 10^{-9}}{0.12 \times 10^{-3}} = 4.833 \times 10^{-3}$
$\theta = \sin^{-1}(4.833 \times 10^{-3}) = 0.277°$
[1] for correct formula and substitution, [1] for correct answer.

(ii) Linear distance: $y = D \tan\theta \approx D\sin\theta$ (small angle) [2]
$y = 2.0 \times 4.833 \times 10^{-3}$
$y = 9.67 \times 10^{-3}$ m $= 9.67$ mm
[1] for correct method, [1] for correct answer with unit.

16. (a) Any two of the following: [2]

In a transverse wave, particles oscillate perpendicular to the direction of wave travel; in a longitudinal wave, particles oscillate parallel to the direction of wave travel. [1]
Transverse waves can be polarised; longitudinal waves cannot. [1]
Transverse waves have crests and troughs; longitudinal waves have compressions and rarefactions. [1]
[1] each, any two valid differences.

(b) Polarisation occurs when the oscillations of a wave are restricted to a single plane. [1]
This phenomenon is only possible for transverse waves, because only transverse waves have oscillation directions that can be selected or filtered. [1]
Since light can be polarised (e.g. using a Polaroid filter), this demonstrates that light is a transverse wave. [1]
Note: Longitudinal waves cannot be polarised because their oscillation is always along the direction of propagation.

17. (a) Using $v = f\lambda$ : [2]
$\lambda = \frac{v}{f} = \frac{340}{850} = 0.40$ m
[1] for correct formula/substitution, [1] for correct answer with unit.

(b) (i) Using Pythagoras' theorem: [3]
$r_1 = \sqrt{(0.50)^2 + (3.0)^2} = \sqrt{0.25 + 9.0} = \sqrt{9.25} = 3.041$ m
$r_2 = \sqrt{(1.10)^2 + (3.0)^2} = \sqrt{1.21 + 9.0} = \sqrt{10.21} = 3.194$ m
Path difference $= r_2 - r_1 = 3.194 - 3.041 = 0.153$ m $\approx 0.15$ m
[1] for correct calculation of $r_1$ , [1] for correct calculation of $r_2$ , [1] for correct path difference.
Note: The distances from each speaker to point P are calculated using the geometry of the setup. S1 is 0.50 m from the central axis on one side, S2 is 1.10 m from the central axis on the other side (since speakers are 1.60 m apart, the midpoint is at 0.80 m from each speaker, so S1 is at 0.80 − 0.30 = 0.50 m and S2 is at 0.80 + 0.30 = 1.10 m from point P's perpendicular).

(ii) Order $n = \frac{\text{path difference}}{\lambda} = \frac{0.15}{0.40} = 0.375$
Since this is not an integer, this is not a maximum of exact constructive interference. However, the question states the first loud sound is detected, so we interpret this as the first maximum near the centre. The path difference of 0.15 m corresponds to approximately $0.375\lambda$ , which is closest to the zeroth order (central maximum at 0 path difference). Given the context of "first loud sound" away from centre, the order is $n = 0$ (central maximum) or the question may intend the nearest integer order.
Re-evaluation: If the first loud sound away from centre is detected, and path difference = 0.15 m = 0.375λ, this does not correspond to a perfect integer order. The question likely expects: order = path difference / λ = 0.15/0.40 ≈ 0.4, which rounds to order 0 (the central maximum is the zeroth order).
Answer: $n = 0$ [1]
Note: If the path difference were exactly $\lambda$ (0.40 m), it would be first order. The value 0.15 m suggests this is a point near the central maximum.

Section C: Data and Application Questions [20 marks]

18. (a) In a pipe closed at one end: [3]

The closed end must be a node (zero displacement, as the air cannot move at the closed end). [1]
The open end must be an antinode (maximum displacement, as the air is free to move). [1]
The simplest standing wave pattern has a node at the closed end and an antinode at the open end, giving $\frac{\lambda}{4} = L$ . The next possible pattern adds half a wavelength, giving $\frac{3\lambda}{4} = L$ , then $\frac{5\lambda}{4} = L$ , etc. Only odd multiples of $\frac{\lambda}{4}$ fit, so only odd harmonics ( $n = 1, 3, 5, \ldots$ ) are present. [1]

(b) For the fundamental (first harmonic) of a closed pipe: $L = \frac{\lambda}{4}$ [2]
$\lambda = \frac{v}{f} = \frac{340}{512} = 0.664$ m
$L = \frac{0.664}{4} = 0.166$ m $\approx 0.17$ m
[1] for correct formula, [1] for correct answer with unit.

(c) For a closed pipe, resonant lengths are $L = \frac{n\lambda}{4}$ where $n = 1, 3, 5, \ldots$ [3]
First resonance: $L_1 = \frac{\lambda}{4} = 0.166$ m (for $n = 1$ )
Next resonance: $L_2 = \frac{3\lambda}{4} = 3 \times 0.166 = 0.498$ m $\approx 0.50$ m (for $n = 3$ )
This is consistent with the theory, as the next resonant length should be $3 \times$ the first resonant length.
[1] for stating the relationship, [1] for correct calculation, [1] for confirming consistency with 0.50 m.

19. (a) $I \propto A^2$ , or $I = kA^2$ where $k$ is a constant of proportionality. [1]
Note: The intensity of a wave is directly proportional to the square of its amplitude.

(b) If amplitude is doubled: $A' = 2A$ [2]
$I' \propto (2A)^2 = 4A^2$
$I' = 4I = 4 \times 1.2 \times 10^{-4} = 4.8 \times 10^{-4}$ W m $^{-2}$
[1] for correct relationship, [1] for correct answer with unit.

(c) When two waves of the same frequency and amplitude but with a phase difference of $\pi$ radians (180°) are superimposed: [3]

The waves are exactly out of phase at every point.
By the principle of superposition, the resultant displacement at every point is the sum of the two individual displacements, which are equal in magnitude but opposite in direction.
Therefore, destructive interference occurs everywhere, and the resultant wave has zero amplitude.
[1] for stating the waves are out of phase, [1] for applying superposition, [1] for resultant amplitude = 0.

20. (a) Using $v = f\lambda$ where $v = c = 3.0 \times 10^8$ m s $^{-1}$ : [2]
$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{1.2 \times 10^{10}} = 2.5 \times 10^{-2}$ m $= 2.5$ cm
[1] for correct formula/substitution, [1] for correct answer with unit.

(b) Using the diffraction grating equation $d \sin\theta = n\lambda$ for $n = 1$ : [3]
$\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 2.5 \times 10^{-2}}{2.5 \times 10^{-2}} = 1.0$
$\theta = \sin^{-1}(1.0) = 90°$
[1] for correct formula, [1] for correct substitution, [1] for correct answer.
Note: The first-order maximum occurs at 90°, meaning the diffracted wave travels parallel to the grating surface.

(c) If $d$ is reduced to 1.0 cm: [3]
$\sin\theta = \frac{\lambda}{d} = \frac{2.5 \times 10^{-2}}{1.0 \times 10^{-2}} = 2.5$
Since $\sin\theta$ cannot exceed 1, no diffraction maximum is possible for any order.
The student's claim is correct. Reducing the slit separation below the wavelength means the condition $d \sin\theta = n\lambda$ cannot be satisfied for any real angle $\theta$ (since $\lambda/d > 1$ ).
[1] for calculating $\sin\theta = 2.5$ , [1] for noting that $\sin\theta > 1$ is impossible, [1] for concluding the claim is correct.

Total: 60 marks