S5 Physics Practice Paper: Waves, Sound And Optics (12 Questions, 100 Marks, 75 Minutes, With Answers)

S5 Physics Practice Paper: Waves, Sound and Optics (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Physics: Waves, Sound and Optics

12 questions, 100 marks, 75 minutes on wave speed, sound, refraction, total internal reflection, the Doppler effect, standing waves, double-slit interference and diffraction gratings.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 35 minutes on the wave and superposition questions in Section A, then use the remaining time for the sound and refraction calculations in Section B and the Doppler, standing-wave and interference questions in Section C.

Section A: Wave Fundamentals

Wave speed, wavelength and frequency, transverse and longitudinal waves, waves on a string, superposition and interference conditions

#	Question	Answer and marking notes
1	Distinguish between transverse and longitudinal waves, and define wavelength, frequency, amplitude and wave speed. State the relationship between wave speed, frequency and wavelength.	(a) In a transverse wave the vibration is perpendicular to the direction of travel, for example waves on a string and light waves (3 marks). (b) In a longitudinal wave the vibration is parallel to the direction of travel, for example sound waves (3 marks). (c) Wavelength is the distance between two consecutive points in phase, frequency is the number of vibrations per second, amplitude is the maximum displacement from equilibrium, and wave speed equals frequency times wavelength (2 marks).
2	A wave on a string has a frequency of 50 Hz and a wavelength of 0.8 m. (a) Find the wave speed. (b) Find the period. (c) Find the time for the wave to travel 12 m. (d) If the speed stays the same but the frequency rises to 100 Hz, find the new wavelength.	(a) The wave speed is 50 x 0.8 = 40 m/s (2 marks). (b) The period is 1/50 = 0.02 s (2 marks). (c) The time needed is 12/40 = 0.3 s (2 marks). (d) With the speed unchanged the new wavelength is 40/100 = 0.4 m (3 marks).
3	A stretched string has a linear density of 0.040 kg/m. (a) Find the wave speed when the tension is 64 N. (b) Find the new speed when the tension is raised to 256 N, and explain how wave speed depends on tension.	(a) The wave speed is the square root of 64/0.040, which is 40 m/s (4 marks). (b) The new speed is the square root of 256/0.040, which is 80 m/s; wave speed is proportional to the square root of the tension, so quadrupling the tension doubles the speed (4 marks).
4	Two transverse waves of the same frequency superpose at a point with amplitudes of 3 cm and 4 cm. (a) Find the resultant amplitude for in-phase and for antiphase superposition. (b) State the path-difference conditions for constructive and destructive interference. (c) Explain why the antiphase resultant amplitude is the difference of the two amplitudes.	(a) The resultant amplitude is 7 cm for in-phase superposition and 1 cm for antiphase superposition (3 marks). (b) Constructive interference needs a path difference of a whole number of wavelengths, and destructive interference needs an odd number of half wavelengths (3 marks). (c) In antiphase the displacements are opposite, so the maximum resultant displacement equals the difference of the amplitudes (2 marks).

Section B: Sound and Light

Echo distance, harmonics of open and closed pipes, refraction and Snell’s law, critical angle and total internal reflection

#	Question	Answer and marking notes
5	The speed of sound in air is about 340 m/s. (a) A person claps and hears the echo from a wall after 0.60 s; find the distance to the wall. (b) Thunder is heard 3.0 s after the lightning is seen; estimate the distance to the strike. (c) Explain why sound cannot travel through a vacuum.	(a) The sound travels to the wall and back, so the distance is 340 x 0.60 / 2 = 102 m (3 marks). (b) The distance is about 340 x 3.0 = 1020 m (3 marks). (c) Sound is a mechanical wave and needs a medium to transfer the vibration; a vacuum contains no particles, so sound cannot travel through it (3 marks).
6	A pipe open at both ends is 0.85 m long and the speed of sound in it is 340 m/s. (a) Find the fundamental frequency. (b) Find the second and third harmonics. (c) If one end is closed, find the new fundamental frequency.	(a) The fundamental frequency is 340 / (2 x 0.85) = 200 Hz (4 marks). (b) The second harmonic is 400 Hz and the third harmonic is 600 Hz (2 marks). (c) With one end closed the fundamental is 340 / (4 x 0.85) = 100 Hz (2 marks).
7	Light passes from air (refractive index 1.00) into glass (refractive index 1.50) at an angle of incidence of 30 degrees. (a) Find the angle of refraction. (b) Find the critical angle for light travelling from glass to air. (c) Find the speed of light in the glass.	(a) By Snell’s law, 1.00 x sin 30 degrees = 1.50 x sin r, so sin r = 0.3333 and the angle of refraction is about 19.5 degrees (3 marks). (b) sin c = 1/1.50 = 0.6667, so the critical angle is about 41.8 degrees (3 marks). (c) The speed in glass is 3.0 x 10^8 / 1.50 = 2.0 x 10^8 m/s (2 marks).
8	Explain the two conditions required for total internal reflection in an optical fibre, and explain why the cladding must have a lower refractive index than the core.	(a) Light must travel from an optically denser medium to an optically less dense medium, that is from the core to the cladding (4 marks). (b) The angle of incidence must exceed the critical angle; because the cladding has a lower refractive index the critical angle at the boundary is well below 90 degrees, so the light is totally internally reflected repeatedly along the core (4 marks).

Section C: Wave Phenomena and Applications

Doppler effect, standing waves and nodes, double-slit fringe spacing, diffraction grating and maximum order

#	Question	Answer and marking notes
9	An ambulance sounding an 800 Hz siren moves towards a stationary observer at 30 m/s. The speed of sound is 340 m/s. (a) Find the frequency heard by the observer. (b) Find the frequency heard after the ambulance has passed and is moving away. (c) Explain the cause of the Doppler effect.	(a) The observed frequency is 800 x 340 / (340 – 30) = 877 Hz (4 marks). (b) As the ambulance moves away the frequency is 800 x 340 / (340 + 30), about 735 Hz (3 marks). (c) A moving source compresses the wavefronts ahead of it and stretches those behind, so the wavelength changes while the wave speed stays the same, and the observed frequency therefore changes (2 marks).
10	A string 1.2 m long is fixed at both ends and the wave speed on it is 48 m/s. (a) Find the fundamental frequency and its wavelength. (b) Find the third harmonic. (c) Find the number of nodes of the third harmonic.	(a) The fundamental wavelength is 2 x 1.2 = 2.4 m and the fundamental frequency is 48 / 2.4 = 20 Hz (3 marks). (b) The third harmonic is 3 x 20 = 60 Hz (3 marks). (c) The third harmonic has 4 nodes (2 marks).
11	In a double-slit experiment the slit separation is 3.0 x 10^-4 m, the screen distance is 2.0 m and the wavelength is 6.0 x 10^-7 m. (a) Find the fringe spacing. (b) Find the distance between the first and the fourth bright fringes. (c) Explain how the fringe spacing changes if red light is used.	(a) The fringe spacing is 6.0 x 10^-7 x 2.0 / (3.0 x 10^-4) = 4.0 x 10^-3 m, that is 4.0 mm (4 marks). (b) The distance is three fringe spacings, 12 mm (3 marks). (c) Red light has a longer wavelength and the fringe spacing is proportional to wavelength, so the spacing becomes larger (2 marks).
12	A diffraction grating has 500 lines per millimetre and is illuminated by light of wavelength 6.0 x 10^-7 m. (a) Find the grating spacing and the first-order diffraction angle. (b) Find the second-order angle and state the highest order that can be observed.	(a) The grating spacing is 1/500 mm = 2.0 x 10^-6 m; sin theta = 0.30, so the first-order angle is about 17.5 degrees (4 marks). (b) sin theta = 0.60, so the second-order angle is about 36.9 degrees; sin theta cannot exceed 1, so the third order is still observable and the fourth is not, giving a maximum order of 3 (4 marks).

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