AS Physics: Waves — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Waves revision notes

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Questions

1. Define amplitude, wavelength, frequency and period. [4]

2. Distinguish between transverse and longitudinal waves, giving one example of each. [3]

3. Explain why only transverse waves can be polarised, and state what this shows about light. [3]

4. A wave has frequency 250 Hz and travels at 340 m s⁻¹.

(a) Calculate the wavelength. [2] (b) Calculate the period. [2] (c) Two points on the wave are 0.34 m apart. Calculate their phase difference in degrees. [3]

5. State the regions of the electromagnetic spectrum in order of increasing wavelength, and give one use of each of any three. [6]

6. Explain what is meant by intensity, and state how it varies with amplitude and with distance from a point source. [3]

7. A source of sound of frequency 480 Hz moves towards a stationary observer.

(a) State and explain the change in the frequency heard. [3] (b) State what happens after the source passes the observer. [1]

8. Plane-polarised light of intensity 800 W m⁻² passes through a polarising filter oriented at 40° to the plane of polarisation.

(a) State the formula (Malus’s law) that gives the transmitted intensity. [1] (b) Calculate the transmitted intensity. [3]

9. An ambulance siren emits sound of frequency 600 Hz and moves at 30 m s⁻¹. The speed of sound in air is 340 m s⁻¹.

(a) Calculate the frequency heard by a stationary observer as the ambulance approaches. [3] (b) Calculate the frequency heard by the same observer once the ambulance has passed and is moving away at the same speed. [2]

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Answers

1. Amplitude — the maximum displacement from the equilibrium position [1]. Wavelength — the distance between two adjacent points in phase [1]. Frequency — the number of complete oscillations per second [1]. Period — the time for one complete oscillation [1].

2. Transverse — oscillations are perpendicular to the direction of energy travel [1], e.g. light [1]. Longitudinal — oscillations are parallel to the direction of travel, e.g. sound [1].

3. Only transverse waves have oscillations in more than one plane perpendicular to travel [1], so those oscillations can be restricted to a single plane [1]. Since light can be polarised, light must be transverse [1].

4. (a) λ = v ÷ f = 340 ÷ 250 [1] = 1.36 m [1]. (b) T = 1 ÷ f = 1 ÷ 250 [1] = 4.0 × 10⁻³ s [1]. (c) 0.34 ÷ 1.36 = 0.25 wavelengths [1] [1] Phase difference = 0.25 × 360 = 90° [1].

5. Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma — in order of increasing wavelength this is reversed: gamma, X-ray, ultraviolet, visible, infrared, microwave, radio [1] [1] [1]. Uses (any three): radio — broadcasting [1]; microwave — cooking or satellite communication [1]; X-ray — medical imaging [1].

6. Intensity is the power per unit area [1]. It is proportional to the square of the amplitude [1], and for an ideal isotropic point source with negligible absorption, it obeys an inverse square law with distance [1].

7. (a) The frequency heard increases [1]. Each successive wavefront is emitted from a position closer to the observer [1], so the wavefronts arrive more frequently and the observed wavelength is shortened [1]. (b) The frequency heard drops below the source frequency [1].

8. (a) I = I₀cos²θ [1], where I₀ is the incident intensity and θ is the angle between the filter’s transmission plane and the light’s plane of polarisation. (b) I = 800 × cos²(40°) [1] = 800 × 0.587 [1] = 470 W m⁻² [1].

9. (a) Approaching, so use the minus sign in the denominator [1]: f = f₀ × v ÷ (v − v_s) = 600 × 340 ÷ (340 − 30) = 600 × 340 ÷ 310 [1] = 658 Hz [1]. (b) Receding, so use the plus sign [1]: f = 600 × 340 ÷ (340 + 30) = 600 × 340 ÷ 370 = 551 Hz [1].

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Where marks are usually lost

• Saying sound can be polarised.
• Giving the EM spectrum in the wrong direction when “increasing wavelength” is specified.
• Saying intensity is proportional to amplitude rather than amplitude squared.
• Explaining the Doppler effect as the source “pushing” the waves.
• Applying Malus’s law to unpolarised light — the cos²θ relationship only applies once light is already plane-polarised; this specification only requires the already-polarised-light case.
• Using the wrong sign in the Doppler denominator — approaching always makes the denominator smaller (v − v_s), which is what raises the frequency; mixing the signs up gives an answer on the wrong side of the source frequency entirely.
• Forgetting that the source frequency itself never changes in the Doppler effect — only the frequency heard by the observer changes, because of how the wavefronts bunch together or spread out.

A useful way to keep the two calculation-heavy parts of this topic straight: Malus’s law is about what fraction of an already-polarised beam gets through a second filter at an angle, while the Doppler equation is about how relative motion compresses or stretches the wavefronts between source and observer — different physical mechanisms, but both frequently tested as two-part numerical questions rather than pure definitions. See the Waves revision notes for the underlying formulae and exam traps in condensed form.