Free Waves questions for AS Level Physics (9702), part of Waves, electricity & particles. Each question has the answer and the marking points an examiner would look for.
Generate a full Waves worksheet →All Waves, electricity & particles questionsAt a distance r from a point source, the intensity of a wave is I. The amplitude of the source is doubled and the detector is moved to a distance 4r from the source.
What is the new intensity?
Answer: D
Which statement about all electromagnetic waves in a vacuum is correct?
Answer: D
Which statement describes intensity?
Answer: B
A sound wave is displayed on an oscilloscope. The time-base is 1 ms per division and the Y-gain is 0.20 V per division. One complete wave occupies 4 divisions and the peak height is 3.0 divisions.
Calculate the frequency of the sound.
State the amplitude of the signal.
The speed of sound is 340 m s−1. Calculate the wavelength.
The amplitude is halved. State the effect on the intensity.
The source moves towards a stationary observer at 20 m s−1. Calculate the observed frequency.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 250 Hz T = 4 × 1 ms = 4.00 ms C1 f = 250 Hz A1 | 2 |
| (b) | 0.60 V 0.60 V B1 | 1 |
| (c) | 1.36 m 1.36 m B1 | 1 |
| (d) | quarter intensity ∝ amplitude2, so it becomes a quarter B1 | 1 |
| (e) | 266 Hz fo = fsv/(v − vs) C1 = 266 Hz A1 | 2 |
The electromagnetic spectrum includes waves with a very wide range of wavelengths.
State the orders of magnitude of the wavelengths of X-rays and microwaves.
State one property common to all electromagnetic waves.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 10−10 m; 10−2 m X-rays about 10−10 m (10−12 to 10−8) B1 microwaves about 10−2 m (10−3 to 10−1) B1 | 2 |
| (b) | speed c in vacuum; transverse travel at 3.00 × 108 m s−1 in a vacuum / transverse / can be polarised B1 | 1 |
A car horn emits sound of frequency 500 Hz. The car moves at 20 m s−1 directly towards, and then away from, a stationary observer. The speed of sound is 340 m s−1.
State the difference between a transverse and a longitudinal wave.
Calculate the wavelength of the sound emitted when the car is stationary.
Calculate the frequency heard by the observer as the car approaches.
Calculate the change in observed frequency as the car passes the observer.
Explain, in terms of wavefronts, why the observed frequency is higher as the car approaches.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | transverse: oscillations perpendicular to energy transfer; longitudinal: parallel transverse: particle vibrations perpendicular to the direction of energy transfer B1 longitudinal: vibrations parallel to it B1 | 2 |
| (b) | 0.680 m 340/500 = 0.680 m B1 | 1 |
| (c) | 531 Hz fo = fsv/(v − vs) C1 531 Hz A1 | 2 |
| (d) | 59.0 Hz receding: 472 Hz C1 change = 59.0 Hz A1 | 2 |
| (e) | source moves towards the waves it has emitted; wavefronts closer, shorter wavelength the source moves forwards between emitting successive wavefronts B1 so the wavefronts are closer together in front of the car (shorter wavelength) and arrive more often B1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in