OxfordAQA International AS and A-level Physics: Progressive waves, longitudinal and transverse waves, polarisation (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow sections 3.5.3 and 3.5.4 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, which is shared AS and A-level content. Take the speed of electromagnetic waves in a vacuum as 3.00 × 10⁸ m s⁻¹. Learn the content from the study guide or the revision notes. The course hub and checklist list the rest of the course.

Questions

1. A buoy makes 24 complete oscillations in 60 s as sea waves pass. (a) Calculate the frequency and the period of the oscillation. [2] (b) The wavelength is 9.0 m. Calculate the speed of the waves. [2]

2. A sound wave of frequency 420 Hz travels at 336 m s⁻¹ in air. (a) Calculate its wavelength. [2] (b) Calculate the phase difference between two points 0.50 m apart along the wave, in degrees and in radians. [3]

3. The table shows the displacement y of one particle in a wave at 1.0 ms intervals.

t / ms	0	1.0	2.0	3.0	4.0	5.0	6.0	7.0	8.0
y / mm	0	2.1	3.0	2.1	0	-2.1	-3.0	-2.1	0

(a) Determine the amplitude, the period and the frequency. [3] (b) A second particle Q further along the wave repeats this motion 2.0 ms later. Calculate the phase difference in degrees and state the time at which Q first reaches maximum displacement. [3]

4. (a) Compare the direction of oscillation in a transverse wave and in a longitudinal wave, relative to the way the wave carries energy. [2] (b) Classify each of the following as transverse or longitudinal: light, sound, ultrasound, waves on a stretched string. [2]

5. A space probe 2.25 × 10¹¹ m from Earth sends a signal of frequency 8.4 GHz. (a) Calculate the time the signal takes to reach Earth. [2] (b) Calculate the wavelength of the signal. [2] (c) A beam of yellow light of wavelength 590 nm is sent along the same path. State its travel time and calculate its frequency. [2]

6. Ultrasound pulses are used to examine a kidney. Take the speed of ultrasound in tissue as 1540 m s⁻¹. Echoes from the front and back surfaces of the kidney return 26 μs and 104 μs after the pulse is emitted. (a) Explain why the depth is half of speed × time delay. [1] (b) Calculate the depth of each surface and the thickness of the kidney. [3] (c) A gel is spread between the probe and the skin. Explain why. [2] (d) The ultrasound frequency is 2.5 MHz. Calculate its wavelength in tissue. [1]

7. Light from a lamp passes through Polaroid A and then through Polaroid B. B is rotated slowly through 360° about the beam. (a) Describe how the intensity of the light leaving B changes. [3] (b) Explain why this is evidence that light is a transverse wave. [3]

8. A vertical rod aerial transmits radio waves with the electric field oscillating vertically. (a) State and explain the orientation of a rod receiving aerial that gives the strongest signal. [2] (b) State the signal when the receiving rod is horizontal, and explain. [1] (c) Rotating a microphone about the line to a loudspeaker does not change the sound heard. Explain why. [1]

9. Ground waves from an earthquake include P-waves (longitudinal, speed 8.0 km s⁻¹) and S-waves (transverse, speed 4.5 km s⁻¹). A station records the S-wave 60 s after the P-wave. (a) State the direction of the ground oscillation relative to the direction of travel for each type of wave. [2] (b) Calculate the distance from the earthquake to the station. [3] (c) An S-wave has frequency 0.50 Hz. Calculate its wavelength. [2]

10. A wave of frequency 15 Hz travels at 6.0 m s⁻¹ along a string from point A to point B, 0.90 m apart. (a) Calculate the wavelength. [2] (b) Calculate the phase difference between A and B as a fraction of a cycle and in radians, and state which point lags. [3] (c) At an instant when A has its maximum positive displacement, state the displacement of B and the direction in which B is moving. Justify your answer. [3] (d) Calculate the time the wave takes to travel from A to B, and express it in periods. [2]

Answers

1. (a) f = 24/60 = 0.40 Hz [1]; T = 1/f = 2.5 s [1] (b) c = fλ = 0.40 × 9.0 [1] = 3.6 m s⁻¹ [1] Examiner insight: write the equation before substituting, and carry the unrounded frequency into part (b) rather than a rounded value.

2. (a) λ = c/f [1] = 336/420 = 0.80 m [1] (b) 0.50/0.80 = 0.625 cycle [1]; × 360 = 225° [1]; × 2π = 5π/4 rad (3.93 rad) [1] Examiner insight: a phase difference should always be built from separation divided by wavelength. A bare angle with no working leaves nothing to credit if it is wrong.

3. (a) Amplitude = 3.0 mm [1]; period = 8.0 ms (back to zero going the same way) [1]; f = 1/(8.0 × 10⁻³) = 125 Hz [1] (b) 2.0/8.0 = 0.25 cycle [1]; = 90° [1]; P peaks at 2.0 ms so Q peaks at 4.0 ms [1] Examiner insight: read the period from one full repeat of the pattern, and do not stop at the first zero, which is only half a cycle.

4. (a) Transverse: oscillations at 90° to the way the energy travels [1]; longitudinal: oscillations along that line [1] (b) Sound and ultrasound are longitudinal [1]; light and waves on a string are transverse [1] Examiner insight: give the comparison in terms of the way the energy travels. “Up and down” is not a reliable description.

5. (a) t = d/c [1] = 2.25 × 10¹¹ / 3.00 × 10⁸ = 750 s [1] (b) λ = c/f [1] = 3.00 × 10⁸ / 8.4 × 10⁹ = 0.036 m (3.6 cm) [1] (c) Same travel time, 750 s, because every electromagnetic wave has one speed in a vacuum [1]; f = c/λ = 3.00 × 10⁸ / 590 × 10⁻⁹ = 5.1 × 10¹⁴ Hz [1] Examiner insight: convert GHz and nm before substituting. Powers of ten are the usual place for errors here.

6. (a) The pulse travels to the surface and back, so the distance travelled is twice the depth [1] (b) Front: 1540 × 26 × 10⁻⁶ / 2 = 0.020 m [1]; back: 1540 × 104 × 10⁻⁶ / 2 = 0.080 m [1]; thickness = 0.060 m (6.0 cm) [1] (c) The gel removes the air between probe and skin [1]; otherwise most of the ultrasound would be reflected at the air-skin boundary and little would enter the body [1] (d) λ = 1540 / 2.5 × 10⁶ = 6.2 × 10⁻⁴ m [1] Examiner insight: in “explain” questions about the gel, name the boundary where reflection happens. “It helps the ultrasound get in” is too vague.

7. (a) The intensity varies between a maximum and a minimum as B rotates [1]; there are two positions of maximum and two of minimum (almost zero) in each full turn [1]; the minima are 90° from the maxima, when the axes are crossed [1] (b) Polaroid A allows only oscillations along its transmission axis, so light leaving A is plane-polarised [1]; B transmits only the component along its own axis, so no light passes when the axes are at 90° [1]; this needs oscillations at right angles to the direction of travel, which longitudinal waves do not have, because rotating a filter would make no difference to them [1] Examiner insight: link the observation to the direction of oscillation. A description of the pattern alone does not explain why it implies transverse waves.

8. (a) Vertical, parallel to the transmitter [1]; a rod responds to the component of the electric field along its length, which is greatest when parallel [1] (b) Almost no signal [1], because the electric field has almost no component along a horizontal rod (c) Sound is longitudinal, so it cannot be polarised and has no direction at right angles to the line of travel for the microphone to select [1] Examiner insight: use “component along the rod” in the explanation. “The signal is blocked” says what happens but not why.

9. (a) P-wave: parallel to the direction of travel [1]; S-wave: perpendicular to the direction of travel [1] (b) Times: d/8.0 and d/4.5 (d in km) [1]; d/4.5 - d/8.0 = 60 [1]; d = 60 / 0.09722 = 617 km [1] (c) λ = c/f = 4.5/0.50 [1] = 9.0 km [1] Examiner insight: set the time difference as the later arrival minus the earlier, and keep units consistent (km and km s⁻¹).

10. (a) λ = c/f [1] = 6.0/15 = 0.40 m [1] (b) 0.90/0.40 = 2.25 wavelengths [1]; phase difference 0.25 cycle = π/2 rad [1]; B lags A because the wave travels from A to B [1] (c) B lags by a quarter cycle, so it has the displacement A had a quarter cycle earlier [1]; a quarter cycle ago A was at equilibrium, moving upward towards its maximum [1]; so B has zero displacement and is moving upward [1] (d) t = 0.90/6.0 = 0.15 s [1]; T = 1/15 = 0.0667 s, so t = 0.15/0.0667 = 2.25 periods [1] Examiner insight: remove whole cycles before reasoning about the motion. Only the 0.25 cycle left over affects the answer in part (c).

Where marks are usually lost

• Using a rounded intermediate value, then giving a final answer to the wrong number of significant figures.
• Forgetting prefixes: GHz, MHz, μs and nm.
• Leaving out the factor of 2 for the there-and-back journey of an ultrasound pulse.
• Describing particle motion as moving along with the wave.
• Reading the period from half a cycle on a displacement-time graph.
• Giving a phase difference without saying which point leads or lags when the question asks for it.
• Explaining the Polaroid observation without referring to the transmission axis or the direction of oscillation.
• Writing “polarisation shows light is a wave” instead of a transverse wave.

Next steps

• Review the revision notes and the study guide
• Check the course hub and the checklist
• Take the free diagnostics: all free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.5.3 and 3.5.4.