Pearson Edexcel International GCSE (9–1) · PHYSICS · Section
Describing waves
17 practice questions
Practice questions
11–17 of 17
Question 11
2 marks · Normal
Answer and see the solution →State what is meant by the time period of a wave and explain how it relates to frequency, including the correct formula.
Question 12
3 marks · Normal
Answer and see the solution →Explain, with reference to energy transfer, how a wave can move along a spring without any part of the spring being carried along with the wave.
Question 13
3 marks · Normal
Answer and see the solution →Describe one example each of a transverse wave and a longitudinal wave, and for each, state clearly how the particles vibrate compared to the direction of wave travel.
Question 14
3 marks · Normal
Answer and see the solution →A wave has a time period of 0.02 seconds. Calculate its frequency.
Question 15
4 marks · Normal
Answer and see the solution →A student makes waves on the surface of a pool by repeatedly dipping her finger up and down in the water.
(a)[1 mark]State what is meant by a transverse wave.
(b)[2 marks]Describe, with reference to this situation, how the direction of vibration of the water relates to the direction in which the wave moves.
(c)[1 mark]Name another example of a transverse wave (other than water waves).
Question 16
4 marks · Normal
Answer and see the solution →The following diagram shows a wave travelling along a rope.
Wave on a rope (a)[2 marks]Label the wavelength and amplitude on the diagram.
(b)[1 mark]Which point on this wave shows the maximum positive displacement from the equilibrium position? State its letter.
(c)[1 mark]What is the name of this maximum point?
Question 17
4 marks · Normal
Answer and see the solution →A loudspeaker produces a sound wave with a frequency of 450 Hz. Each wave has a time period of 0.0022 seconds.
(a)[1 mark]State the equation relating frequency and time period for a wave.
(b)[2 marks]Using the equation, calculate the frequency expected if the time period is 0.0022 s.
(c)[1 mark]Suggest why the calculated frequency and measured frequency might not match exactly.