Edexcel IGCSEPhysicsGeneral Applications10 Marks
A student investigates the speed of sound.
- (a)Figure 10 shows the student making a sound at a distance from a wall. [Figure 10 shows a student clapping in front of a wall, with a sound wave travelling to the wall and reflecting back as an echo. The distance between the student and the wall is labeled 'd'.][3]
- (a)(i)State the relationship between speed, distance and time.[1]
- (a)(ii)Explain why the student must stand a large distance from the wall to make an accurate measurement.[2]
- (b)The student stands 250 m from the wall and claps. He hears the echo 1.5 s later. Calculate the speed of sound.[3]
- (c)The student then uses an oscilloscope and two microphones to measure the speed of sound. Figure 11 shows the oscilloscope trace for two microphones placed 0.50 m apart. [Figure 11 shows an oscilloscope display with two waveforms, representing the sound detected by two microphones. The horizontal axis is time, and the vertical axis is amplitude. The waveforms are identical but shifted in time. The time difference between the peaks of the two waveforms is labeled 'time difference = 1.5 divisions'. The time base setting is 0.5 ms/division.] (i) Calculate the time difference between the sound reaching the two microphones.[2]
- (c)(ii)Calculate the speed of sound.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme. The mark or mark type is shown in brackets.
- Question 19(a)(i)
- speed = distance / time / v = d / t (1)
- Question 19(a)(ii)
- to allow enough time between the clap and echo (1)
- so that the time can be measured more accurately (1)
- Allow to overcome reaction time / human error
- Question 19(b)
- distance = 2 × 250 = 500 (m) (1)
- substitution into v = d/t (1)
- 500 / 1.5 = 333 (m/s) (1)
- Accept 333.3 (m/s)
- Correct answer without working gains full marks.
- Allow 250 / 1.5 for 1 mark if distance not doubled.
- Question 19(c)(i)
- time difference = 1.5 × 0.5 (ms) (1)
- 0.75 (ms) / 0.00075 (s) (1)
- Correct answer without working gains full marks.
- Question 19(c)(ii)
- v = d/t (1)
- 0.50 / 0.00075 = 667 (m/s) (1)
- Accept 666.6 (m/s)
- Correct answer without working gains full marks.
- Allow ecf from (c)(i)
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