O-LevelPhysicsWaves5054/21 · Oct/Nov 2014 · Q54 Marks
The device shown in Fig. 5.1 uses the reflection of ultrasound to measure distances. [Figure 5.1]
- (a)State what is meant by ultrasound.[2]
- (b)Fig. 5.2 shows a builder using the ultrasound device to measure the width of a room. [Figure 5.2] The ultrasound device is placed against one wall and it emits an ultrasound wave that reflects back from the opposite wall. The time between sending out the ultrasound wave and receiving the reflection is 0.030 s. The speed of ultrasound in air is 340 m/s. Calculate the distance between the device and the opposite wall.[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 (Oct/Nov 2014). The mark or mark type is shown in brackets.
- [2] (a) longitudinal/pressure/sound (wave) or compressions and rarefactions
- (frequency) greater than 15 – 25kHz/above limits of audibility
- [2] (b) (x = ) vt/2 or 340 × 0.030/2 or 340 × 0.015 or 10.2
- 5.1 m
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