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Lo 7(2)

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Catherinee Lin

The document describes a physics problem where an object emits a sound at the surface of the sea. The sound wave has a wavelength of 1m in air and 4.5m in sea water, and the speed of sound in air is 340m/s. The echo is received 0.5 seconds later. Using the equations relating wavelength, frequency and speed, and that frequency does not change between mediums, the document calculates that the depth of the sea is 382.5m.

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Yiying Lin 15621148 LO 7/8 Short answer question: The sound produced by an object has a wavelength 1m in the air; the wave speed is 340 meters per second. It has a wavelength 4.5m in the sea. This object emits a sound at the surface of the sea, and the echo can be received after 0.5 second. Therefore, how deep is the sea is? Answer: use both v=了f and wave propagation to solve this question. In different mediums, same sound waves will have same frequency. Frequency= vair/了air In the sea: vsea=vair(了sea/了air)=1530m/s h = vsea(t/2)=1530x0.5x0.2 = 382.5m

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Lo 7(2)

  • 1. Yiying Lin 15621148 LO 7/8 Short answer question: The sound produced by an object has a wavelength 1m in the air; the wave speed is 340 meters per second. It has a wavelength 4.5m in the sea. This object emits a sound at the surface of the sea, and the echo can be received after 0.5 second. Therefore, how deep is the sea is? Answer: use both v=了f and wave propagation to solve this question. In different mediums, same sound waves will have same frequency. Frequency= vair/了air In the sea: vsea=vair(了sea/了air)=1530m/s h = vsea(t/2)=1530x0.5x0.2 = 382.5m