I need your help plz! A submarine (sub A) travels through water at a speed of 8.

Question

I need your help plz!

A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. a) What frequency is detected by an observer riding on sub B as the subs approach each other? b) The subs barely miss each other and pass. What frequency is directed by an observer riding on a sub B as the si.bs recede from each other?

Explanation / Answer

Va = speed of submarine A = 8 m/s

Vb = speed of submarine B = 9 m/s    towards A

Vs = speed of sound in water = 1533 m/s

f = frequency emitted by source A = 1400 Hz

a)

since source A and observer B are moving towards each other , from doppler's formula , the observed frequency is given as ::

f' = (Vs + Vb) (f) / (Vs - Va)

f' = (Vs + Vb) (f) / (Vs - Va)

f' = (1533 + 9) (1400) / (1533 - 8)

f' = 1415.61 Hz

b)

now the source A and observer B move away from each other

so

f' = (Vs - Vb) (f) / (Vs + Va)

f' = (1533 - 9) (1400) / (1533 + 8)

f' = 1384.56 Hz

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