A stationary submerged submarine tracks an approaching submarine with sonar. A s

Question

A stationary submerged submarine tracks an approaching submarine with sonar. A short pulse of ultrasound with a frequency of 400 kHz is sent toward the sub and returns 10 s later with an observed frequency of 412 kHz.

(a) What is the speed of the approaching sub? Take the speed of sound in seawater to be 1200 m/s.
(b) What is the approximate range of the approaching sub at the time of reflection?

Explanation / Answer

a) f = ((v+vr)/(v+vs)) f0 here f0 = 400 kHz; f = 412 kHz ; v=1200 ; vs=0 substituting these, 412 = (1200+vr)/(1200) * (400) --> vr = 36 m/s b) signal is received after 10 sec. Assuming signal took 5 sec to reach the other submarine and 5 more sec to come back ..... distace = time * velocity of sound = 5*1200 = 6000 m/s = 6 km/sec

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