A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward one another. The second submarine is moving

Question

A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1 400 Hz. The speed of sound in the water is 1 533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward one another. 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 detected by an observer riding on sub B as the subs recede from each other?

Accepted Answer

(A) We use Equation 17.13 to find the Doppler-shifted frequency.
As the two submarines approach each other, the observer in sub B hears the frequency

[latex] f^{\prime}=\left(\frac{v+v_{O}}{v-v_{S}}\right) f [/latex] (17.13)

[latex] f^{\prime}=\left(\frac{v+v_{O}}{v-v_{S}}\right) f [/latex].

[latex] =\left(\frac{1533 m / s +(+9.00 m / s )}{1539 m / s -(+8.00 m / s )}\right)(1400 Hz )=1416 Hz [/latex].

(B) As the two submarines recede from each other, the observer in sub B hears the frequency

[latex] f^{\prime}=\left(\frac{v+v_{O}}{v-v_{S}}\right) f [/latex].

[latex] =\left(\frac{1533 m / s +(-9.00 m / s )}{1533 m / s -(-8.00 m / s )}\right)(1400 Hz )=1385 Hz [/latex].

Question 17.6: A submarine (sub A) travels through water at a speed of 8.00...

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