Question

A submarine moves away from an undersea vertical cliff at a speed v_s = 10.0 m/s. It emits a sound wave at a frequency f_e = 20,000. Hz, which travels to the cliff and reflects off it at a frequency f_r. The reflected wave travels back to the submarine and generates beats with the emitted wave ( f_e ) that are detected by the sub’s sensors. For a speed of sound in seawater v = 1,174 m/s, the frequency of the beats are (Enter your answer in units of Hz. Only enter the number using 3 significant figures.

Answer 1

a.)

from Doppler's effect, when Submarine (Source) is moving away from cliff (Stationary observer)

f' = f*(V + Vo)/(V - Vs)

here, f= original frequency = 20000 Hz

V = velocity of sound wave in seawater = 1174 m/s

Vo = velocity of observer(cliff) = 0 m/s

Vs = velocity of source(submarine) = +10 m/s

So, frequency detected bt cliff = f' = 20000*(1174 + 0)/(1174 + 10.0)

f' = 19831.08 Hz = frequency detetced by cliff

now cliff reflects back the wave, So it acts as a source of waves with frequency of 19831.08 Hz

then, Frequency detect by submarine will be, when source is stationary and observer is moving away from the source:

f'' = f'*(V - Vo)/(V - Vs)

here, Vo = velocity of submarine = 10.0 m/s

Vs = velocity of cliff = 0 m/s

So, f'' = 19831.08*(1174 - 10.0)/(1174 - 0)

f'' = 19662.16 Hz = reflected frequency measured by submarine

So beat frequency will be:

df = f - f''

df = 20000 - 19662.16

df = 337.84

In three significant figures:

df = beat frequency = 338 Hz

Let me know if you've any query.