Answer :
Answer:
a) The frequency detected by the submarine B, f' = 1595 Hz
b) The frequency of the reflected wave detected by the A is, f'' = 6021 Hz
Explanation:
Given data,
The speed of submarine A, v = 15.8 m/s
The speed of submarine B, v' = 8.63 m/s
The SONAR wave travels at a speed of V = 1522 m/s
The formula for Doppler effect is,
[tex]f' = \frac{V+v'}{V-v} f[/tex]
Substituting the given values above,
[tex]f' = \frac{1522+8.63}{1522-15.8} 1570[/tex]
f' = 1595 Hz
The reflected SONAR has the frequency of, f' = 1595 Hz
Therefore, the frequency of the reflected wave at A is given by the formula,
[tex]f'' = \frac{V+v}{V-v'} f'[/tex]
[tex]f'' = \frac{1522+15.8}{1522-8.63} 1595[/tex]
f'' = 6021 Hz
Hence, the reflected wave has the frequency of f'' = 6021 Hz