Question

Two submarines are underwater and approaching each other head-on. Sub A has a speed of 15.8 m/s and sub B has a speed of 8.63 m/s. Sub A sends out a 1570-Hz sonar wave that travels at a speed of 1522 m/s.
(a) What is the frequency detected by sub B?
(b) Part of the sonar wave is reflected from B and returns to A.
What frequency does A detect for this reflected wave?

Answer 1

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,

                              $f' = \frac{V+v'}{V-v} f$

Substituting the given values above,

                              $f' = \frac{1522+8.63}{1522-15.8} 1570$

                             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,

                             $f'' = \frac{V+v}{V-v'} f'$

                             $f'' = \frac{1522+15.8}{1522-8.63} 1595$

                             f'' = 6021 Hz

Hence, the reflected wave has the frequency of f'' = 6021 Hz