51. We denote the speed of the French submarine by u

1

and that of the U.S. sub by u

2

.

(a) The frequency as detected by the U.S. sub is

f

1

= f

1

v + u

2

v

− u

1



= (1000 Hz)

5470 + 70

5470

− 50



= 1.02

× 10

3

Hz .

(b) If the French sub were stationary, the frequency of the reflected wave would be f

r

= f

1

(v +u

2

)/(v

−

u

2

). Since the French sub is moving towards the reflected signal with speed u

1

, then

f

r

=

f

r

v + u

1

v



= f

1

(v + u

1

)(v + u

2

)

v(v

− u

2

)

=

(1000 Hz)(5470 + 50)(5470 + 70)

(5470)(5470

− 70)

=

1.04

× 10

3

Hz .