Here, υ=40.0‌kHz ;
Speed of sound wave in water,
v=1,450ms–1
The frequency of the waves from SONAR system will undergo a change in the following two steps :
(i) Frequency of waves from SONAR system as received by the enemysubmarine moving towards the system :
Here, v0=360‌km‌h−1=

360×1,000

60×60

=100ms−1
Now, υ′=

v+v0

v

υ =

1,450+100

1,450

×40.0=42.76‌kHz
(ii) Frequency of waves from the enemy’s submarine asreceived by SONAR system :
The enemy submarine will reflect the waves of frequency υ′′=42.756‌kHz and will thus act as a source of waves moving with a speed; vs=100ms–1 toward the SONAR system (listener). If υ′′ is the apparent frequency as received by SONAR system, then
υ′′=

v

v−vs

υ′ =

1,450

1,450−100

×42.76=45.93‌kHz