A SONAR system fixed in a submarine operates at a frequency 40.0kHz. An enemy submarine moves towards the SONAR with a speed of 360 km`h^(-1)`. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 `ms^(-1)`.

Options

A. 38.21 kHz
B. 48.16 kHz
C. 51.17 kHz
D. 45.93 kHz

Answer

D

Explanation

Here, `v = 40.0" kHz"`,
Speed of sound wave in water, `v =1, 450"m s"^(-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 enemy submarine moving towards the system:
Here, `v_0 = 360" km h"^(-1)`
`=(360 xx 1,000)/(60 xx 60) = 100"m s"^(-1)`
Now, `v. = (v+v_0)/(v) v = (1,450 + 100)/(1,450) xx 40.0 = 42.76" kHz"`
(ii) Frequency of waves from the enemy.s submarine as received by SONAR system :
The enemy submarine will reflect the waves of frequency `v. = 42.76` kHz and will thus act as a source of waves moving with a speed, `v_s = 100" m s"^(-1)` toward the SONAR system (listener). If `v"` is the apparent frequency as received by SONAR system, then
`v.. = (v)/( v- v_s) v. = (1,450)/(1,450 - 100) xx 42.76`
`= 45.93" kHz"`