A rocket is moving at a speed of `200 ms^(-1)` towards a stationary target. While moving , it emits a wave of frequency 1000 Hz.Some of the sound reaching the target gets reflected back to the rocket as an echo.Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket.

Explanation

(1) The observer is at rest and the source is moving with a speed of `200 ms^(-1)` . Since this is comparable with the velocity of sound `330 ms^(-1)`,we must use Eq.
`v= v_(0) [ (1+v_(s))/(v)]^(-1)` and not the approximate
`v= v_(0) [ (1+v_(s))/(v)]`
Since the source is approaching a stationary target,`v_(0)= 0` , and `v_(s)` mustbe replaced by `-v_(s)`.Thus,we have
`v= v_(0) (1-(v_(s))/(v))^(-1)`
`v= 1000 Hzxx[ 1-200 ms^(-1)//330 ms^(-1)]^(-1)`
` ~= 2540Hz`
(2) The target is now the source ( because it is the source of echo ) and the rocket's detector is now the detector or observer ( because it detects echo ) . Thus, `v_(s) = 0 ` and `v_(0)` has a positive value. The frequency of the sound emitted by the source ( the target) is v, the frequency intercepted by the target and not `v_(0)`.Therefore , the frequency as registered by the rocket is
`v^(')= v((v+v_(0))/(v))`
`= 2540Hz xx((200 ms^(-1) + 330 ms^(-1))/(330 ms^(-1)))`
`~= 4080 Hz`