A rocket is moving at a speed of `200ms^(-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 racket as an echo. Calculate the frequency of sound as detected by the person at the position of target and frequency of echo as detected by the rocket . Given velocity of sound `=330ms^(-1)`.

Explanation

(1) The observer is at rest and the source is moving with a speed of 200 m `s^(-1)`. Since this is comparable with the velocity of sound, 330 m `s^(-1)`, we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, `v_(o) = 0`, and `v_(s)` must be replaced by `–v_(s)`. Thus, we have
`v=v_(0)(1-(v_(s))/(v))^(-1)`
`v=1000 Hz xx [1-200 m s^(-1)//300 m s^(-1)]^(-1)`
`~= 2540 Hz`
(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_(o)` 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_(o)`. Therefore, the frequency as registered by the rocket is
`v'=v((v+v_(0))/(v))`
`=2540 Hz xx ((200 m s^(-1)+330 m s^(-1))/(330 m s^(-1)))`
`~= 4080 Hz`