1.

A rocket is moving at a speed of 200 m s^(-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.

Answer» <html><body><p></p>Solution :(1) The observer is at rest and the source is moving with a speed of `200 m s^(-1)`. <a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> this is <a href="https://interviewquestions.tuteehub.com/tag/comparable-7359202" style="font-weight:bold;" target="_blank" title="Click to know more about COMPARABLE">COMPARABLE</a> with the <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of sound, `300 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 <br/> `v = v_(o) (1 - (v_(s))/(v))^(-1)` <br/> `v = 1000 Hz xx [1 - 200 m s^(-1)//330 m s^(-1)]^(-1)` <br/> `~= 2540 Hz` <br/> (2) The target is now the source (because it is the source of echo) and the rocket.s <a href="https://interviewquestions.tuteehub.com/tag/detector-949853" style="font-weight:bold;" target="_blank" title="Click to know more about DETECTOR">DETECTOR</a> id now the detector or observer (because it detects echo). thus , `v_(s) = 0` and `v_(o)` has a positive value. The drequency of the sound emitted by the source (the target) is v. Therefore , the frequency as registered by the rocket is <br/> `v. = v ((v + v_(o))/(v))` <br/> `= 2540 Hz xx ((200 m s^(-1) + 330 m s^(-1))/(330 m s^(-1)))` <br/> `~= 4080 Hz`</body></html>


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