A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s-1 at an angle of 60 with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m s-1), is [figure]
- 106 Hz
- 97 Hz
- 100 Hz
- 103 Hz
Answer
(4) 103 Hz
Solution
Only the component of the source velocity along the source–observer line causes the Doppler shift.
⇒ vs = 19.4 60 = 19.4 12 = 9.7 m s-1 (towards the observer)
⇒ For a source approaching a stationary observer:
f' = f ( vv - vs )
= 100 330330 - 9.7
= 100 330320.3
103 Hz
Answer: (4)