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Q. A whistle of frequency 500 Hz tied to the end of a string of length 1.2 m revolves at 400 revolutions per minute. If the velocity of sound is 340 m/sec, a listener standing some distance away in the plane of rotation of the whistle hears the frequencies in the range

AMUAMU 1999

Solution:

: Velocity of source $ {{v}_{s}}=R\omega $ $ =R\times 2\pi n $ $ =1.2\times 2\times 3.14\times \frac{400}{600} $ = 50 approximately $ \therefore $ Maximum apparent frequency $ =\frac{vn}{(v-{{v}_{s}})} $ $ =\frac{340\times 500}{(340-50)}=\frac{170000}{290}=586\,Hz $ Minimum apparent frequency $ =\frac{vn}{v+{{v}_{s}}} $ $ =\frac{340\times 500}{(340-50)}=\frac{17000}{390}=436\,Hz $ $ \therefore $ Range is 436 to 586 Hz.