1.

A source producing a sound of frequency 90 Hz is approaching a stationary listener with a speed equal to (1/10) of the speed of sound. What will be the frequency heard by the listener?

Answer»

Solution :When the SOURCE is MOVING towards the STATIONARY listener, the expression for apparent frequency is
`N.=((v)/(v-v_(s)))n`
`=((v)/(v-((1)/(10))v))n=((10)/(9))n`
`=((10)/(9))xx90=100Hz`


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