A whistle producing sound waves of frequencies `9500 Hz` and above is approaching a stationary person with speed `vms^(-1)`. The velocity of sound in air is `300 ms^(-1)`. If the person can hear frequencies upto a maximum of `10,000 Hz`.The maximum value of `v` upto which he can hear whistle is

Options

A. `30ms^(-1)`
B. `15sqrt(2)ms^(-1)`
C. `(15)/(sqrt(2))ms^(-1)`
D. `15ms^(-1)`

Answer

A

Explanation

Velocity of sound in air, `v=300ms^(-1)`
As the source is moving towards the stationary person with velocity `v_(s)`, the apparent frequency is
`upsilon.=(upsilonv)/(v-v_(s))orupsilon.(v-v_(s))=upsilonv`
`therefore (10000)/(9500)=(300)/(300-v_(s))`
or `300-v_(s)=285 or v_(s)=15ms^(-1)`
Thus the maximum value of v upto which the person can hear the whistle is `15ms^(-1)`.