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. `15 ms^(-1)`
B. `30 ms^(-1)`
C. `15 sqrt2 ms^(-1)`
D. `(15)/(sqrt2) ms^(-1)`

Answer

A

Explanation

`V = 300 m//s, n = 9500 Hz, n_(A) = 10,000`
As per Doppler's principle, the apparent frequency `(n_(A))` of sound heard by the stationary observer will be more than the actual frequency (n), when the source approaches the observer.
`n_(A) = ((V)/(V - V_(s)))n`
`10000 = ((300)/(300 - V_(s))) xx 9500`
`20 = 19 [(300)/(300 - V_(s))]`
`:. 6000 - 20V_(s) = 5700`
`300 = 20V_(s) " " :. V_(s) = 15m//s`