A statinary observer receives a sound of frequency 2000 Hz. The variation of apparent frequency and time is shown. Find the speed of source, if velocity of sound is `300(m)/(s)`

Options

A. The minimum value of apparent frequency is 889 Hz
B. The natural frequency of souce is 1000 Hz
C. The frequency time curve corresponds to a source moving at an angle to the stationary observer.
D. The maximum value of apparent frequency is 1111 Hz

Answer

A

Explanation

This frequency-time curve corresponds to a source moving at an angle to a stationary observer.
In the region SN, the source is moving towards the observer, i.e., the apparent frequency
`n'=n_0((v)/(v-v_Scostheta))`
`n'=n_0((300)/(300-30costheta))`
When `theta=(pi)/(2)`,i.e, at `N'`
`n'=n_0=1000Hz`, i.e., natural frequency of source, in the region `NS'` the source is moving away from the observer, i.e., apparent frequency
`n'=n_0((300)/(300-30costheta))`
when `theta=0`,i.e., `costheta=1`,
`n_(max)=n_0(v)/(v-v_S)=((1000Hz)(300(m)/(s)))/((300(m)/(s)-30(m)/(s)))`
`=(10)/(9)xx1000Hz=1111Hz`
`n_(min)=n_0(v)/(v+v_S)=(1000xx300)/(330)=909Hz`