A stationary observer receives a sound from a sound of freqeuency `v_0` moving with a constant velocity `v_S=30(m)/(s)` The apparent frequency varies with time as shown in figure. Velocity of sound `v=300(m)/(s)`. Then which of the following is incorrect?

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`