A source of sonic oscillation with frequency `n_0=600Hz` moves away and at right angles to a wall with velocity `u=30(m)/(s)`. A stationary reciever is located on the line of source in succession wall`rarr`source`rarr`receiver. If velocity of osund propagation is `v=330(m)/(s)`, then
Q. The beat frequency recorded by the receiver is

Options

A. `110Hz`
B. `210Hz`
C. `150Hz`
D. `220Hz`

Answer

A

Explanation

Since source moves away from the wall, it means that its velocity is towards the receiver as shown in the figure. Hence frequency of direct sound received by it is greater than natural frequency of the source.
Frequency of direct waves
`n_d=n_0((v)/(v-u))=660Hz`
Frequency of reflected sound is equal to frequency received by the wall. since source is moving away from the wall, therefore, frequency received by the wall is less than natural frequency of the source which is equal to
`n_0((v)/(v+u))`
Therefore, the frequency of reflected sound is
`n_r=n_0((v)/(v+u))=550Hz`
Hence, the beat frequency recorded by the receiver is
`n_r-n_d=110hz`
Since the reciever is stationary, therefore velocity of both direct and reflected sound relative to the receiver is equal to u. Hence, wavelength of direct waves is
`(v)/(n_d)=0.5m=50cm`
and wavelength of reflected waves is
`(v)/(n_r)=0.6m=60cm`