A source of sound of frequency 256 Hz is moving towards a wall with a velocity of 5 m/s. How many beats per second will be heard by an observer O standing in such a positoin that the source S is between O and wall? `(c=330(m)/(s)`)

Explanation

Direct sound `V_s=-5(m)/(s)` because positive direction is from S to O.
`f=f_0((c-V_0)/(c-V_s))=256((330-0)/(330+5))=252.2Hz`
i. Frequency received at wall `(f_1)` is given by
`(V_s=+5(m)/(s)`)
`f=(f_0(c-V_(wall)))/(c-V_S)=(f_0(c-0))/(c-V_s)=(256xx330)/(330-5)Hz`
ii. Frequency received by observer `(f_2)` is given by
`f_2=(f_1(c-V_0))/(c-V_(wall))=(f_1(c-0))/(c-0)=f_1`
`impliesf_2=(256xx330)/(325)=259.9Hz`
Beats per second is `f_2-f=(259.9-252.2)Hz=7.7Hz`