A source of sonic oscillations with frequency `n=1700Hz` and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity `u=6.0(m)/(s)`. Find the beat frequency registered by the receiver. The velocity of sound is `v=340(m)/(s)`.

Answer

`f = (2Vv_(0)^(2))/(2v_(0)sqrt(V^(2)+2aVt+a)) cong (V_(v_(0)))/(sqrt(V^(2)+2aVt)) cong 1.35 khz`

Explanation

Firste detector `(D)` and source `(s)`

are at `O`. Now source
start recending with acceleration `a` and in time `t_(1)` it reaches at `B` and sound emitted at that moment reaches to the detector in time `t_(2)` here in equation `t_(1) + t_(2) = 10` second.
At `B` velocity of source `V_(S) = at_(1)`
If we consider two consecutive sound wave reaching to `D` starting from `B`
first wave will take `t_(2) = (X)/(V)`
next wave `t^(11) = T + (X + at_(1)T + (1)/(2)aT^(2))/(V)`
time period `= t^(11) - t_(2) = (VT + at_(1)T+(1)/(2)aT^(2))/(V)`
frequency `= (2V)/(2VT + 2at_(1)T + aT^(2)) = (2Vf^(2))/(2Vf + 2 at_(1)f + a)`
`f^(1) = (2Vff^(2))/(2f(V + at_(1)) + a)` .......(1)
Now `X = Vt_(2) = V(t - t_(1)) , X = (1)/(2)a(t - t_(2))^(2) = (1)/(2)at_(1^(2))`
`(1)/(2)at_(1^(2)) = Vt - Vt_(1) , at_(1^(2)) + 2Vt_(1) - 2Vt = 0`
`t_(1) = (-2V+-sqrt(4V^(2)+4axx2Vt))/(2a)`
`t_(1)` is positive so
`t_(1) = (sqrt(V^(2)+2aVt))/(a) - (V)/(a)`
Put this value of `t_(1)` in equaion (1)
`f' = (2Vf^(2))/(2fsqrt(V^(2)+2aVt)+a) equiv (Vf)/(sqrt(V^(2) + 2aVt)) = 1.35 khz`.