A bus is moving towards a huge wall with a velocity of `5m//s^(-1)`. The driver sounds a horn of frequency `200 Hz`. The frequency of the beats heard by a passenger of the bus will be ………… `Hz` (Speed of sound in air `= 342m//s^(-1))`

Explanation

The first frequency the driver of bus hears is the original frequency `200 Hz`. The second frequency the driver hears is the frequency of sound reflected from the wall. The two frequencies of sound heard by the driver are
(i) Original frequency `(200 Hz)`
(ii) Frequency of sound reflected from the wall `(v')`

The frequency of sound reflected form the wall
`v' = v[(u+u_(0))/(u-u_(s))]`
`v' =200[(342 +5)/(342-5)] = 205.93 Hz`
`:.` Frequency of beats
=`v' - v = 205.93 - 200 = 5.93 = 6 Hz`
`8. x = 0.04 m, KE = 0.5 J` and `PE = 0.4 J`
Also `v = (25)/(pi) Hz`
Now, `TE = (1)/(2) m omega^(2) a^(2) = (1)/(2)m xx 4pi ^(2) v^(2) a^(2)`
`rArr 0.9= (1)/(2) xx 0.2 xx 4 pi^(2) xx (25)/(pi) xx(25)/(pi) xx a^(2)`
`rArr a = (3)/(50) = 0.06 m`.
Alternate Method :
`(1)/(2)mv^(2) =0.3rArr (1)/(2)(0.2)v^(2) = 0.5 rArr v^(2) = 5`
Now `v^(2) = omega^(2)(A^(2)-x^(2))`
`rArr 5 = (2pi v)^(2)[A^(2) -(0.04)^(2)]`
`rArr 5 = 4pi^(2) xx (25)/(pi) xx(25)/(pi)[A^(2)-(0.04)^(2)]`
`rArr A = 0.06 m`.