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

As the source (horn of bus) is approaching stationary wall (say, listener), therefore, apparent frequency striking the wall is
`v^(')=(upsilonxxv)/(upsilon-upsilon_(s))` …(i)
Sound of this frequency is reflected by the wall (nnow the source ). The bus passenger is now the listener moving towards source. Therefore, frequency heard by the listener is
`v^('')=((upsilon+upsilon_(L))v^('))/(upsilon)`
using (i) , we get
`v^('')=((upsilon+upsilon_(L)))/(upsilon)(upsilonv)/(upsilon-upsilon_(s))`
`v^('')=((upsilon+upsilon_(L))v)/(upsilon-upsilon_(s))=((342+5)200)/(342-5)`
`v^('')=(347)/(337)xx200=205.9Hz`
Beat frequency `=(v^('')-v)`
`=205.9-200=5.9Hz`