Consider two tuning forks with natural frequency 250 Hz. One is moving away and another is moving towards a stationary observer at same speed. If the observer hears beats of frequency 5 Hz, then the speed of the tuning fork is : ( Given, speed of sound wave is 350 m/s . )

Options

A. 2.5 m/s
B. 3.5 m/s
C. 5.0 m/s
D. 2.0 m/s

Answer

B

Explanation

Given, speed of sound, `v=350m//s`, actual frequency, `n_(0)=250` Hz and number of beats heared, `x=5`
As source is moving towards the observer
therefore, the apparent frequency,
`n_(1)=n_(0). (v)/(v-v_(s))` ... (i)
Similarly, as source is moving away from the observer, therefore the apparent frequency,
`n_(2)=(n_(0)v)/(v+v_(s))` ... (ii)
Beats heared by the observer, `x=n_(1)-n_(2)`
Hence, from the Eq. (i) and (ii), we get,
`x=n_(0)v[(1)/(v-v_(s))-(1)/(v+v_(s))]`
`x=n_(0)v[(v+v_(s)-v+v_(s))/(v^(2)-v_(s)^(2))]`
`rArr x=n_(0)v[(2v_(s))/(v^(2)-v_(s)^(2))]`
`x(v^(2)-v_(s)^(2))=n_(0)v(2v_(s))`
where, `v_(s)=` speed of the tuning fork
`rArr v_(s)^(2)+(2n_(0)v)/(5).v_(s)-v^(2)=0`
Now, putting the values, in above Eqs. we get
`v_(s)^(2)+35000v_(s)-122500=0`
It is a quadratic equation, so the positive value of `v_(s)` is
`v_(s)=7/2=3.5 m//s`
Hence, the correct option is (b).