A train moves towards a stationary observer with speed `34 m//s`. The train sounds a whistle and its frequency registered by the observer is `f_(1)`. If the train's speed is reduced to `17 m//s`, the frequency registered is `f_(2)`. If the speed of sound of `340 m//s`, then the ratio `f_(1)//f_(2)`is

Options

A. `(19)/(18)`
B. `(18)/(19)`
C. 2
D. `1//2`

Answer

A

Explanation

According to Doppler.s effect, the approximate frequency heard by stationary observer, `v=(V)/(v-v_(s))v_(0)`
Case (i) `v_(s)= 34m//s`
where, v= speed of sound in air, `u_(s)`= speed of source
and `v_(0)=` frequency of the source.
`therefore v_(1)= (340)/(340-34)v_(0)`
`=(340)/(306)v_(0)` ...(i)
Case (ii) `v_(s)= 17m//s`
`therefore v_(2)= (340)/(340-17) v_(0)= (340)/(323)v_(0)` ...(ii)
From Eqs (i) and (ii), we get
`therefore (v_(1))/(v_(2))= (340//306)/(340//323)rArr (323)/(306) ~~ (19)/(18)`