Question

Two trains are approaching each other on parallel tracks with same velocity. The whistle sound produced by one train is heard by a passenger in another train. If actual frequency of whistle is 620Hz and apparent increase in its frequency is 100Hz, the velocity of one of the two trains is (Velocity of sound in air $=335m{s}^{-1}$)

• A. 90kmph
• B. 72 kmph
• C. 54kmph
• D. 36 kmph

Answer 1

Answer: (A)

90kmph

Given : $\delta =620H_z$

${\delta }^{{}^{\prime }}=(620+100)H_z=720H_z$

$v_0=-v_s=-vm/s$

$v=335m/s$

By doppler effect formula

${\delta }^{{}^{\prime }}=\delta \left(\frac{v-v_0}{v-v_3}\right)$

$720=620\left(\frac{335+v}{335-v}\right)$

$335\left(720\right)-720v=620\left(335\right)+620v$

$v=25m/s$

$=25\times \frac{3600}{1000}km/h$

$=90km/h$