A train blowing its whistle moves with constantspeed on a straight tra

1.	A train blowing its whistle moves with constantspeed on a straight track towards observer andthen crosses him. If the ratio of difference betweenthe actual and apparent frequencies be 3:2 in thetwo cases, then the speed of train is [v is speed of the sound]1. 2v/3 2. v/5 3. v/3 4. 3v/2
Answer» Answer: LET VELOCITY of train be v_{t} and velocity of SOUND be v. 1st CASE (when train is approaching observer) : $f2 = \bigg(\dfrac{v - 0}{v - v_{t} } \bigg)f$ $= > f2 = \bigg(\dfrac{v }{v - v_{t} } \bigg)f$ Now difference between apparent and actual frequency : $\bigstar \: \: f - f2=-(f2 - f)$ $= \: \: -\bigg\{ \bigg( \dfrac{v}{v - v_{t}} \bigg)f- f\bigg\}$ $= \: \: - \bigg( \dfrac{v_{t}}{v - v_{t}} \bigg)f$ 2nd case : (When train is going away from observer) $f3= \bigg(\dfrac{v - 0}{v + v_{t} } \bigg)f$ $= > f3= \bigg(\dfrac{v }{v + v_{t} } \bigg)f$ Now difference between apparent frequency and actual frequency : $\bigstar \: \: f- f3 = -( f3 - f)$ $= \: \: -\bigg\{ f - \bigg( \dfrac{v}{v + v_{t}} \bigg)f\bigg\}$ $= \: \: - \bigg( \dfrac{v_{t}}{v + v_{t}} \bigg)f$ Taking ratio as per question in both the cases : $\bigstar \: \: \dfrac{-(v + v_{t})}{-(v -v_{t}) } = \dfrac{3}{2}$ $= > 2v + 2v_{t} = 3v - 3v_{t}$ $= > 5v_{t} = v$ $= > v_{t} = \dfrac{v}{5}$ So final answer : $\boxed{ \red{ \huge{ \bold{v_{t} = \dfrac{v}{5} }}}}$