Question

A train blowing its whistle moves with constant
speed on a straight track towards observer and
then crosses him. If the ratio of difference between
the actual and apparent frequencies be 3:2 in the
two 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 1

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 :

$\boxed{ \red{ \huge{ \bold{v_{t} = \dfrac{v}{5} }}}}$

Answer 2

Solution :

⏭ Concept:

✏ This question is completely based on concept of Doppler effect. (Apparent frequency of a sound heard by listener)

⏭ Formula:

$\bigstar \: \boxed{ \sf{ \pink{f = { \huge{( }}\frac{V \pm \: V_L}{V \pm \: V_s} { \huge{)}}fo}}} \: \bigstar$

⏭ Terms indication:

• f denotes apparent frequency
• fo denotes actual frequency
• V$_L$ denotes velocity of listener
• V$_s$ denotes velocity of source (train)
• V denotes velocity of sound

⏭ Calculation:

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1) First case

✒ Source of sound(train) is moving toward stationary listener, then

$\implies \sf \: f_1 = { \huge{( }}\dfrac{V}{V - V_s}{ \huge{ )}} \: fo \\ \\ \therefore \sf \: fo - f_1 = fo{ \huge{(}}1 - \frac{V}{V - V_s} { \huge{)}}$

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2) Second case

✒ Source of sound(train) is moving awaybfrom stationary listener, then

$\implies \sf \: f_2 = { \huge{(}} \dfrac{V}{V + V_s} { \huge{)}}fo \\ \\ \therefore \sf \: fo - f_2 = fo{ \huge{(}} 1 - \frac{V}{V + V_s} { \huge{)}}$

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✒ As per given question...

$\circ \sf \: \: \dfrac{fo - f_1}{fo - f_2} = \dfrac{3}{2} \\ \\ \circ \sf \: \: \dfrac{1 - \dfrac{V}{V - V_s} }{1 - \dfrac{V}{V + V_s} } = \frac{3}{2} \\ \\ \circ \sf \: \: \dfrac{(V - V_s - V)(V + V_s)}{(V + V_s - V)(V - V_s)} = \frac{3}{2} \\ \\ \circ \sf \: \: - \dfrac{V + V_s}{V - V_s} = \dfrac{3}{2} \\ \\ \circ \sf \: \:5V_s = V \\ \\ \circ \: \: \boxed{ \tt{ \red{ \large{V_s = \frac{V}{5}}}}} \: \orange{ \bigstar}$

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