Question

A locomotive engine approaches a railway station and whistles at a frequency of 400Hz. A stationary observer on the platform observes a change of 40Hz as the engine passes across him. If the velocity of sound is $330{m}/{s}$, the speed of the engine is
A.$33{m}/{s}$
B.$18{m}/{s}$
C.$16.5{m}/{s}$
D.$24{m}/{s}$

Answer 1

Hint: Obtain the apparent frequency when the engine is approaching the railway station. Then, obtain the apparent frequency when the railway station is approaching the engine. Change in the frequency is given, So, subtracting the apparent frequencies will be equal to the change in frequency. Substitute the values and calculate the speed of the engine.
Formula used:
${ n }^{ ' }=\left( \dfrac { v }{ v-{ v }_{ 0 } } \right) n$
${ n }^{ '' }=\left( \dfrac { v }{ v+{ v }_{ 0 } } \right) n$

Complete answer:
When engine is approaching railway station, apparent frequency is given by,
${ n }^{ ' }=\left( \dfrac { v }{ v-{ v }_{ 0 } } \right) n$ …(1)
Where, n is the actual frequency
             v is the velocity
When railway station is approaching engine, apparent frequency is given by,
${ n }^{ '' }=\left( \dfrac { v }{ v+{ v }_{ 0 } } \right) n$ …(2)
It is mentioned in the question that change in the frequency is 40Hz. Therefore,
${ n }^{ 1 }-{ n }^{ '' }=40$
Now, substituting values in above equation from equation. (1) and (2) we get,
$\left( \dfrac { v }{ v-{ v }_{ 0 } } \right) n-\left( \dfrac { v }{ v+{ v }_{ 0 } } \right) n=40$
$\Rightarrow \left( \dfrac { 330 }{ 330-{ v }_{ 0 } } \right) 400-\left( \dfrac { 330 }{ 330+{ v }_{ 0 } } \right) 400=40$
$\Rightarrow \left( \dfrac { 1 }{ 300-{ v }_{ 0 } } \right) -\left( \dfrac { 1 }{ 330+{ v }_{ 0 } } \right) =\dfrac { 40 }{ 330\times 400 }$
$\Rightarrow \dfrac { 330+{ v }_{ 0 }-300+{ v }_{ 0 } }{ (330-{ v }_{ 0 })(330+{ v }_{ 0 }) } =\dfrac { 40 }{ 330\times 400 }$
$\Rightarrow \dfrac { 2{ v }_{ 0 } }{ { 330 }^{ 2 }-{ v }_{ 0 }^{ 2 } } =\dfrac { 1 }{ 3300 }$
$\Rightarrow 2{ v }_{ 0 }\times 3300=108900-{ v }_{ 0 }^{ 2 }$
$\Rightarrow { v }_{ 0 }^{ 2 }+6600{ v }_{ 0 }-108900=0$
This is a quadratic equation. Solving this equation we get,
${ v }_{ 0 }=16.5{ m }/{ s }$
Thus, the speed of engine is $16.5{ m }/{ s }.$

Hence, the correct answer is option C i.e. $16.5{m}/{s}.$

Note:
Take write while solving the problem that you do not miss any term as there are many terms used in the equation. This equation for apparent frequency is derived from Doppler effect. This formula is used when one object is in motion while the other is stationary. Doppler effect is the apparent change in the frequency when the source, observer and medium are in a motion relative to each other.