Question

A motor car cycle from rest and accelerates along a straight path at $2m/{s^2}.$ At the starting point of the motor cycle there is a stationary electric siren. How far has the motorcycle gone when the driver hears the frequency of the siren at $94\% $ of its value when the motorcycle was at rest? (Speed of sound$ = 330m/{s^{ - 1}}$)
(A) $98m$
(B) $147m$
(C) $196m$
(D) $49m$

Answer 1

Hint: Use the equation of motion to calculate the velocity of the car. Then substitute it in the formula of Doppler’s frequency to solve the question.

Formula used:
${v^2} - {u^2} = 2as$
Where,
$v$ is final velocity
\[u\] is initial velocity
$s$ is the distance covered
$a$ is acceleration
$f' = f\left( {\dfrac{{v - {v_0}}}{v}} \right)$
Where,
$f'$ is Doppler frequency
$v$ is speed of wave
${v_0}$ is the speed of observer

Complete step by step answer:
In this equation we will use Doppler’s effect. Doppler’s effect is the change in frequency or wavelength of a wave for an observer moving relative to the source.
Doppler’s effect of frequency of source when the source is stationary and observer is moving, is given by
$f' = f\left( {\dfrac{{v - {v_0}}}{v}} \right)$ . .. (1)
Where,
The frequency of source is $f$
$f'$ is Doppler frequency
$v$ is speed of wave
${v_0}$ is the speed of observer
Now, from the law of motion
${v^2} - {u^2} = 2as$
Where,
$v$ is final velocity
$u$ is initial velocity
$s$ is the distance covered
$a$ is acceleration
The car starts from the rest. Therefore, initial velocity of the car is $u = 0$
$ \Rightarrow {v^2} = 2as$
$ \Rightarrow v = \sqrt {2as} $
$ \Rightarrow v = \sqrt {2 \times 2s} $
$ \Rightarrow v = 2\sqrt s $
Also, it is given that the frequency at $s$ distance is $94\% $ of the initial frequency.
Now, by substituting this value and other given information in equation (1), we get
$\dfrac{{94}}{{100}}f = f\left( {\dfrac{{330 - 2\sqrt s }}{{330}}} \right)$
By cancelling the common terms and rearranging the equation, we get
$\dfrac{{94}}{{100}} = \left( {\dfrac{{330 - 2\sqrt s }}{{330}}} \right)$
By rearranging if further, we get
$\sqrt s = \dfrac{{330 \times 100 - 330 \times 94}}{{200}}$
Simplifying it, we get
$\sqrt s = \dfrac{{330 \times 6}}{{200}}$
$ \Rightarrow \sqrt s = 9.9$
$ \Rightarrow s = 98.01m$
Therefore, from the above explanation, the correct answer is, option (A) $98m$

Note:The general formula of Doppler’s effect is
$f' = f\left( {\dfrac{{v - {v_0}}}{{v + {v_s}}}} \right)$
We get the formula used in the question if we put ${v_s} = 0$ in this formula as the source is at rest.
One of the uses of Doppler’s effect is to calculate the speed of the stars or galaxies relative to the earth.