Question

A policeman on duty detects a drop of $10\%$ in the pitch of the horn of motion of a car as it crosses him. If the velocity of sound is $330m/s$. Calculate the speed of the car.
(A) $17.4m/s$
(B) $20.4m/s$
(C) $18.6m/s$
(D) $16.4m/s$

Answer 1

To solve this question, we need to use the Doppler Effect formula for the apparent frequency as heard by a listener. We have to substitute the speeds with the proper sign convention for the two cases. On solving the two equations formed, we will get the required value of the speed of the car.

Formula used: The formula used to solve this question is given by
$\nu = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}$, here $\nu$ is the apparent frequency as listened by a listener moving with a speed of ${v_L}$ due to a source of sound moving with a speed of ${v_S}$ and emitting a sound of frequency ${\nu _0}$. The velocity of sound is $v$.

Complete step-by-step solution:
Let the original frequency of the horn of the car be ${\nu _0}$.
We know from the Doppler Effect formula, that the apparent frequency received by a listener is given by
$\nu = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}$
According to the question, the listener is the policeman, and the source is the car. The policeman is at rest, so we have ${v_L} = 0$. Also, the speed of sound is given to be $v = 330m/s$. Substituting these in the above formula we get
$\nu = \left( {\dfrac{{330 - 0}}{{330 - {v_S}}}} \right){\nu _0}$
$\Rightarrow \nu = \left( {\dfrac{{330}}{{330 - {v_S}}}} \right){\nu _0}$............(1)
Now, when the car has crossed the policeman, the direction of the speed of the car with respect to the policeman will get reversed. So we have to replace ${v_S}$ with $- {v_S}$. Also, according to the question, the frequency is dropped by $10\%$. So now the frequency of the horn listened by the policeman becomes $0.9\nu$. Making these replacements in (1) we get
$\Rightarrow 0.9\nu = \left( {\dfrac{{330}}{{330 + {v_S}}}} \right){\nu _0}$...........(2)
Dividing (1) by (2) we have
$\dfrac{1}{{0.9}} = \left( {\dfrac{{330 + {v_S}}}{{330 - {v_S}}}} \right)$
Applying componendo and dividendo, we get
$\dfrac{{1 + 0.9}}{{1 - 0.9}} = \left( {\dfrac{{330 + {v_S} + 330 - {v_S}}}{{330 + {v_S} - \left( {330 - {v_S}} \right)}}} \right)$
$\Rightarrow \dfrac{{330}}{{{v_S}}} = \dfrac{{1.9}}{{0.1}}$
By cross multiplying we get
${v_S} = \dfrac{{330}}{{19}}m/s$
On solving we get
${v_S} = 17.36m/s \approx 17.4m/s$
Thus, the velocity of the car is equal to $17.4m/s$.

Hence, the correct answer is option A.

Note: The direction of velocity is from the source to the listener. According to the sign convention, all the speeds which are parallel to the velocity of sound are taken to be positive and those in the opposite direction are taken to be negative. Do not apply the concept of relative velocity while substituting the speeds of the source and the listener.