Questions & Answers

Question

Answers

a) 20m/s

b) 36.6m/s

c) 25m/s

d) None of these

Answer
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To begin with let us define Doppler’s effect.

It states that whenever there is a relative motion between the source of sound and the observer, the frequency of sound received as by the observer is different from the actual frequency of the source.

The apparent frequency of sound heard by the observer when the source is moving away from the observer is given by ${{\text{V}}_{\text{A}}}\text{=}\dfrac{v}{v-{{v}_{S}}}\text{V}....\text{(1)}$ where ${{\text{V}}_{\text{A}}}$ is the apparent frequency heard by the observer, V is the actual frequency of the source $v$ is the speed of sound in air and ${{v}_{S}}$ is the speed of the source.

In the above case the policeman detects a drop in the frequency by 10 percent of the actual frequency. Hence the frequency of sound heard by him is 90 percent of the actual frequency i.e. ${{\text{V}}_{\text{A}}}$= 0.9V where V is the actual frequency of the motor car and ${{\text{V}}_{\text{A}}}$ is the frequency of sound heard by the policeman.

Now using equation 1 let us determine the speed of the source ${{v}_{S}}$ i.e. of the car.

Using equation 1,

$\begin{align}

& {{\text{V}}_{\text{A}}}\text{=}\dfrac{v}{v-{{v}_{S}}}\text{V} \\

& \text{0}\text{.9V=}\dfrac{v}{v-{{v}_{S}}}\text{V} \\

& \text{0}\text{.9(}v-{{v}_{S}})=v \\

\end{align}$

$\begin{align}

& \text{0}\text{.9(}v-{{v}_{S}})=v \\

& 0.9{{v}_{S}}=v-0.9v \\

& {{v}_{S}}=\dfrac{0.1}{0.9}v=\dfrac{1}{9}\text{v m/s} \\

\end{align}$

It is given to us that the speed of sound in air i.e. v= 330m/s. Hence substituting in the above equation we get ${{v}_{S}}$ i.e. speed of the car as,

${{v}_{S}}=\dfrac{1}{9}\text{v m/s}=\dfrac{330}{9}m/s=36.6m/s$.