Question

A stationary source emits sound of frequency \[{f_0} = 492\,{\text{Hz}}\]. The sound is reflected by a large car approaching the source with a speed of \[2\,{\text{m}}{{\text{s}}^{ - 1}}\].The reflected signal is received by the source and superimposed with the original. What will be the beat frequency of the signal in \[{\text{Hz}}\]? (Given that the speed of sound in air is \[{\text{330}}\, {\text{m}}{{\text{s}}^{ - 1}}\] and the car reflects the sound at the frequency it has received).

Answer 1

Hint:The frequency of beats is the frequency difference of two waves. That is because of constructive interference and negative interference.
Use the formulae:
\[{f_1} = {f_0} \times \dfrac{{{u_{{\text{sound}}}} + {u_{{\text{car}}}}}}{{{u_{{\text{sound}}}}}}\] and
\[{f_2} = {f_1} \times \dfrac{{{u_{{\text{sound}}}}}}{{{u_{{\text{sound}}}} - {u_{{\text{car}}}}}}\]
Finally, find the beat frequency by the formula:
\[{f_{\text{B}}} = \left| {{f_1} - \left. {{f_2}} \right|} \right.\]

Complete step by step answer:
In sound, we hear such beat frequency as the intensity at which the sound's loudness changes while we hear the ordinary wave frequency as the sound's pitch.
Given,
\[{f_0} = 492\,{\text{Hz}}\] ,\[{u_{{\text{car}}}} = 2\,{\text{m}}{{\text{s}}^{ - 1}}\]
And \[{u_{{\text{sound}}}} = {\text{330}}\,{\text{m}}{{\text{s}}^{ - 1}}\]
We know, by the Doppler’s effect,
Apparent frequency which is perceived by a person sitting inside the car is:
\[{f_1} = {f_0} \times \dfrac{{{u_{{\text{sound}}}} + {u_{{\text{car}}}}}}
{{{u_{{\text{sound}}}}}}\] …… (1)
Where,
\[{f_0}\] indicates the frequency of the sound emitted by the source.
\[{u_{{\text{source}}}}\] indicates the velocity of the sound in air.
\[{u_{{\text{car}}}}\] indicates the velocity of the car at which it is moving.
Again,
the apparent frequency which is perceived by the source is given by the formula
\[{f_2} = {f_1} \times \dfrac{{{u_{{\text{sound}}}}}}{{{u_{{\text{sound}}}} -{u_{{\text{car}}}}}}\] …… (2)
Now, we substitute the values,
\[{f_0} = 492\,{\text{Hz}}\],
\[\Rightarrow{u_{{\text{car}}}} = 2\,{\text{m}}{{\text{s}}^{ - 1}}\] and
\[\Rightarrow{u_{{\text{sound}}}} = {\text{330}}\,{\text{m}}{{\text{s}}^{ - 1}}\] in equation (1), as follows:
\[
{f_1} = {f_0} \times \dfrac{{{u_{{\text{sound}}}} + {u_{{\text{car}}}}}}{{{u_{{\text{car}}}}}} \\
\Rightarrow{f_1}= 492 \times \dfrac{{330 + 2}}{{330}}\,{\text{Hz}} \\
\Rightarrow{f_1}= 492 \times \dfrac{{332}}{{330}}\,{\text{Hz}} \\
\]
Now, we use this value of \[{f_1}\] in equation (2):
\[{f_2} = {f_1} \times \dfrac{{{u_{{\text{sound}}}}}}{{{u_{{\text{sound}}}} - {u_{{\text{car}}}}}} \\
\Rightarrow{f_2}= 492 \times \dfrac{{332}}{{330}} \times \dfrac{{330}}{{330 - 2}} \\
\Rightarrow{f_2}= 492 \times \dfrac{{332}}{{328}}\,{\text{Hz}} \\
\Rightarrow{f_2}= {\text{498}}\,{\text{Hz}} \\
\]
The apparent frequency which is perceived by the source after reflection comes out to be \[{\text{498}}\,{\text{Hz}}\]
The beat frequency is given by the formula:
\[{f_{\text{B}}} = \left| {{f_1} - \left. {{f_2}} \right|} \right.\] …… (3)
Now, substituting the values respectively:
\[
{f_{\text{B}}} = \left| {{f_1} - \left. {{f_2}} \right|} \right. \\
\Rightarrow{f_{\text{B}}}= \left| {492 - \left. {498} \right|\,{\text{Hz}}} \right. \\
\Rightarrow{f_{\text{B}}}= \left| { - \left. 6 \right|\,{\text{Hz}}} \right. \\
\therefore{f_{\text{B}}}= 6\,{\text{Hz}} \\
 \]
Hence, the beat frequency is \[6\,{\text{Hz}}\].

Note: In this problem, we are given a scenario of the source of sound and the car and asked to find the beat frequency. For this, apply the doppler’s formula to find the different frequencies. While finding the beat frequency, there is a modulus sign, because it is taken as absolute value, which cannot be negative in nature. Absence of modulus sign, will affect the result.