Question

A police siren emits a sinusoidal wave with frequency ${f_{\text{s}}} = 300\,{\text{Hz}}$ . The speed of the sound is $340\,{\text{m}}\,{{\text{s}}^{ - 1}}$ . If the siren is moving at $30\,{\text{m}}\,{{\text{s}}^{ - 1}}$ , find the wavelength of the waves ahead and behind the source.

Answer 1

First of all, we will use the Doppler’s effect, when the source is approaching the listener and the source going away from the listener. Wavelength is shorter when the source is approaching the listener.

Formula used:
The formula which gives the apparent wavelength of the waves, when the source approaches the listener a rest:
$\lambda ' = \dfrac{{v - {v_{\text{s}}}}}{{{f_{\text{s}}}}}$ …… (1)
The formula which gives the apparent wavelength of the waves, when the source is moving away from the listener a rest:
$\lambda ' = \dfrac{{v + {v_{\text{s}}}}}{{{f_{\text{s}}}}}$ …… (2)
Where,
$\lambda '$ indicates the apparent wavelength.
$v$ indicates velocity of sound.
${v_{\text{s}}}$ indicates the velocity of the siren.
${f_{\text{s}}}$ indicates the frequency of the siren.

Complete step by step answer:
To begin with, we have the speed of the sound and the source. So, we can apply the formula which gives the apparent wavelength when the source is approaching the stationary listener and the apparent wavelength, when the source is moving away from the stationary listener.
Let us proceed to solve the numerical. We use the equation (1) in order to find the apparent wavelength of the waves, when the source approaches the listener a rest:
By substituting the required values in the above equation, then we get:
\lambda ' = \dfrac{{v - {v_{\text{s}}}}}{{{f_{\text{s}}}}} \\\ \Rightarrow \lambda ' = \dfrac{{340 - 30}}{{300}} \\\ \Rightarrow \lambda ' = \dfrac{{310}}{{300}} \\\ \therefore \lambda ' = 1.03\,{\text{m}} \\\
Hence, the wavelength is $1.03\,{\text{m}}$ .
Again, we apply the equation (2) in order to find the apparent wavelength when the source is moving away from the listener a rest:
$\lambda ' = \dfrac{{v + {v_{\text{s}}}}}{{{f_{\text{s}}}}} \\\ \Rightarrow \lambda ' = \dfrac{{340 + 30}}{{300}} \\\ \Rightarrow \lambda ' = \dfrac{{370}}{{300}} \\\ \therefore \lambda ' = 1.23\,{\text{m}}$

Hence, the wavelength is $1.23\,{\text{m}}$ .

Note: While solving the problem, most of the students seem to have some confusion regarding the formulas needed to be used for the two different situations. It is important to note that wavelength of the waves ahead of the source is associated with the situation when the source approaches the listener at rest. The wavelength gets increased when the source moves away from the listener.