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An observer moves towards a stationary source of sound with a speed \(\frac{1}{5}^{\text {th }}\) of the speed of sound. The wavelength and frequency of the waves emitted by the source are \(\lambda\) and \(f\) respectively. The apparent frequency and wavelength heard by the observer are respectively,
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The correct answer is:
\(1.2 f, \lambda\)
When an observer moves towards on a stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation.
\(f^{\prime}=\left(\frac{v+v_0}{v-v_s}\right) f,\) (As source is stationary)
Hence, \(v_s=0\)
\(\begin{aligned}
& \therefore \quad f^{\prime}=\left(\frac{v+v_0}{v}\right) f \\
& \text{Given, } v_0=\frac{v}{5} \\
& f^{\prime}=\left(\frac{v+\frac{v}{5}}{v}\right) f=\frac{6}{5} f=1.2 f \\
\end{aligned}\)
Motion of the observer does not affect the wavelength reaching the observer, hence wavelength remains \(\lambda\).
Hence, the apparent frequency and wavelength heard by the observes are respectively, \(1.2 f\) and \(\lambda\).
\(f^{\prime}=\left(\frac{v+v_0}{v-v_s}\right) f,\) (As source is stationary)
Hence, \(v_s=0\)
\(\begin{aligned}
& \therefore \quad f^{\prime}=\left(\frac{v+v_0}{v}\right) f \\
& \text{Given, } v_0=\frac{v}{5} \\
& f^{\prime}=\left(\frac{v+\frac{v}{5}}{v}\right) f=\frac{6}{5} f=1.2 f \\
\end{aligned}\)
Motion of the observer does not affect the wavelength reaching the observer, hence wavelength remains \(\lambda\).
Hence, the apparent frequency and wavelength heard by the observes are respectively, \(1.2 f\) and \(\lambda\).
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