An observer moves towards a stationary source | vedclass

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(A) The speed of sound is $v$. The speed of the observer is $v_0 = v/5$. The source is stationary,so $v_s = 0$.The apparent frequency $f'$ is given by

Vedclass · Accepted Answer

$1.2f, \lambda$

Vedclass · Answer

(A) The speed of sound is $v$. The speed of the observer is $v_0 = v/5$. The source is stationary,so $v_s = 0$.The apparent frequency $f'$ is given by the Doppler effect formula: $f' = f \left( \frac{v + v_0}{v} \right)$.Substituting the values: $f' = f \left( \frac{v + v/5}{v} \right) = f \left( \frac{6v/5}{v} \right) = 1.2f$.Since the source is stationary,the wavelength $\lambda$ of the sound waves in the medium does not change for the observer. Therefore,the apparent wavelength is $\lambda' = \lambda$.Thus,the apparent frequency is $1.2f$ and the apparent wavelength is $\lambda$.

An observer moves towards a stationary source of sound with a speed $1/5^{th}$ of the speed of sound. The wavelength and frequency of the source emitted are $\lambda$ and $f$ respectively. The apparent frequency and wavelength recorded by the observer are respectively:

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