An observer moves towards a stationary source | vedclass

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(A) When an observer moves towards a stationary source of sound,the apparent frequency heard by the observer increases.The general formula for the Dop

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$1.2 f, \lambda$

Vedclass · Answer

(A) When an observer moves towards a stationary source of sound,the apparent frequency heard by the observer increases.The general formula for the Doppler effect is $f^{\prime} = f \left( \frac{v + v_0}{v - v_s} \right)$.Since the source is stationary,$v_s = 0$.Given that the observer's speed $v_0 = \frac{v}{5}$,the apparent frequency $f^{\prime}$ is:$f^{\prime} = f \left( \frac{v + \frac{v}{5}}{v} \right) = f \left( \frac{1.2v}{v} \right) = 1.2 f$.The wavelength of the sound waves depends only on the source and the medium. Since the source is stationary and the medium is unchanged,the wavelength reaching the observer remains $\lambda$.Therefore,the apparent frequency is $1.2 f$ and the wavelength is $\lambda$.

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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