Question

An observer moves towards a stationary source of sound with a speed of $1/5th$ 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:

(A) $1.2f,0.8\lambda$

(B) $0.8f,0.8\lambda$

(C) $1.2f,1.2\lambda$

(D) $0.8f,1.2\lambda$

Answer 1

The frequency increases for when the observer is moving towards the stationary source. The wavelength decreases for when the observer is moving towards a stationary source.

Formula used: In this solution we will be using the following formulae;

$f' = f\left( {\dfrac{{v + {v_L}}}{v}} \right)$ where $f'$ is the frequency of a wave as observed by an observer when he is moving towards a stationary source, $f$ is the actual frequency of the wave from the source, $v$ is the speed of sound and ${v_L}$ is the velocity of the observer.

Complete step by step solution:

An observer is said to be moving towards a stationary source of sound at a speed which is a fifth of the speed of sound. We are to determine the frequency and wavelength observed by the observer with respect to frequency and wavelength of the actual sound.

To do this, we shall recall the formula for the frequency observed by the observer when moving towards a stationary source. This is given as

$f' = f\left( {\dfrac{{v + {v_L}}}{v}} \right)$ where $f$ is the actual frequency of the wave from the source, $v$ is the speed of sound and ${v_L}$ is the velocity of the observer.

Hence, by substituting known values

$f' = f\left( {\dfrac{{v + \dfrac{1}{5}v}}{v}} \right) = f\left( {\dfrac{{\dfrac{6}{5}v}}{v}} \right)$

Hence, by cancelling $v$, we have

$f' = \dfrac{6}{5}f = 1.2f$

For the wavelength, the frequency needs only to be inverted.

Hence, $\lambda ' = \dfrac{5}{6}\lambda = 0.8\lambda$

Thus, the correct option is (A).

Note: For clarity, the wavelength is simply inverted because frequency and wavelength inversely proportional to each other, and since the exact value is not required the constant (which is actually the speed of the wave) will cancel out, as in;

$\lambda ' = \dfrac{k}{{f'}}$ and $\lambda = \dfrac{k}{f}$

Then $\dfrac{{\lambda '}}{\lambda } = \dfrac{k}{{f'}} \times \dfrac{f}{k} = \dfrac{f}{{f'}}$

$\Rightarrow \dfrac{{\lambda '}}{\lambda } = {\left( {\dfrac{{f'}}{f}} \right)^{ - 1}}$

Then the wavelength observed can be

$\lambda ' = {\left( {\dfrac{{f'}}{f}} \right)^{ - 1}}\lambda$

Also, in general, knowing that the frequency observed is higher than actual frequency, then without calculation the wavelength must be lower.