Question

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:
A. 1.2f, $\lambda $
B. 0.8f, 0.8$\lambda $
C. 1.2f, 1.2$\lambda $
D. 0.8f, 1.2$\lambda $

Answer 1

Hint: Doppler effect is known as, when the source and observer are moving relative to each other, the frequency observed by the observer ‘ ${f_a}$’ is different from the actual frequency produced by the source ‘ ${f_ \circ }$’. The general formula of doppler effect is given as, $f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$

Formula Used:
$f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$

Complete answer:
Frequency is defined as the estimation of the number of times that a repeated event occurs per unit of time. The frequency of wave-like patterns such as the electromagnetic waves i.e. radio waves or light waves, sound waves, electrical signals, or other waves, expresses the number of cycles of the repetitive waveform per second. Wavelength ($\lambda $) of a wave is defined as the distance between two successive crests or troughs.
According to the concept of doppler effect, when an observer moves towards a stationary source of sound then the apparent frequency heard by the observer increases. So, the apparent frequency heard in this situation is given as,
$f' = \left( {\dfrac{{v + {v_ \circ }}}{{v - {v_s}}}} \right)f$
As we see the source is stationary, so ${v_s} = 0$
$f' = \left( {\dfrac{{v + {v_ \circ }}}{v}} \right)$
It is given that, ${v_ \circ } = \dfrac{v}{5}$
Substituting the values in $f'$ we get,
$f' = \left( {\dfrac{{v + v/5}}{v}} \right)f = \dfrac{6}{5} = 1.2f$
Motion of observer does not affect the wavelength reaching the observer hence wavelength remains unchanged i.e. $\lambda $.
Thus, the apparent frequency and wavelength recorded by the observer are 1.2f, $\lambda $ respectively.

Hence, option (A) is the correct answer.

Note:
Wavelength is known to be inversely proportional to the frequency of the wave. Which means that the longer is the wavelength, lower will be the frequency of the wave. In the similar manner, shorter is the wavelength, higher will be the frequency of the wavelength.