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An observer moves towards a stationary source of sound with a speed $\dfrac{1}{5}$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
$1.2f$, $1.2\lambda$
B
$1.2f$, $\lambda$
C
$f$, $1.2\lambda$
D
$0.8f$, $0.8\lambda$
Detailed Solution
The source is stationary, so the wavelength of the waves in the medium does not change: $\lambda' = \lambda$. Only the rate at which the observer meets the wave crests changes.
For an observer moving towards a stationary source with speed $v_o$: $f' = \left(\dfrac{v + v_o}{v}\right)f = \left(1 + \dfrac{v_o}{v}\right)f$
Here $v_o = \dfrac{v}{5}$: $f' = \left(1 + \dfrac{1}{5}\right)f = 1.2f$
So the apparent frequency is 1.2f and the wavelength remains $\lambda$.
For an observer moving towards a stationary source with speed $v_o$: $f' = \left(\dfrac{v + v_o}{v}\right)f = \left(1 + \dfrac{v_o}{v}\right)f$
Here $v_o = \dfrac{v}{5}$: $f' = \left(1 + \dfrac{1}{5}\right)f = 1.2f$
So the apparent frequency is 1.2f and the wavelength remains $\lambda$.