Question

An observer move towards a stationary source of sound with a speed equal to one-fourth of the speed of sound. If the wavelength and frequency of the sound produced are $\lambda$ and $f$ respectively, then what will be the apparent frequency and wavelength recorded by the observer respectively?

• A. $1.25f$ and $\lambda$
• B. $f$ and $1.25\lambda$
• C. $1.25f$ and $1.25\lambda$
• D. $f$ and $\lambda$

Answer 1

The correct option is A $1.25f$ and $\lambda$

From Doppler's effect, the apparent frequency heard by the observer is,
$f^{{}^{\prime }}=f\left(\frac{v\pm v_0}{v\pm v_s}\right)$, here,
$f$ is the original frequency produced by the source, $v$ is the speed of sound, $v_0$ is the speed of observer and $v_s$ is the speed of source.
Given, $v_0=\frac{v}{4}$
$\Rightarrow f^{{}^{\prime }}=f\left(\frac{v+\frac{v}{4}}{v-0}\right)$
[as separation is decreasing, so frequency will increase]
$\Rightarrow f^{{}^{\prime }}=\frac{5}{4}f=1.25f$
As the source is stationary, hence the wavelength will remain the same.
$\Rightarrow {\lambda }^{{}^{\prime }}=\lambda$