Question

When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?

• A. More frequency, less wavelength
• B. More frequency, more wavelength
• C. Less frequency, less wavelength
• D. More frequency, constant wavelength

Answer 1

Answer: (A)

According to Doppler's effect, whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source. Let S be source of sound and L the listener of sound. Let v be the actual frequency of sound emitted by the source and $\lambda $ be the actual wavelength of the sound emitted.
If v is velocity of sound in still air, then
$\lambda =\frac{v}{V}$
If velocity of listener is $v_{L}$ and velocity of source is $v_{s}$ , then apparent frequency of sound waves heard by the listener is
$v^{′ ′}=\frac{v - v_{L}}{v - s_{s}}\times V$
Here, both source and listener are approaching each other.



Then $v_{s}$ is positive and $v_{L}$ is negative.
$\therefore v^{′ ′}=\frac{v - \left(- v_{L}\right)}{v - v_{s}}v=\left(\frac{v + v_{L}}{v - v_{s}}\right)v$
$i.e., \, v^{′ ′}>v$
Also,
$\lambda ^{′ ′} < \lambda $
So, listener listens more frequency and observes less wavelength.