### Theory:

There will be no Doppler effect in the following situations, and the apparent frequency heard by the listener will be the same as the source frequency.
• When both the source ((\S\)) and the listener ((\L\)) are at rest.
• When ((\S\)) and ((\L\)) move in such a way that their distance from each other remains constant.
• When the source ((\S\)) and the destination ((\L\)) are moving in opposite directions.
• If the source is in the centre of the circle that the listener moves around in.
Numerical based on apparent change in frequency due to doppler effect:

Example 1:

A source producing a sound of frequency $$90$$ $$Hz$$ is approaching a stationary listener with a speed equal to ($$1/10$$) of the speed of sound. What will be the frequency heard by the listener?

Given:

Frequency(\n\)) $$=$$ $$90$$ $$Hz$$

Speed(\v\)) $$=$$ ($$1/10$$) of the speed of sound

When the source is moving towards the stationary listener, the expression for apparent frequency is

$\begin{array}{l}{n}^{\prime }=\left(\frac{V}{V-{V}_{S}}\right)n\\ \mathit{Apply}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{known}\phantom{\rule{0.147em}{0ex}}\mathit{values},\\ {n}^{\prime }=\left(\frac{V}{V-\left(\frac{1}{10}\right)V}\right)×90\\ {n}^{\prime }=\left(\frac{V}{V\left(1-\frac{1}{10}\right)}\right)×90\\ {n}^{\prime }=\left(\frac{10}{9}\right)×90\\ {n}^{\prime }=10×10\\ {n}^{\prime }=100\phantom{\rule{0.147em}{0ex}}\mathit{Hz}\end{array}$

Example 2:

A source producing a sound of frequency $$500$$ $$Hz$$ is moving towards a listener with a velocity of $$30$$ $$m/s$$. The speed of the sound is $$330$$ $$m/s$$. What will be the frequency heard by the listener?

Given:

Frequency of the sound ($$n$$) $$=$$ $$500$$ $$Hz$$

Velocity of the listener (${V}_{L}$) $$=$$ $$30$$ $$m/s$$

Speed of the sound($$V$$) $$=$$ $$330$$ $$m/s$$

When the source is moving towards the stationary listener, the expression for apparent frequency is

$\begin{array}{l}{n}^{\prime }=\left(\frac{V}{V-{V}_{S}}\right)n\\ \mathit{Apply}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{known}\phantom{\rule{0.147em}{0ex}}\mathit{values},\\ {n}^{\prime }=\left(\frac{330}{330-30}\right)×500\\ {n}^{\prime }=\left(\frac{330}{300}\right)×500\\ {n}^{\prime }=110×5\\ {n}^{\prime }=550\phantom{\rule{0.147em}{0ex}}\mathit{Hz}\end{array}$