Question

The apparent frequency of sound heard by a listener is $10\%$ more than the actual frequency of the note emitted by the source when the source moves towards the stationary listener with velocity$v$. When the source moves with a velocity$2v$, the apparent frequency will be more than the actual frequency by
a) $17.5\%$
b) $20\%$
c) $22.2\%$
d) $24.5\%$

Answer 1

It is given that the apparent sound heard by the listener is a percentile of the original when the source is moving towards the person at a velocity. In this case the source is approaching the listener. Use the apparent frequency formula for the given condition and find out the velocity value and use it in the second case to find actual frequency percentile.
Formula Used:
${n_a} = (\dfrac{{v \pm {v_s}}}{{v \mp {v_2}}})n$

Complete step by step solution:
It is given that the source is said to move towards the listener with a fixed velocity v. This change in frequency from the source to the listener when the object is moving is said to be as a consequence of Doppler’s effect. So, using the Doppler’s effect equation, we can get a suitable equation for apparent frequency. In general conditions, the apparent frequency equation is given as ,
${n_a} = (\dfrac{{v \pm {v_L}}}{{v \mp {v_s}}})n$
Where ${n_a}$ apparent frequency is listened, $n$is actual frequency of the source, ${v_s}$is the speed of the source and ${v_L}$is the speed of the listener.
When the listener is said to be stationary and the object is said to move, we can conclude that ${v_L} = 0$. Rephrasing the equation we get,
$\Rightarrow {n_a} = (\dfrac{v}{{v \mp {v_s}}})n$
Substituting our first condition, where apparent frequency is $10\%$more than actual frequency. Pn substituting this in the above equation we get,
$\Rightarrow (0.1n + n) = (\dfrac{v}{{v - {v_s}}}) \times n$(since apparent frequency is given to be $10\%$more than actual)
Cancelling out common terms we get,
$\Rightarrow (1.1) = (\dfrac{v}{{v - {v_s}}})$
On rearranging, we get
$\Rightarrow \dfrac{1}{{1.1}} = (\dfrac{{v - {v_s}}}{v})$
$\Rightarrow \dfrac{1}{{1.1}} = 1 - \dfrac{{{v_s}}}{v}$
$\Rightarrow \dfrac{{0.1}}{{1.1}} = \dfrac{{{v_s}}}{v}$
Now, in the second condition, it is said that the speed is doubled up, so changing the equation accordingly we get,
$\Rightarrow {n_a} = (\dfrac{v}{{v - 2{v_s}}})n$
Rearranging the equation, we get.
$\Rightarrow {n_a} = (\dfrac{1}{{\dfrac{{v - 2{v_s}}}{v}}})n$
On further simplification, we get,
$\Rightarrow {n_a} = (\dfrac{1}{{1 - \dfrac{{2{v_s}}}{v}}})n$
Substituting for the known variable, we get
$\Rightarrow {n_a} = (\dfrac{1}{{1 - \dfrac{{0.2}}{{1.1}}}})n$
$\Rightarrow {n_a} = (\dfrac{{1.1}}{{1.1 - 0.2}})n$
On further simplification we get,
$\Rightarrow {n_a} = 1.222n$
Which can also be written as
$\Rightarrow {n_a} = (1 + 0.222)n$
Which can be said as $22.2\%$ more than the actual frequency.

Hence, option (c) is the right answer

Note: Doppler’s effect is given when there is an increase in frequency of sound waves or other waveforms when there is a mutual movement from the source and observer or listener towards each other at a specified speed.