Question

An ambulance driving with velocity $\dfrac{{{v}_{1}}}{8}$ where ‘${{v}_{1}}$ ’ is the speed of sound, emits a siren with a frequency of f. What is the frequency heard by a stationary observer toward whom the ambulance is driving?
A. $\dfrac{f}{8}$
B. $\dfrac{7f}{8}$
C. $f$
D. $\dfrac{8f}{7}$
E. $8f$

Answer 1

Hint: Recall the expression for observed frequency under Doppler Effect. Note that the observer is stationary and hence the speed of the observer is zero. Substitute other values as given in the question. Note that the source is approaching the observer, so, assign the sign accordingly.

Formula used:
Expression for observed frequency under Doppler Effect,
${{f}_{0}}={{f}_{s}}\left( \dfrac{v\pm {{v}_{o}}}{v\mp {{v}_{s}}} \right)$

Complete step-by-step answer:
Before going into solving the given question, let us recall our experience with a moving ambulance emitting a siren. We had observed two characteristic changes in the sound of the siren, one in its loudness and other in its pitch. If you were stationary, then you might have observed that loudness and frequency of sound is higher when the ambulance is approaching and lower when it moves away. This alteration is termed as Doppler Effect. This effect is more pronounced when the observer is stationary.
Doppler Effect occurs due to the relative motion between the source and the observer. That is, it can also be observed when the observer is moving with the source being stationary. Also when, both are in motion but with different velocity.
The observed frequency under Doppler Effect is given by,
${{f}_{0}}={{f}_{s}}\left( \dfrac{v\pm {{v}_{o}}}{v\mp {{v}_{s}}} \right)$ ……………………. (1)
Where, ${{f}_{o}}$ = observed frequency
${{f}_{s}}$ = source frequency
${{v}_{o}}$ = speed of the observer
${{v}_{s}}$ = speed of the source
$v$ = speed of the sound
Also the top sign is for approaching and the bottom sign is for departing.
Now let us modify equation (1) for solving the given question.
We are given that the observer is stationary, so, the speed of the observer is zero. That is,
${{v}_{o}}=0$
We are also given:
Speed of sound,
$v={{v}_{1}}$
Speed of source,
${{v}_{s}}=\dfrac{{{v}_{1}}}{8}$
Source frequency,
${{f}_{s}}=f$
We are supposed to find ${{f}_{o}}$. Since the source is approaching the observer we take the sign of ${{v}_{s}}$ as negative. So, equation (1) now becomes,
${{f}_{o}}=f\left( \dfrac{v}{v-{{v}_{s}}} \right)=f\left( \dfrac{{{v}_{1}}}{{{v}_{1}}-\dfrac{{{v}_{1}}}{8}} \right)$
${{f}_{o}}=f\left( \dfrac{8{{v}_{1}}}{8{{v}_{1}}-{{v}_{1}}} \right)=f\left( \dfrac{8{{v}_{1}}}{7{{v}_{1}}} \right)$
${{f}_{o}}=\dfrac{8}{7}f$
Therefore, we get the frequency heard by a stationary observer as $\dfrac{8}{7}$ times that of the source frequency.

So, the correct answer is “Option D”.

Note: Since this effect is the result of the relative motion between the source and the observer, there are many cases that come under this effect. Such as, (1) observer stationary (2) source stationary (3) both observer and source in motion with different speed. So, instead of memorizing the equation for each of these cases, it is better that you remember the equation given in the above solution and then substitute values and assign signs accordingly to avoid confusion.