Question

The number of beats heard per second if there are three sources of sound of frequencies ($n - 1$), $n$ and ($n + 1$) of equal intensities sounded together is:
A. $2$
B. $1$
C. $4$
D. $3$

Answer 1

Hint: First, we will express the displacement equation of each sound wave and find the expression for the resultant displacement. Using the conditions for the maxima and minima of the sound waves, beats per second can be found out.

Complete step by step answer:
Since the intensities of the sounds heard from three different sources are given to be equal, we can say that their amplitudes will also be equal.
First, we can write the displacement equation corresponding to the source producing a sound of frequency ($n - 1$) as
\[{y_1} = a\sin 2\pi \left( {n - 1} \right)t\]
Here, $t$ is the time at the instant and $a$ is the amplitude of sound.
The displacement equation corresponding to the source producing a sound of frequency $n$ is written as,
${y_2} = a\sin 2\pi nt$
The displacement equation corresponding to the source producing sound of frequency ($n + 1$) is written as,
${y_3} = a\sin 2\pi \left( {n + 1} \right)t$
Now, we can write the resultant displacement equation for the three sound waves as
\[
y = {y_1} + {y_2} + {y_3}\\
 = a\sin 2\pi \left( {n - 1} \right)t + a\sin 2\pi nt + a\sin 2\pi \left( {n + 1} \right)t\\
 = a\left( {\sin 2\pi \left( {n - 1} \right)t + \sin 2\pi \left( {n + 1} \right)t} \right) + a\sin 2\pi nt
\]
Since $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$, we can write
$
\sin 2\pi \left( {n - 1} \right)t + \sin 2\pi \left( {n + 1} \right)t = 2\sin \left( {2\pi t\left( {\dfrac{{2n}}{2}} \right)} \right)\cos \left( {2\pi t\left( {\dfrac{{ - 2}}{2}} \right)} \right)\\
 = 2\sin 2\pi nt\cos 2\pi t
$
Now, we will put the above result in the equation for the resultant displacement.

$
y = 2a\sin 2\pi nt\cos 2\pi t + a\sin 2\pi nt\\
 = a\left( {1 + 2\cos 2\pi t} \right)\sin 2\pi nt
$
In the above equation, the term $a\left( {1 + 2\cos 2\pi t} \right)$ represents amplitude. Hence, the resultant amplitude is
$A = a\left( {1 + 2\cos 2\pi t} \right)$
The resultant amplitude will have a maximum value when
$
2\pi t = 2\pi q\\
t = q
$
where $q = 0,1,2,3..$
Hence, $t = 0,1,2,3,....$
Thus, we can say that there is an interval of $1\,{\rm{s}}$ between two adjacent maxima.
In the same way, the resultant amplitude will have a minimum value when
$
2\pi t = 2\pi q + \dfrac{{2\pi }}{3}\\
t = q + \dfrac{1}{3}
$
where $q = 0,1,2,3....$
Hence, $t = \dfrac{1}{3},\dfrac{4}{3},\dfrac{7}{3},\dfrac{{10}}{3},....$
Thus, we can say that there is an interval of $1\,{\rm{s}}$ between two adjacent minima also.
The time interval for both the maxima and the minima are equal to $1\;{\rm{s}}$. Since frequency is the reciprocal of the time interval, the frequency of the beats heard is $1\,{\rm{Hz}}$. Therefore, one beat per second will be heard if there are three sound sources.

So, the correct answer is “Option B”.

Note:
Beats are produced when the intensity of the sound varies as a result of the superimposing of sound waves of nearby frequencies.
The displacement equation corresponding to the source producing a sound of frequency ($n - 1$).
The displacement equation corresponding to the source producing sound of frequency ($n + 1$).
We can find the beats per second or the beat frequency of two sound waves by finding the difference between the frequencies of the waves.