Question

Three identical sources of sound are placed on a straight line as shown in the figure. A detector D is placed at a distance d from \[{S_1}\]. The wavelength of sound is \[\lambda \].
Determine the value of d so that the intensity of the wave at the detector may be
(a) zero.
(b) nine times the intensity of each source.

Answer 1

Hint:We will use the concept for the interference of three sound waves which is equal to the summation of sound from each wave. We will also utilize the concept of constructive and destructive interference to find the relation between wavelength and distance from the source \[{S_1}\].

Complete step by step answer:
It is given that all the sources of sound are identical, which means their phase difference is zero, their wavelength, frequency and time is also the same, and hence, the amplitude will also be the same.
We can express the excess pressure of the sound wave with respect to the pressure at equilibrium pressure for the source \[{S_1}\] can be expressed as below:
\[{P_1} = {P_0}\sin \left( {k{x_1} - \omega t} \right)\]……(1)
Here \[{P_0}\] is the excess pressure, k is the wavenumber, \[{x_1}\] is the distance between the source \[{S_1}\] and detector D, \[\omega \] is the angular frequency and t is time period of the wave.
We know that the expression for the distance of the sound source \[{S_1}\] from the detector D can be written as:
\[{x_1} = d\]
We will substitute d for \[{x_1}\] in equation (1).
\[{P_1} = {P_0}\sin \left( {kd - \omega t} \right)\]
Let us write the expression for excess pressure of the sound wave obtained from the source \[{S_2}\].
\[{P_2} = {P_0}\sin \left( {k{x_2} - \omega t} \right)\]……(2)
Here \[{P_2}\] is the excess pressure for source \[{S_2}\] and \[{x_2}\] is the distance between the sound source \[{S_2}\] and detector D.
We know that the expression for the distance of the sound source \[{S_2}\] from the detector D can be written as:
\[{x_2} = 2d\]
We will substitute \[2d\] for \[{x_2}\] in equation (2).
\[
\Rightarrow{P_2} = {P_0}\sin \left( {k \cdot 2d - \omega t} \right)\\
\Rightarrow{P_2} = {P_0}\sin \left( {2kd - \omega t} \right)
\]
We can write the expression for excess pressure for the sound wave obtained from the source \[{S_3}\].
\[{P_3} = {P_0}\sin \left( {k{x_3} - \omega t} \right)\]……(3)
Here \[{P_3}\] is the excess pressure for source \[{S_3}\] and \[{x_3}\] is the distance between the sound source \[{S_3}\] and detector D.
We know that the expression for the distance of the sound source \[{S_3}\] from the detector D can be written as:
\[{x_3} = 3d\]
We will substitute \[3d\] for \[{x_3}\] in equation (3).
\[
\Rightarrow{P_3} = {P_0}\sin \left( {k \cdot 3d - \omega t} \right)\\
\Rightarrow{P_3} = {P_0}\sin \left( {3kd - \omega t} \right)
\]
Let us write the expression for the resultant wave of interference when all the sources are producing sound together.
\[P = {P_1} + {P_2} + {P_3}\]
Here P is the express pressure for the interference of all the sound sources.
We will substitute \[{P_0}\sin \left( {kd - \omega t} \right)\] for \[{P_1}\], \[{P_0}\sin \left( {2kd - \omega t} \right)\] for \[{P_2}\] and \[{P_0}\sin \left( {3kd - \omega t} \right)\] for \[{P_3}\] in the above expression.
\[
\Rightarrow P = {P_0}\sin \left( {kd - \omega t} \right) + {P_0}\sin \left( {2kd - \omega t} \right) + {P_0}\sin \left( {3kd - \omega t} \right)\\
\Rightarrow P = 2{P_0}\sin \left( {2kd - \omega t} \right)\cos \left( {kd} \right) + {P_0}\sin \left( {2kd - \omega t} \right)\\
\Rightarrow P = {P_0}\sin \left( {2kd - \omega t} \right)\left[ {2\cos \left( {kd} \right) + 1} \right]
\]……(4)
For destructive interference, we can write the above expression of the resultant wave equal to zero.
\[{P_0}\sin \left( {2kd - \omega t} \right)\left[ {2\cos \left( {kd} \right) + 1} \right] = 0\]
On closely observing the above equation, we can see that the term \[\left[ {2\cos \left( {kd} \right) + 1} \right]\] should be zero for the resultant wave to be zero. Therefore we can write:
\[
\Rightarrow\left[ {2\cos \left( {kd} \right) + 1} \right] = 0\\
\Rightarrow\cos kd = - \dfrac{1}{2}\\
\Rightarrow kd = \left( {2n + 1} \right)\pi \pm \dfrac{\pi }{3}
\]……(5)
Here n is a positive integer.
We know that the wave number can be expressed as:
\[k = \dfrac{{2\pi }}{\lambda }\]
We will substitute \[\dfrac{{2\pi }}{\lambda }\] for k in equation (5).
\[
\Rightarrow\dfrac{{2\pi }}{\lambda }d = \left( {2n + 1} \right)\pi \pm \dfrac{\pi }{3}\\
\Rightarrow d = \left[ {\dfrac{{\left( {2n + 1} \right)}}{2} \pm \dfrac{1}{6}} \right]\lambda
\]
For constructive interference, we can write:
\[
\Rightarrow\left[ {2\cos \left( {kd} \right) + 1} \right] = 1\\
\Rightarrow\cos \left( {kd} \right) = 0\\
\Rightarrow kd = 2n\pi
\]
We will substitute \[\dfrac{{2\pi }}{\lambda }\] for k in the above expression.
\[
\Rightarrow\dfrac{{2\pi }}{\lambda }d = 2n\pi \\
\Rightarrow d = n\lambda
\]
Therefore, we can say that the distance d is an integral multiple of the wavelength of the wave.
On substituting 1 for \[\cos \left( {kd} \right)\] in equation (4), we get:
\[
\Rightarrow P = {P_0}\sin \left( {2kd - \omega t} \right)\left[ {2\left( 1 \right) + 1} \right]\\
\therefore P = 3{P_0}\sin \left( {2kd - \omega t} \right)
\]
From the above expression, we can see that the amplitude of the resultant wave is \[3{P_0}\] and we know that intensity is square of the amplitude; therefore, the intensity is nine times the initial amplitude of each wave.

Note: We can note that for destructive interference, the resultant wave is equal to zero, and for constructive interference, the resultant wave is equal to unity. We can also remember the generalized form of cosine when it is equal to zero and unity.