14.5 The principle of superposition of waves

Resultant Amplitude and Intensity due to Coherent Sources

Two or more waves can travel simultaneously in a medium without affecting the motion of one another. The resultant displacement of each particle of the medium at any instant is equal to the vector sum of the individual displacements produced by the two waves. This principle is called principle of superposition.
If \(\mathbf{y}_1, \mathbf{y}_2, \mathbf{y}_3, \ldots\) are the displacements of particles at a particular time due to individual waves, then the resultant displacement is given by
\(
\mathbf{y}=\mathbf{y}_1+\mathbf{y}_2+\ldots
\)

Resultant Amplitude

Consider the superposition of two sinusoidal waves of the same frequency at a point. Let us assume that the two waves are travelling in the same direction with same velocity. The equation of the two waves reaching at a point can be written as
\(
\begin{array}{ll}
& y_1=A_1 \sin (k x-\omega t) \\
\text { and } & y_2=A_2 \sin (k x-\omega t+\phi)
\end{array}
\)

Assume a vector \({A}_1\) of length \(A_1\) to represent the amplitude of first wave. Another vector \(\mathbf{A}_2\) of length \(A_2\), making an angle \(\phi\) with \({A}_1\) represent the amplitude of second wave. The resultant of \({A}_1\) and \({A}_2\) represent the amplitude of resulting function \(y\). The angle \(\theta\) represents the phase difference between the resulting function and the first wave. The resultant displacement of the point, where the waves meet is

\(
\begin{aligned}
y= & y_1+y_2=A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t+\phi) \\
= & A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t) \cos \phi \\
& \quad+A_2 \cos (k x-\omega t) \sin \phi \\
= & \left(A_1+A_2 \cos \phi\right) \sin (k x-\omega t)+A_2 \sin \phi \cos (k x-\omega t)
\end{aligned}
\)
Substituting \(A_1+A_2 \cos \phi=A \cos \theta\) and \(A_2 \sin \phi=A \sin \theta\) in above equation, we get
\(
\begin{aligned}
& y=A \cos \theta \sin (k x-\omega t)+A \sin \theta \cos (k x-\omega t) \\
& y=A \sin (k x-\omega t+\theta)
\end{aligned}
\)
\(
\begin{aligned}
& A^2=\left(A_1+A_2 \cos \phi\right)^2+\left(A_2 \sin \phi\right)^2 \\
& A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi} \dots(i)
\end{aligned}
\)
and \(\tan \theta=\frac{A \sin \theta}{A \cos \theta}=\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi}\)

Example 1: Two harmonic waves are represented in SI units by
\(
y_1(x, t)=0.2 \sin (x-3.0 t)
\)
and \(y_2(x, t)=0.2 \sin (x-3.0 t+\phi)\)
(i) Write an expression for the sum \(y=y_1+y_2\) for \(\phi=\pi / 2 \mathrm{rad}\).
(ii) Suppose the phase difference \(\phi\) between the waves is unknown and the amplitude of their sum is 0.32 m, what is \(\phi\)?

Solution: (i) \(y=y_1+y_2=0.2 \sin (x-3.0 t)+0.2 \sin \left(x-3.0 t+\frac{\pi}{2}\right)\)
\(
=A \sin (x-3.0 t+\theta)
\)

Here, \(\quad A=\sqrt{(0.2)^2+(0.2)^2}=0.28 \mathrm{~m}\) and \(\theta=\frac{\pi}{4}\)
\(
\therefore \quad y=0.28 \sin \left(x-3.0 t+\frac{\pi}{4}\right)
\)
(ii) Since, the amplitude of the resulting wave is 0.32 m and
\(
\begin{aligned}
& A=0.2 \mathrm{~m}, \text { we get } \\
& 0.32=\sqrt{(0.2)^2+(0.2)^2+(2)(0.2)(0.2) \cos \phi}
\end{aligned}
\)
Solving this, we get \(\phi= \pm 1.29 \mathrm{rad}\)

Resultant Intensity

we have read that the intensity of a wave is given by
\(
I=\frac{1}{2} \rho \omega^2 A^2 v \quad \text { or } \quad I \propto A^2
\)
So, if \(\rho, \omega\) and \(v\) are same for the both interfering waves, then Eq. (i) can also be written as
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \dots(ii)
\)
Here, the proportionality constant \(\left(I \propto A^2\right)\) cancels out on right-hand side and left-hand side.

Proof: Principle of Superposition:
When two waves of the same frequency \((\omega)\) and speed \((v)\) interfere, their resultant amplitude \(A\) is given by the vector sum of their individual amplitudes \(A_1\) and \(A_2\), accounting for the phase difference \(\phi\) :
\(
A^2=A_1^2+A_2^2+2 A_1 A_2 \cos \phi \dots(1)
\)
Relation Between Intensity and Amplitude:
As you noted, the intensity \(I\) of a wave is:
\(
I=\frac{1}{2} \rho \omega^2 v A^2
\)
If the medium \((\rho, v)\) and the frequency \((\omega)\) are the same for both waves, we can define a proportionality constant \(k\) such that:
\(
k=\frac{1}{2} \rho \omega^2 v
\)
Thus, \(I=k A^2\), or \(A^2=\frac{I}{k}\).
By taking the square root, we also find that:
\(
A=\sqrt{\frac{I}{k}}
\)
Substituting into the Amplitude Equation:
Now, we substitute the expressions for \(A^2, A_1^2\), and \(A_2^2\) into Equation (1):
\(
\frac{I}{k}=\frac{I_1}{k}+\frac{I_2}{k}+2\left(\sqrt{\frac{I_1}{k}}\right)\left(\sqrt{\frac{I_2}{k}}\right) \cos \phi
\)
Simplify the product in the third term:
\(
\frac{I}{k}=\frac{I_1}{k}+\frac{I_2}{k}+2 \frac{\sqrt{I_1 I_2}}{k} \cos \phi
\)
Final Result:
Since the constant \(k\) (containing the density, frequency, and velocity) is common to every term on both the left-hand and right-hand sides, it cancels out completely:
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
\)

Key Observations:

• Constructive Interference: When \(\phi=0,2 \pi, \ldots(\cos \phi=1)\), the intensity is maximum:
  \(
  I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
  \)
• Destructive Interference: When \(\phi=\pi, 3 \pi, \ldots(\cos \phi=-1)\), the intensity is minimum:
  \(
  I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
  \)
• Energy Conservation: While intensity varies at different points in the interference pattern, the average intensity across the whole pattern remains \(I_1+I_2\), meaning energy is redistributed, not lost.

Note: The special case of the above two equations is when the individual amplitudes (or intensities) are equal. or
\(
A_1=A_2=A_0 \text { (say) } \Rightarrow I_1=I_2=I_0 \text { (say) }
\)
In this case, Eqs. (i) and (ii) become
\(
\begin{aligned}
& A=2 A_0 \cos \frac{\phi}{2} \dots(iii) \\
& I=4 I_0 \cos ^2 \frac{\phi}{2} \dots(iv)
\end{aligned}
\)

Interference of waves

When two coherent waves of same frequency propagate in same direction and superimpose on each other, then the intensity of resultant wave becomes maximum at some points and minimum at some other points, this phenomenon of intensity variation is called interference.

If two sinusoidal waves \(S_1\) and \(S_2\) of same angular frequency \(\omega\) meet at a point \(B\), where the phase difference between them is \(\phi\) and path difference is \(\Delta x\) as shown in the figure, then according to the principle of superposition, the resulting amplitude is given by

\(
A^2=A_1^2+A_2^2+2 A_1 A_2 \cos \phi \dots(i)
\)
As the intensity of a wave is given by
\(
I=\frac{1}{2} \rho A^2 \omega^2 v, \text { i.e. } I \propto A^2
\)
So, if \(\rho, \omega\) and \(v\) are the same for both interfering waves, Eq. (i) can also be written as
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi \dots(ii)
\)

Constructive interference

From Eqs. (i) and (ii), we see that the resulting amplitude \(A\) and intensity \(I\) depends on the phase difference \(\phi\) between the interfering waves. If the resultant wave has maximum amplitude, then the interference is called constructive interference (these are the points where the resultant amplitude or intensity is maximum).

The Condition for Maximums:

For the amplitude and intensity to be at their maximum, the phase difference \(\phi\) must be such that the waves do not cancel each other out. This happens when:
\(
\cos \phi=+1
\)
This occurs at specific phase angles:
\(
\phi=0,2 \pi, 4 \pi, \ldots, 2 n \pi
\)
Using the relationship between phase difference \((\phi)\) and path difference \((\Delta x)\) :
\(
\Delta x=\frac{\lambda}{2 \pi} \phi
\)
Substituting \(\phi=2 n \pi\), we get the condition for path difference:
\(
\Delta x=n \lambda \quad(n=0,1,2, \ldots)
\)
This means the waves must travel distances that differ by a whole number of wavelengths.

Resultant Amplitude (\(\boldsymbol{A}_{\text {max }}\)):

From the general formula for resultant amplitude:
\(
A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}
\)
When \(\cos \phi=1\) :
\(
\begin{gathered}
A_{\max }=\sqrt{A_1^2+A_2^2+2 A_1 A_2}=\sqrt{\left(A_1+A_2\right)^2} \\
A_{\max }=A_1+A_2
\end{gathered}
\)

Resultant Intensity (\(I_{\text {max }}\)):

Using the relation \(I \propto A^2\), we substitute the maximum amplitude into the intensity equation:
\(
I_{\text {max }}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\)
Expanding this gives \(I_1+I_2+2 \sqrt{I_1 I_2}\), which is the highest possible energy concentration for those two waves.
The Special Case: Identical Waves
If the two interfering waves are identical (same source), then:
\(A_1=A_2=a\)
\(I_1=I_2=I_0\)
Maximum Amplitude:
\(
A_{\max }=a+a=2 a
\)
(Note: The \(\pm\) in your text indicates that the displacement can be \(2 a\) at a crest or \(-2 a\) at a trough).
Maximum Intensity:
Using the identity \(I=k A^2\) :
\(
I_{\max } \propto(2 a)^2=4 a^2
\)
Since \(I_0 \propto a^2\), we find:
\(
I_{\max }=4 I_0
\)

Destructive interference

If the resultant wave has minimum amplitude, then the interference is called destructive interference.
The Condition for Minimums:
The resultant amplitude \(A\) of two interfering waves is given by:
\(
A=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}
\)
For the amplitude to be at its minimum, the term \(\cos \phi\) must be at its most negative value:
\(
\cos \phi=-1
\)
This occurs when the phase difference \(\phi\) is an odd multiple of \(\pi\) :
\(
\phi=\pi, 3 \pi, 5 \pi, \ldots,(2 n-1) \pi
\)
Path Difference (\(\Delta x\)):
Using the relationship between phase difference and path difference:
\(
\Delta x=\frac{\lambda}{2 \pi} \phi
\)
Substituting \(\phi=(2 n-1) \pi\) :
\(
\Delta x=\frac{\lambda}{2 \pi}(2 n-1) \pi=(2 n-1) \frac{\lambda}{2}
\)
This means destructive interference occurs when the path difference is an odd multiple of half-wavelengths (e.g., \(\lambda / 2,3 \lambda / 2, \ldots\)). At these points, the crest of one wave meets the trough of another.
Resultant Amplitude (\(\boldsymbol{A}_{\text {min }}\)):
Substituting \(\cos \phi=-1\) into the amplitude formula:
\(
\begin{gathered}
A_{\min }=\sqrt{A_1^2+A_2^2-2 A_1 A_2} \\
A_{\min }=\sqrt{\left(A_1-A_2\right)^2} \\
A_{\min }=\left|A_1-A_2\right|
\end{gathered}
\)
Resultant Intensity (\(I_{\text {min }}\)):
Since intensity \(I \propto A^2\), we use the amplitude relationship to find:
\(
I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\)
The Special Case: Identical Waves
If the two waves have the same amplitude (\(A_1=A_2=a\)) and the same initial intensity (\(I_1= I_2=I_0\)):
Minimum Amplitude:
\(
A_{\min }=a-a=0
\)
Minimum Intensity:
\(
I_{\min }=\left(\sqrt{I_0}-\sqrt{I_0}\right)^2=0
\)

The graph shown below represents the variation of resultant intensity, \(I\) with \(\left(I_1+I_2\right)\), i.e. sum of individual intensities.

Points to remember: 

In interference
\(
\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2 \\
& =\left(\frac{A_1 / A_2+1}{A_1 / A_2-1}\right)^2=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2=\left(\frac{A_{\max }}{A_{\min }}\right)^2
\end{aligned}
\)

Example 2: Two waves of equal frequencies have their amplitudes in the ratio of \(3: 5\). They are superimposed on each other. Calculate the ratio of maximum and minimum intensities of the resultant wave.

Solution: Given, \(\frac{A_1}{A_2}=\frac{3}{5}\)
\(
\therefore \quad \sqrt{\frac{I_1}{I_2}}=\frac{3}{5} \quad\left(\text { As, } I \propto A^2\right)
\)
Maximum intensity is obtained, where
\(
\cos \phi=1 \text { and } I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2
\)
Minimum intensity is obtained, where
\(
\cos \phi=-1 \text { and } I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\)
Hence, \(\quad \frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\right)^2\)
\(
=\left(\frac{3 / 5+1}{3 / 5-1}\right)^2=\frac{64}{4}=\frac{16}{1}
\)

Example 3: Two coherent sound sources are at distances \(x_1=0.2 \mathrm{~m}\) and \(x_2=0.48 \mathrm{~m}\) from a point. Calculate the intensity of the resultant wave at that point, if the frequency of each wave is \(f=400 \mathrm{~Hz}\) and the velocity of the wave in the medium is \(v=448 \mathrm{~ms}^{-1}\). The intensity of each wave is \(I_0=60 \mathrm{Wm}^{-2}\).

Solution: Path difference, \(\Delta x=x_2-x_1=0.48-0.2=0.28 \mathrm{~m}\)
Now, phase difference, \(\phi=\frac{2 \pi}{\lambda} \Delta x=\left(\frac{2 \pi f}{v}\right) \Delta x\)
\(
=\frac{2 \pi(400)(0.28)}{448}=\frac{\pi}{2}
\)
Now, the intensity of the resultant wave,
\(
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
\)
\(
\begin{aligned}
I & =I_0+I_0+2 I_0 \cos \left(\frac{\pi}{2}\right) \\
& =2 I_0=2(60)=120 \mathrm{Wm}^{-2}
\end{aligned}
\)

Example 4: The figure shows a tube structure in which a sound signal is sent from one end and received at the other end. The semicircular part has a radius of 20.0 cm. The frequency of the sound source can be varied electronically between 1000 and 4000 Hz. Find the frequencies at which maxima of intensity are detected. The speed of sound in air \(=340 \mathrm{~m} / \mathrm{s}\).

Solution: According to the question,

Path difference, \(\Delta x=\pi r-2 r=(\pi-2) r\) For maximum intensity,
\(
\Delta x=n \lambda
\)
\(
\Rightarrow \quad(\pi-2) r=n \lambda=\frac{n v}{f}
\)
\(
\begin{aligned}
\Rightarrow \quad f & =\frac{n v}{(\pi-2) r}=\frac{n \times 340}{(3.14-2) \times 0.2} \\
& =1491 n \mathrm{~Hz}
\end{aligned}
\)
\(
\text { If } n=1 \text {, then } f=1491 \mathrm{~Hz}
\)
If \(n=2\), then \(f=2982 \mathrm{~Hz}\)
If \(n=3\), then \(f=4473 \mathrm{~Hz}\)
The frequencies in the given range are 1491 Hz and 2982 Hz.

Example 5: Two sources are placed from a person \(P\) as shown in figure. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). If sound signal is continuously varied from 500 Hz to 2500 Hz, for which frequency listener will hear minimum sound intensity?

Solution: Path difference, \(\Delta x=S_2 P-S_1 P=6.4-6.0=0.4 \mathrm{~m}\)
For minimum sound intensity,
\(
\Delta x=(2 n+1) \frac{\lambda}{2}=\frac{(2 n+1)}{2} \frac{v}{f}
\)
\(
\Rightarrow \quad f=\frac{(2 n+1) v}{2 \Delta x}=\frac{(2 n+1) 320}{2(0.4)}=400(2 n+1) \mathrm{Hz}
\)
where, \(n=0,1,2, \ldots\)
If \(\quad n=0\), then \(f=400 \mathrm{~Hz}\)
If \(\quad n=1\), then \(f=1200 \mathrm{~Hz}\)
If \(\quad n=2\), then \(f=2000 \mathrm{~Hz}\)
If \(n=3\), then \(f=2800 \mathrm{~Hz}\)
The frequencies in the given range are 1200 Hz and 2000 Hz.

Example 6: In interference, two individual amplitudes are \(A_0\) each and the intensity is \(I_0\) each. Find the resultant amplitude and intensity at a point, where
(a) the phase difference between two waves is \(60^{\circ}\)
(b) path difference between two waves is \(\frac{\lambda}{3}\).

Solution: (a) Substituting \(\phi=60^{\circ}\) in the equations,
\(
A=2 A_0 \cos \frac{\phi}{2}
\)
and \(I=4 I_0 \cos ^2 \frac{\phi}{2}\)
We get, \(A=\sqrt{3} A_0 \text { and } I=3 I_0\)

(b) Given, \(\Delta x=\frac{\lambda}{3}\)
\(\phi\) or \(\Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \cdot \Delta x\)
\(
=\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{3}\right)=\frac{2 \pi}{3} \text { or } 120^{\circ}
\)
\(
\begin{aligned}
&\text { Now, substituting } \phi=120^{\circ} \text { in the above two equations we get }\\
&A=A_0 \text { and } I=I_0
\end{aligned}
\)

Example 7: In interference, \(\frac{I_{\max }}{I_{\min }}=\alpha\), find
(a) \(\frac{A_{\max }}{A_{\min }}\)
(b) \(\frac{A_1}{A_2}\)
(c) \(\frac{I_1}{I_2}\)

Solution: (a) \(\frac{A_{\max }}{A_{\min }}=\sqrt{\frac{I_{\max }}{I_{\min }}}=\sqrt{\alpha}\)
(b) \(\frac{A_{\max }}{A_{\min }}=\sqrt{\alpha}=\frac{A_1+A_2}{A_1-A_2}=\frac{A_1 / A_2+1}{A_1 / A_2-1}\)
Solving this equation, we get
\(
\frac{A_1}{A_2}=\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}
\)
(c) \(\frac{I_1}{I_2}=\left(\frac{A_1}{A_2}\right)^2\left(\frac{\sqrt{\alpha}+1}{\sqrt{\alpha}-1}\right)^2\)