So far we have considered a single wave passing on a string. What happens when two wave pulses travelling in opposite directions cross each other (Fig. 15.9)? Figure 15.9 shows the situation when two pulses of equal and opposite shapes move towards each other. When the pulses overlap, the resultant displacement is the algebraic sum of the displacement due to each pulse. This is known as the principle of the superposition of waves.

According to this principle, each pulse moves as if others are not present. The constituents of the medium, therefore, suffer displacements due to both and since the displacements can be positive and negative, the net displacement is an algebraic sum of the two. Fig. 15.9 gives graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the displacements due to the two pulses have exactly cancelled each other and there is zero displacement throughout.

The principle of superposition states that when two or more waves simultaneously pass through a point, the disturbance at the point is given by the sum of the disturbances each wave would produce in absence of the other wave(s).

To put the principle of superposition mathematically, let \(y_1(x, t)\) and \(y_2(x, t)\) be the displacements due to two wave disturbances in the medium. If the waves arrive in a region simultaneously, and therefore, overlap, the net displacement \(y(x, t)\) is given by
\(
y(x, t)=y_1(x, t)+y_2(x, t) \dots(15.25)
\)
If we have two or more waves moving in the medium the resultant waveform is the sum of wave functions of individual waves. That is if the wave functions of the moving waves are
\(
\begin{aligned}
& y_1=f_1(x-v t), \\
& y_2=f_2(x-v t), \\
& \cdots \cdots \cdots \\
& \cdots \cdots \cdots \\
& y_n=f_n(x-v t)
\end{aligned}
\)
then the wave function describing the disturbance in the medium is
\(
\begin{aligned}
y & =f_1(x-v t)+f_2(x-v t)+\ldots+f_n(x-v t) \\
& =\sum_{i=1}^n f_t(x-v t) \dots(15.26)
\end{aligned}
\)
The principle of superposition is basic to the phenomenon of interference.

The resultant of two harmonic waves of equal amplitude and wavelength

For simplicity, consider two harmonic travelling waves on a stretched string, both with the same \(\omega\) (angular frequency) and \(k\) (wave number), and, therefore, the same wavelength \(\lambda\). Their wave speed will be identical. Let us further assume that their amplitudes are equal and they are both travelling in the positive direction of \(x\)-axis. The waves only differ in their initial phase. The two waves are described by the functions:
\(
y_1(x, t)=a \sin (k x-\omega t) \dots(15.27)
\)
and \(y_2(x, t)=a \sin (k x-\omega t+\phi) \dots(15.28)\)
The net displacement is then, by the principle of superposition, given by
\(
y(x, t)=a \sin (k x-\omega t)+a \sin (k x-\omega t+\phi) \dots(15.29)
\)
\(
=a\left[2 \sin \left[\frac{(k x-\omega t)+(k x-\omega t+\phi)}{2}\right] \cos \frac{\phi}{2}\right] \dots(15.30)
\)
where we have used the familiar trignometric Identity for \(\sin A+\sin B\). We then have
\(
y(x, t)=2 a \cos \frac{\phi}{2} \sin \left(k x-\omega t+\frac{\phi}{2}\right) \dots(15.31)
\)
Eq. (15.31) is also a harmonic travelling wave in the positive direction of \(x\)-axis, with the same frequency and wavelength. However, its initial phase angle is \(\frac{\phi}{2}\). The significant thing is that its amplitude is a function of the phase difference \(\phi\) between the constituent two waves:
\(
A(\phi)=2 a \cos 1 / 2 \phi \dots(15.32)
\)
For \(\phi=0\), when the waves are in phase,
\(
y(x, t)=2 a \sin (k x-\omega t) \dots(15.33)
\)
i.e., the resultant wave has amplitude \(2 \mathrm{a}\), the largest possible value for \(A\). For \(\phi=\pi\), the waves are completely, out of phase and the resultant wave has zero displacement everywhere at all times
\(y(x, t)=0 \dots(15.34)\)

Eq. (15.33) refers to the so-called constructive interference of the two waves where the amplitudes add up in the resultant wave. Eq. (15.34) is the case of destructive interference where the amplitudes subtract out in the resultant wave. Fig. 15.10 shows these two cases of interference of waves arising from the principle of superposition.

Constructive and Destructive Interference

If we assume the two wave equations have different amplitudes, the two equations can be written as
\(
\begin{array}{ll}
& y_1=A_1 \sin (k x-\omega t) \\
\text { and } & y_2=A_2 \sin (k x-\omega t+\phi) .
\end{array}
\)
According to the principle of superposition, the resultant wave is represented by
\(
\begin{aligned}
y= & y_1+y_2=A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t+\phi) \\
= & A_1 \sin (k x-\omega t)+A_2 \sin (k x-\omega t) \cos \phi \\
& \quad+A_2 \cos (k x-\omega t) \sin \phi \\
= & \sin (k x-\omega t)\left(A_1+A_2 \cos \phi\right)+\cos (k x-\omega t)\left(A_2 \sin \phi\right) .
\end{aligned}
\)
We can evaluate it by combining the two simple harmonic motions.
\(
A_1+A_2 \cos \phi=A \cos \theta \dots(i)
\)
and
\(
A_2 \sin \phi=A \sin \theta \dots(ii)
\)
we get
\(
\begin{aligned}
y & =A[\sin (k x-\omega t) \cos \theta+\cos (k x-\omega t) \sin \theta] \\
& =A \sin (k x-\omega t+\theta) .
\end{aligned}
\)
Thus, the resultant is indeed a sine wave of amplitude \(A\) with a phase difference \(\theta\) with the first wave. By (i) and (ii),
\(
\begin{aligned}
A^2 & =A^2 \cos ^2 \theta+A^2 \sin ^2 \theta \\
& =\left(A_1+A_2 \cos \phi\right)^2+\left(A_2 \sin \phi\right)^2 \\
& =A_1^2+A_2^2+2 A_1 A_2 \cos \phi \\
\text { or, } A & =\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi} \dots(iii) \\
\text { Also, } \tan \theta & =\frac{A \sin \theta}{A \cos \theta}=\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi} \dots(iv)
\end{aligned}
\)
We draw a vector of length \(A_1\) to represent \(y_1=A_1 \sin (k x-\omega t)\) and another vector of length \(A_2\) at an angle \(\phi\) with the first one to represent \(y_2=A_2 \sin (k x-\omega t+\phi)\). The resultant of the two vectors then represents the resultant wave \(y=A \sin (k x-\omega t+\theta)\).

We see from equation (iii) that the resultant amplitude \(A\) is maximum when \(\cos \phi=+1\), or \(\phi=2 n \pi\) and is minimum when \(\cos \phi=-1\), or \(\phi=(2 n+1) \pi\), where \(n\) is an integer. In the first case, the amplitude is \(A_1+A_2\) and in the second case, it is \(\left|A_1-A_2\right|\). The two cases are called constructive and destructive interferences respectively. The conditions may be written as,
Constructive interference \(: \phi=2 n \pi\)
Destructive interference \(: \phi=(2 n+1) \pi\)

Example 1: Two waves are simultaneously passing through a string. The equations of the waves are given by
\(
y_1=A_1 \sin k(x-v t)
\)
and
\(
y_2=A_2 \sin k\left(x-v t+x_0\right) \text {, }
\)
where the wave number \(k=6.28 \mathrm{~cm}^{-1}\) and \(x_0=1.50 \mathrm{~cm}\). The amplitudes are \(A_1=5.0 \mathrm{~mm}\) and \(A_2=4.0 \mathrm{~mm}\). Find the phase difference between the waves and the amplitude of the resulting wave.

Solution:

The phase of the first wave is \(k(x-v t)\) and of the second is \(k\left(x-v t+x_0\right)\).
The phase difference is, therefore,
\(
\phi=k x_0=\left(6.28 \mathrm{~cm}^{-1}\right)(1.50 \mathrm{~cm})=2 \pi \times 1 \cdot 5=3 \pi .
\)
The waves satisfy the condition of destructive interference. The amplitude of the resulting wave is given by
\(
\left|A_1-A_2\right|=5.0 \mathrm{~mm}-4.0 \mathrm{~mm}=1.0 \mathrm{~mm} \text {. }
\)