## Introduction to the Superposition of Waves

You all like listening to musical instruments, such as guitar, piano. If you are keen on physics, then you may have a question about what waves come out from these instruments?

A human voice has a wave nature.

On throwing a stone in the lake, ripples produced in it pass through each other.

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The films produced in water droplets have a wave nature.

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So, what’s happening here is, the waves in all the examples mentioned above pass through each other. So, this passing of waves is known as the principle of superposition of waves.

Now, let’s understand what does the superposition of waves means.

### State the Principle of Superposition of Waves

Let’s say, two friends, A and B are standing at a distance from each other, each having a string.

They both start to pluck a string. We notice one thing that oscillations (Harmonic waves) of these strings pass through each other. So, these two waves have interference.

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Now, what happens next is, suddenly, their friend (particle P) comes in the middle of these strings. He gets confused in deciding though which wave, he should make a displacement along, either A or B.

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Let’s say the wave equation of string A (String of the person A) be y1, and of string B is y2.

Now, the principle of superposition says, if a particle P gets stuck in deciding the displacement it should take along, then its net displacement is given as;

ynet ⇾ = y1⇾ + y2⇾

Here, ynet can be y1 + y2 or y2 - y1.

This means that the displacement of particle P due to wave A remains y1 though it is along with wave B.

Due to B, it is y2, though it is along with wave A, the net displacement of a particle will always remain y1⇾ + y2⇾.

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This is the principle of superposition of waves.

From here, we understood that meeting of these waves is the interference of waves, and for finding the displacement of this particle P, we use the concept of superposition of waves.

Now, let’s understand the mathematics behind this concept.

### Derivation of Superposition of Waves

Let’s consider two coherent wave sources, namely S1 and S2.

The light from these two sources focus at a point, i.e., P, as shown below:

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The distances, x1 and x2, traveled by S1 and S2 are varying.

So, the displacement in S1 is:

y1 = A Sin (⍵t + k * x)....(1)

Where,

A = Maximum displacement of the wave

⍵ = Angular frequency = 2πf

k = Force constant = \[\frac{2 \pi}{\lambda}\]

Similarly, the displacement in S2 is:

y2 = A Sin (⍵t + k * x).…(2)

Equations (1) and (2) are wave equations.

Here, f, ⍵, are the same in S1 and S2.

Since these two waves are in the same medium, so their speed is also the same.

⇒ λ is also the same for both S1 and S2.

The frequency f is the same for both, so that’s why S1 and S2 are called coherent sources.

Let’s say these travel distances x1 and x2 and reach particle P in time t.

The wave emitting from the source S1 gives the displacement to particle P, which is given by

y1 = A1 (⍵t + k * x1)...(a)

This is the wave equation at distance x1.

Similarly, at distance r2, the wave equation is:

y2 = A2 (⍵t + k * x2)...(b)

These two SHM equations (a and b) represent the displacement made by a particle P during oscillation caused by the two waves on it.

This means two SHMs are happening on particle P.

Now, put the value of (a) and (b) in ynet⇾ = y1⇾ + y2⇾

= ynet⇾ = A1 (⍵t + k * x1) + A2 (⍵t + k * x2)

Here, we can find the resultant SHM because of two oscillations caused on P

= Anet (⍵t + Ө)

Here, Anet = net amplitude, and

Ө = Phase change occurred, and

The phase difference, ф = k (x2 - x1)

We know that, k = \[\frac{2 \pi}{\lambda}\] , and path difference between two waves, Δx = x2 - x1

So, we get the phase difference because of path difference as;

Now, using the parallelogram law of vector addition, Anet is:

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Anet = A1² + A2² + 2A1A2Cosф

Where Anet = Net amplitude of particle due to eq (a) and (b) amplitudes.

фbetween the two waves decide the particle’s amplitude, where Δx decides ф.

Resultant angle, tanӨ = \[\frac{A2Sin \Phi}{A1+A2Cos \Phi}\]

We know that more is the amplitude of a wave of light, more is its brightness.

So, we can say that the intensity of light is directly proportional to the square of the amplitude, i.e.,

I ∝ Anet² ⇒ k Anet² , I1 = kA1², and I2 = kA2²

= k(A1² + A2² + 2A1A2Cosф)

= kA1² + kA2² + 2(\[\sqrt{kA1}\] \[\sqrt{kA2}\] Cosф) = I1 + I2 + 2\[\sqrt{I1I2}\]Cosф

Let’s take two cases here:

1. For constructive interference, I = Imax, when Cosф = +1.

So, ф = 2nπ, putting n = 0, 1, 2,.. We get, ф = 0, 2π, 4π,..

Then, Δx = \[\frac{\lambda \Phi}{2 \pi}\] = \[\frac{\lambda}{2 \pi}\] (2nπ) = nλ

Imax = I1 + I2 + 2\[\sqrt{I1I2}\] = (\[\sqrt{I1}\] + \[\sqrt{I2}\])² ⇒ k(A1 + A2)²

2. For destructive interference, I = Imin

Cosф = -1 when ф = (2n -1)π

On putting n = 1, 2,3, we get ф = π, 3π, 5π,....

So, Δx = \[\frac{\lambda}{2 \pi}\] (2n - 1)π ⇒ \[\frac{(2n - 1)}{2}\]λ

Therefore, Imin = I1 + I2 - 2\[\sqrt{I1I2}\] = (\[\sqrt{I1}\] - \[\sqrt{I2}\])²

Q1: Write Examples of the Principle of Superposition.

Ans: We may find the application of the principle of superposition of waves in real-life, which are:

Light

Sound

Water waves

Earth waves

Q2: Two Waves of the Same Frequency, Having an Amplitude in the Ratio of 5: 4. Find the Ratio of I_{max }: I_{min}.

Ans: Here, A_{1} = 5, A_{2} = 4, I_{max }: I_{min} = ?

We know that I1 = kA1² = k(5)² = 25k , and

I2 = k(A2)² = k(4)² = 16k

So, the ratio is I_{max}/I_{min} = 25/16 = 25: 16.

Q3: How Do You Know if Two Waves are in Phase?

Ans: We know that two sources are said to be coherent only when their frequency is the same. If these two waves align perfectly in a way that their phase difference is zero, then they are said to be in phase.

Being in the same phase, they produce a sound with an amplitude equal to the sum of their amplitudes.

Q4: What Happens When Two Waves Have a Phase Difference of 90°?

Ans: This is the case of destructive interference when the maxima of two waves are 180° out of phase.

This happens when a positive displacement of one wave is canceled out by a negative displacement of another wave, such that ф = 90°, i.e., Cosф = 0.