Principle of Superposition of Waves

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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. 

                           

The films produced in water droplets have a wave nature.     

                               

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.

 

Example of 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.

 

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.       

                                                                  

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;

\[\overrightarrow{y_{net}}  =  \overrightarrow{y_1} +  \overrightarrow{y_2}\]

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 \[\overrightarrow{y_1} +  \overrightarrow{y_2}\].        

 

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:

 

(Image will be Uploaded soon)           

              

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 \[\overrightarrow{y_{net}}  =  \overrightarrow{y_1} +  \overrightarrow{y_2}\]

\[= \overrightarrow{y_{net}}\] = 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;              

   

\[\phi = \frac{2 \pi}{\lambda} \Delta x \]

 

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

Anet = A1² + A2² + 2A1A2Cosф 

 

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

 

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

 

Resultant angle, \[tan \theta = \frac{A_2 Sin\theta}{A_1 + A_2 Cos \theta}\]

 

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., 

 

\[I \alpha A_{net}^2 \Rightarrow  k A_{net}^2, I_1 = k A_1^2, \] and \[I_2 = k A_2^2, \]

 

= \[k (A_1^2 + A_2^2 + 2A_1A_2 Cos \theta)\] = \[kA_1^2 + kA_2^2 + 2(\sqrt{kA_1^2} \sqrt{kA_2^2} Cos \theta)\]

 

=\[ I_1 + I_2 + 2 \sqrt{I_1I_2} cos \theta \]

 

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, \[\triangle x = \frac{\lambda \phi}{2\pi} = \frac{\lambda}{2\pi} (2n\pi) = n\lambda\]

 

\[I_{max} = I_1 + I_2 + 2\sqrt{I_1I_2} = (\sqrt{I_1} + \sqrt{I_2})^2  \Rightarrow k (A_1 + A_2)^2\]

 

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, \[\triangle x = \frac{\lambda \phi}{2\pi} (2n - 1)\pi = \frac{(2n - 1)\lambda}{2} \]

 

Therefore, \[I_{max} = I_1 + I_2 - 2\sqrt{I_1I_2} = (\sqrt{I_1} - \sqrt{I_2})^2  \Rightarrow k (A_1 - A_2)^2\]

 

Consider the following two coherent wave sources: S1 and S2.

The light from these two sources converges on a single point, P.

S1 and S2 go different distances, x1 and x2, respectively.

So, in S1, the displacement is:

y1  = A Sin (⍵t + k * x)....(1)

A = Maximum displacement

⍵= angular frequency of the wave

K= force constant

Likewise, displacement in s2 is, 

 y2  = A Sin (⍵t + k * x).…(2)  

Equation 1 and 2 represents the wave equation

Because these two waves are in the same medium, they have the same speed.

Both S1 and S2 have the same value.

Because the frequency f is the same for both, S1 and S2 are referred to as coherent sources.

Assume that these travel distances x1 and x2 are equal and that particle P is reached in time t.

The movement of particle P is determined by the wave emitted from source S1.

y1  = A1 (⍵t + k * x1)...(a) [wave equation at distance x1}

Likewise, wave equation at distance x2 will be- 

y2  = A2 (⍵t + k * x2)...(a)

Now putting the value of equation (a) and (b) in\[\overrightarrow{y_{net}}  =  \overrightarrow{y_1} +  \overrightarrow{y_2}\]

= \[\overrightarrow{y_{net}}\[ = A1 (⍵t + k * x1) +  A2 (⍵t + k * x2)

Because of two oscillations on P, the resultant SHM can be found here.

Anet  (⍵t + Ө)

Anet = net amplitude of the wave

Ө = Phase change occurred

We get the phase difference as-

\[\phi = \frac{2 \pi}{\lambda} \Delta x \]

Constructive Interference- When two waves are superimposed, constructive interference occurs, resulting in a wave with a cumulatively larger amplitude.

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\[\sqrt{I_1I_2}\] = (\[\sqrt{I_1}\] + \[\sqrt{I_2}\])² ⇒ k(A1 + A2

Destructive Interference- The two waves combine in destructive interference but cancel out (like adding a positive and negative number).

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

 

The majority of waves do not appear to be straightforward. A simple harmonic oscillation can produce simple waves, which have a sinusoidal shape. Complex waves are more intriguing, even beautiful, but they appear to be intimidating. The majority of waves appear complex because they are made up of multiple simple waves that have been added together. Fortunately, the principles for adding waves are straightforward.


When multiple waves arrive at the same location at the same time, they superimpose on each other. When waves collide, the disturbances of the waves are superimposed, a phenomenon known as superposition. Each disruption is associated with a force, which adds up. If the disturbances all follow the same path, the final wave is simply the sum of the individual waves' disturbances—that is, their amplitudes add up.


Understanding the Principle of superposition with Overlapping of Waves

The principle of superposition states that overlapping waves combine algebraically to form a resultant wave. According to the principle, overlapping waves (with frequencies f1,f2,...,fn) do not interfere with each other's motion or journey. As a result, the wave function (y) that labels the medium disturbance can be written as:

y= f1(x - vt) + f2(x - vt)… + fn(x - vt)

Let us say that the wave functions of moving waves are as follows:

y1 = f1(x – vt),

y2 = f2(x – vt)

……….

yn = fn (x–vt)

The wave function characterizing the medium disturbance can then be written as

y= f1(x - vt) + f2(x - vt)… + fn(x - vt)

or, y =\[\sum_{i=0}^n f_i (x - vt)\].

FAQs on Principle of Superposition of Waves

1. Describe the phenomenon of interference.

Interference is the interaction of waves that are correlated or coherent (i.e. "interfere"), either because they emanate from the same source or because they have the same or nearly the same frequency.


Interference effects can be seen in all forms of waves, including light, radio, acoustic, and surface water waves. When two waves collide, their amplitudes (highest absolute value) mix, resulting in interference.

2. What are the conditions required for superposition of waves?

When two or more waves arrive at the same location at the same time, they superimpose on each other. More exactly, when waves collide, their disruptions are overlaid (a phenomenon called superposition). A force, or amplitude, is assigned to each disturbance (and the forces add). If the disturbances all follow the same path, the resulting wave is simply the sum of the individual waves' disturbances.

3. What are the real life applications of superimposed waves?

The use of reflector and director rods in a television aerial; the fading of v.h.f. radio when an aircraft passes overhead and direct and reflected waves are superposed; the blooming of lenses to reduce reflections; and the acoustical design of concert halls are all examples of superposition (it is important to avoid creating dead spots where certain frequencies, or notes, are obliterated by destructive interference; good design will also prevent constructive interference which can locally increase the volume of a particular note).

4. Write Examples of the Principle of Superposition.

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

  1. Light

  2. Sound

  3. Water waves

  4. Earth waves

5. Two Waves of the Same Frequency, Having an Amplitude in the Ratio of 5: 4. Find the Ratio of Imax : Imin.

Here, A1 = 5, A2 = 4, Imax : Imin = ?


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


I2 = k(A2)² = k(4)² = 16k


So, the ratio is \[\frac{I_{max}}{I_{min}} = \frac{25}{16} = 25: 16.\]

6. How Do You Know if Two Waves are in Phase?

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.

7. What Happens When Two Waves Have a Phase Difference of 90°?

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.

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