# Suppose two linear waves of equal amplitude and frequency have a phase difference \phi as they...

## Question:

Suppose two linear waves of equal amplitude and frequency have a phase difference {eq}\phi {/eq} as they travel in the same medium. They can be represented by:

{eq}D_1 = A \sin (kx - \omega t) {/eq}

{eq}D_2 = A \sin (kx - \omega t + \phi) {/eq}

(a) Use trigonometric identity {eq}\sin\theta_1 + \sin\theta_2=2\cdot \sin (\frac{1}{2}(\theta_1 + \theta_2)) \cdot \cos (\frac{1}{2}(\theta_1 - \theta_2)) {/eq} to show that the resultant wave is given by:

{eq}D = 2\cdot A\cdot \cos \frac{\phi}{2}\cdot \sin (kx - \omega t + \frac{\phi}{2}) {/eq}

(b) What is the amplitude of this resultant wave? Is the wave purely sinusoidal, or not?

(c) Show that constructive interference occurs if {eq}\phi = 0, 2\pi, 4\pi, {/eq} and so on. Show that destructive interference occurs if {eq}\phi = \pi, 3\pi, 5\pi, {/eq} and so on.

(d) Describe the resultant wave, by equation and in words, if {eq}\phi = \frac{\pi}{2}. {/eq}

## Superposition of Waves:

The principle of superposition of waves states that if there 2 or more waves traveling in a region of space, then the effect of each wave at any point in space and time, simply add up like vectors.

{eq}y=y_1+y_2+y_3... {/eq}

Here,

- {eq}y_1 {/eq} is the displacement caused by wave 1.

- {eq}y_2 {/eq} is the displacement caused by wave 2.

- {eq}y_3 {/eq} is the displacement caused by wave 3 and so on.

## Answer and Explanation:

The equations of individual waves are:

{eq}d_1=A\sin\left ( kx-\omega t \right )\\ d_2=A\sin\left ( kx-\omega t+\phi \right ) {/eq}

According to the principle of superposition, the net displacement due to two or more waves is equal to the vector sum of the individual displacements caused by the waves.

{eq}d=d_1+d_2 {/eq}

Here,

- {eq}d_1 {/eq} is the displacement of the first wave.

- {eq}d_2 {/eq} is the displacement of the second wave.

Therefore,

{eq}\begin{align*} d&=d_1+d_2\\ &=A\sin\left ( kx-\omega t \right )+A\sin\left ( kx-\omega t+\phi \right )\\ &=A\left ( \sin\left ( kx-\omega t \right )+\sin\left ( kx-\omega t+\phi \right ) \right )\\ &=A\left (2\sin\left (\dfrac{ kx-\omega t+kx-\omega t+\phi}{2} \right )\cos\left (\dfrac{ kx-\omega t-kx+\omega t-\phi}{2} \right ) \right )\\ &=2A\sin\left ( kx-\omega t+\dfrac{\phi}{2} \right )\cos\left (\dfrac{ -\phi}{2} \right ) \\ &=2A\sin\left ( kx-\omega t+\dfrac{\phi}{2} \right )\cos\left (\dfrac{ \phi}{2} \right ) \end{align*} {/eq}

**b)**

The amplitude of the resultant wave is:

{eq}A'=2A\cos\left (\dfrac{ \phi}{2} \right ) {/eq}

In terms of new amplitude, the resultant wave becomes:

{eq}d=A'\sin\left ( kx-\omega t+\dfrac{\phi}{2} \right ) {/eq}

which is purely **sinusoidal.**

**c)**

Constructive interference occurs when the amplitude of the resultant wave is maximum i.e

{eq}A'=2A\cos\left (\dfrac{ \phi}{2} \right ) {/eq}

is maximum.

This occurs when {eq}\cos\left (\dfrac{ \phi}{2} \right ) {/eq} is maximum.

The maximum values of {eq}\cos\,\theta {/eq} occurs when {eq}\theta=0,\pi,2\pi,3\pi... {/eq}

Therefore,

{eq}\phi/2=0,\pi,2\pi,3\pi...\\ \therefore \phi=0,2\pi,4\pi,6\pi ... {/eq}

Similarly, desctructive interference occurs when {eq}A' {/eq} is minimum. This occurs when {eq}\cos\left (\dfrac{ \phi}{2} \right ) {/eq} is zero.

Since {eq}\cos\,\theta=0 {/eq} for {eq}\theta=\pi/2,3\pi/2,5\pi/2...\\ \therefore \phi/2=\pi/2,3\pi/2,5\pi/2...\\ \therefore \phi=\pi, 3\pi, 5\pi {/eq}

**d)**

For {eq}\phi=\dfrac{\pi}{2} {/eq}, the amplitude of the resultant wave becomes:

{eq}A'=2A\cos\left (\dfrac{ \pi}{4} \right ) =\dfrac{A}{\sqrt 2} {/eq}

The equation of the resultant wave becomes:

{eq}d=\dfrac{A}{\sqrt 2}\sin\left ( kx-\omega t+\dfrac{ \pi}{4} \right ) {/eq}

Therefore, the resultant wave has an amplitude lower than either of the original waves. The angular frequency of the resultant wave is the same as that of the original waves.

#### Learn more about this topic:

from MEGA Physics: Practice & Study Guide

Chapter 15 / Lesson 10