# Phase of a Wave

## What is meant by Phase of a wave?

Phase of a wave is the quantity which gives complete information of the medium particle at any instant of time at a particular position of wave motion.

A harmonic wave equation is represented as –

y = A \sin ( \omega t - k x + \phi_0 )

In this equation, the term ( \phi_0 ) is the phase angle or initial phase of a particle.

Therefore, phase of a wave is the phase angle of a medium particle at position ( x )  and at time ( t ) .

Hence, phase of a wave is given by –

\phi = ( \omega t - k x + \phi _ 0 )

Therefore, phase of a wave is a periodic function in time basis and position basis both.

1. At a given position, ( x ) becomes constant and the phase changes with time ( t ) .
2. At a given instant, ( t ) becomes constant and the phase changes with position i.e. distance ( x ) .

### Phase change of a wave with time

The phase of wave at position ( x )  and time ( t )  is given by –

\phi = ( \omega t - k x + \phi _ 0 )

Taking ( x ) as constant, differentiating this equation with respect to time ( t ) , we will get –

\left ( \frac { \Delta \phi }{ \Delta t } \right ) = \omega

Thus, the phase change at a given position in time ( \Delta t ) will be –

\Delta \phi = \omega \Delta t = \left ( \frac { 2 \pi }{ T } \right ) \Delta t

Hence, we can define the time period of a wave as the time in which the phase of a medium particle changes by ( 2 \pi ) .

### Phase change of a wave with position

Taking ( t ) as constant, differentiating the equation with respect to position ( x ) , we will get –

\left ( \frac { \Delta \phi }{ \Delta x } \right ) = - k

Thus, at any instant ( t ) , the phase difference between two particles separated by a distance ( \Delta x ) will be –

\Delta \phi = - k \Delta x = - \left ( \frac { 2 \pi }{ \lambda } \right ) \Delta x

Hence, we can define the wavelength of a wave as the distance between two points having a phase difference of \left ( 2 \pi \right ) at a given instant.

## Particle velocity

The velocity with which the particles of medium vibrate about their mean position is called particle velocity.

Particle velocity ( V ) in a wave motion is different from the wave velocity ( v ) . It is obtained by differentiating the wave equation with respect to time. A harmonic wave equation is represented by –

y = A \sin ( \omega t - k x )

Taking ( x )  as a constant, differentiating this equation with respect to time ( t ) , we will get the particle velocity. Therefore, particle velocity at a given position ( x ) will be –

V = \left ( \frac { d y }{ d t } \right ) = \omega A \cos ( \omega t - k x )

But, \quad \sin \left ( \frac {\pi}{2} + \theta \right ) = \cos \theta

So, \quad V = \omega A \sin \left [ \left ( \omega t - k x \right ) + \frac { \pi }{ 2 } \right ]

Following points must be noted about particle velocity –

1. The wave velocity ( v = \nu \lambda ) remains constant but the particle velocity ( V )  changes harmonically.
2. Phase of particle velocity ( V )  is \left ( \frac { \pi }{ 2 } \right ) radians ahead of the phase of displacement ( x ) .
3. The maximum particle velocity is called velocity amplitude.

Therefore, value of velocity amplitude will be –

V _ 0 = \omega A = \left ( \frac { 2 \pi }{ T } \right ) A

\text {Velocity amplitude} = \left ( \frac { 2 \pi }{ T } \right ) \ \times \ \text {Displacement amplitude}

### Particle acceleration

The acceleration with which the particles of medium vibrate about their mean position is called particle acceleration.

Particle acceleration is obtained by differentiating the wave equation with respect to time. A harmonic wave equation is represented by –

y = A \sin ( \omega t - k x )

Taking ( x )  as constant, differentiating this equation with respect to time ( t ) , we will get the particle acceleration.

Therefore, particle acceleration at a given position ( x ) will be –

a = \left ( \frac { d V }{ d t } \right )

Therefore, \quad a = - \omega ^ 2 A \sin ( \omega t - k x ) = - \omega ^ 2 y

But, \quad \sin \left ( \pi + \theta \right ) = - \sin \theta

Therefore, \quad a = \omega ^ 2 A \sin [ \left ( \omega t - k x \right ) +  \pi ]

Following points must be noted about particle acceleration –

1. Phase of particle acceleration is ahead of the phase of particle displacement by ( \pi ) radians.
2. The maximum value of particle acceleration is called acceleration amplitude.

Therefore, value of acceleration amplitude will be –

a _ 0 = \omega ^ 2 A = \left ( \frac { 2 \pi }{ T } \right ) ^ 2 A

\text {Acceleration amplitude} = \left ( \frac { 2 \pi }{ T } \right ) ^ 2 \ \times \ \text {Displacement amplitude}

### Slope of wave curve

Particle velocity at a given position ( x ) of wave is given by –

V = \left ( \frac { d y }{ d t } \right ) = \omega A \cos ( \omega t - k x ) ….. (1)

Taking time ( t ) as constant and differentiating the wave equation [ y = A \sin ( \omega t - k x ) ] with respect to position ( x ) , we will get –

\left ( \frac {dy}{dx} \right ) = - k A \cos ( \omega t - k x ) …… (2)

Now, dividing equation (1) with equation (2), we will get –

\left ( \frac { V }{ d y / d x } \right ) = \left [ \frac { \omega A \cos \left ( \omega t - k x \right ) }{ - k A \cos \left ( \omega t - k x \right )} \right ]

= - \left ( \frac { \omega }{ k } \right )

But, angular velocity ( \omega = 2 \pi \nu ) and propagation constant ( k = \frac {2 \pi}{\lambda} )

Therefore, \quad \left ( \frac { V }{ d y / d x } \right ) = - \frac { 2 \pi \nu }{ 2 \pi / \lambda } = - \lambda \nu = - v

Or, \quad V = - v \left ( \frac { d y }{ d x } \right )

Or, \quad \left ( \frac {dy}{dx} \right ) = - \left ( \frac {V}{v} \right )

Therefore, \quad \text {Slope of displacement curve} = - \left ( \frac { \text {Particle velocity}}{ \text {Wave velocity}} \right )