## 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

**of a particle.**

*initial phase**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.*

- At a given position, ( x ) becomes constant and
- At a given instant, ( t ) becomes constant and

__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 –

- The wave velocity ( v = \nu \lambda ) remains constant but the particle velocity ( V )
- Phase of particle velocity ( V )
*.* - 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 –

- Phase of particle acceleration
- 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 )

See numerical problems based on this article.