# Energy in Simple Harmonic Motion

## How energy change occurs in Simple Harmonic Motion?

The total energy in Simple Harmonic Motion is consisting of two parts –

### Kinetic energy in Simple Harmonic Motion

Velocity of a particle in simple harmonic motion is given by –

v = - \omega A \sin ( \omega t + \phi _ { 0 } )

Hence, kinetic energy of the particle at any position will be –

K = \left ( \frac { 1 }{ 2 } \right ) m v ^ 2

= \left ( \frac {1}{2} \right ) m [ - \omega A \sin ( \omega t + \phi _ { 0 } ) ]^2

= \left ( \frac { 1 }{ 2 } \right ) m \omega ^ 2 A ^ 2 \sin ^ 2 ( \omega t + \phi _ { 0 } )

But, \quad A ^ 2 \sin ^ 2 ( \omega t + \phi _ { 0 } ) = ( A ^ 2 - x ^ 2 )

Therefore, from equation (1), we will get –

K = \left ( \frac {1}{2} \right ) m \omega^2 ( A ^ 2 - x ^ 2 )

For restoring force or force constant in simple harmonic motion –

\omega ^2 = \left ( \frac { k }{ m } \right )

Or, \quad K = \left ( \frac {1}{2} \right ) k ( A ^ 2 - x ^ 2 )

Therefore, kinetic energy of particle in simple harmonic motion has following characteristics –

1. Variation of kinetic energy is periodic in nature. But it is not a simple harmonic activity.
2. Kinetic energy varies from minimum value ( at extreme positions ) to maximum value ( at mean position ) and vice versa. It happens twice in one complete cycle.
3. Therefore, kinetic energy completes 2 cycles in duration of 1 cycle of simple harmonic motion.

### Potential energy in Simple Harmonic Motion

For a particle in simple harmonic motion, the restoring force is given by –

F = - k x

Let, the particle is further displaced through a small distance ( dx ) . Then work done against the restoring force will be –

d W = - F d x = (+ k x ) d x

Therefore, total work done in moving the particle from mean position to a point of displacement ( x )  is given by –

W = \int d W = \int\limits _ { 0 }^{ x } (+ k x) d x

Or, \quad W = k \left [ \frac { x ^ 2 }{ 2 } \right ] _ { 0 }^{ x } = \left ( \frac { 1 }{ 2 } \right ) k x ^ 2

This work is stored as potential energy in the particle.

Therefore, \quad U = \left ( \frac { 1 }{ 2 } \right ) k x ^ 2

= \left ( \frac { 1 }{ 2 } \right ) m \omega ^ 2 x ^ 2

= \frac { 1 }{ 2 } m \omega ^ 2 A ^ 2 \cos ^ 2 ( \omega t + \phi _ { 0 } )

Therefore, for a particle in simple harmonic motion, potential energy has following characteristics –

1. Variation of potential energy is periodic in nature but it is not a simple harmonic activity.
2. Potential energy varies from minimum value ( at mean position ) to maximum value ( at extreme positions ) and vice versa. It happens twice in one cycle of simple harmonic motion.
3. Therefore, potential energy completes 2 cycles in duration of 1 cycle of simple harmonic motion.

### Total energy in Simple Harmonic Motion

At any position of displacement, the total energy of simple harmonic motion is given by –

E = ( K + U )

= \left [ \left ( \frac { 1 }{ 2 } \right ) k ( A ^ 2 - x ^ 2 ) + \left ( \frac { 1 }{ 2 } \right ) k x ^ 2 \right ]

= \left ( \frac {1}{2} \right ) k A^2

= \left ( \frac { 1 }{ 2 } \right ) m \omega ^ 2 A ^ 2

= \left ( \frac {1}{2} \right ) m \left ( \frac {2 \pi}{T} \right )^2 A^2

= 2 \pi ^ 2 m \nu ^ 2 A ^ 2

Because, \quad \left [ \left ( \frac {1}{T} \right ) = \nu \right ]

Therefore, in simple harmonic motion

1. Total mechanical energy is directly proportional to the mass ( m )  of the particle.
2. It is proportional to the square of frequency ( \nu ) .
3. Total mechanical energy is directly proportional to the square of amplitude.
4. It is independent of time ( t ) and displacement ( x ) .

## Variation in total energy along a cycle

Following table shows the variation of kinetic energy ( K ) and potential energy ( U ) at different displacement positions ( x ) .

 Displacement ( x ) ( x = - A ) ( x = 0 ) ( x = + A ) Kinetic energy ( K ) 0 \left ( \frac { 1 }{ 2 } \right ) k A ^ 2 0 Potential energy ( U ) \left ( \frac { 1 }{ 2 } \right ) k A ^ 2 0 \left ( \frac { 1 }{ 2 } \right ) k A ^ 2 Total energy ( E ) \left ( \frac { 1 }{ 2 } \right ) k A ^ 2 \left ( \frac { 1 }{ 2 } \right ) k A ^ 2 \left ( \frac { 1 }{ 2 } \right ) k A ^ 2

The variations of ( K ) and ( U ) with respect to displacement ( x ) is plotted as shown in figure.

From figure, it is evident that –