__What are equations of Simple Harmonic Motions?__

*An equation governing a simple harmonic motion and representing its properties is called a simple harmonic motion equation.*

*Simple harmonic motion equation gives displacement of particle executing SHM at any instant after time ( t ) from the mean position.*

If a particle executes a *uniform circular motion*, its projection on a fixed diameter will perform a *simple harmonic motion.* Hence, simple harmonic motion equation is easily obtained from the basics of a *uniform circular motion* of a particle.

Consider about the figure shown below. Particle P is describing a uniform circular motion with constant *angular velocity* ( \omega ) in anti-clockwise direction in a circle of radius ( A ) . Suppose –

- At time ( t = 0 )
*,*the particle is at point A and ( \angle { XOA } = \phi _{ 0 } ) - After time ( t = t ) , the particle reaches at the point P and ( \angle { AOP } = \omega t ) .
- A perpendicular PN is drawn on fixed diameter XX' .

When the particle moves on the circumference of the circular path, point N will move “to & fro” on the diameter XX' .

*Hence, motion of the foot of perpendicular N will be a simple harmonic motion.*

__Displacement in a Simple Harmonic Motion__

*Displacement* of point N from the mean position O in time ( t ) will be –

x = ON

In right angle triangle \triangle {ONP} –

\angle { PON } = ( \omega t + \phi _ { 0 } )

Therefore, \quad \left ( \frac { ON }{ OP } \right ) = [ \cos ( \omega t + \phi _{ 0 } ) ]

Or, \quad \left ( \frac { x }{ A } \right ) = [ \cos ( \omega t + \phi _{ 0 } ) ]

Or, \quad x = [ A \cos ( \omega t + \phi _ { 0 } ) ] …….. (1)

*This equation is called simple harmonic motion equation.*

This equation gives displacement of a particle executing SHM at any instant after time ( t ) from initial position.

Quantity ( \omega t + \phi _ { 0 } ) is called phase of the particle and ( \phi _{ 0 } ) is called ** initial phase **or

*phase constant.*Quantity ( A ) is called **amplitude **of the motion. When the particle is at extreme position, the maximum displacement is ( \pm A ) .

__Velocity in Simple Harmonic Motion__

Simple harmonic motion equation is –

x = A \cos ( \omega t + \phi _ { 0 } ) ……… (1)

Differentiating this equation, we will get –

\left ( \frac { dx }{ dt } \right ) = - \omega A \sin ( \omega t + \phi _ { 0 } )

From *kinematic equations* of motion, the *velocity* of any particle is –

v = \left ( \frac { dx }{ dt } \right )

Therefore, \quad v = - \omega A \sin ( \omega t + \phi _ { 0 } ) ……… (2)

Equation (2) can also be written as –

v = - \omega \sqrt { A ^ 2 \sin ^ 2 ( \omega t + \phi _ { 0 } )}

= - \omega \sqrt { A ^ 2 [ 1 - \cos ^ 2 ( \omega t + \phi _ { 0 } ) ]}

= - \omega \sqrt { [ A ^ 2 - A ^ 2 \cos ^ 2 ( \omega t + \phi _ { 0 } )]}

= - \omega \sqrt { ( A ^ 2 - x ^ 2 )} …….. (3)

From this equation, we have –

When velocity is maximum, displacement is minimum. But minimum value of displacement is zero at mean position.

So putting ( x = 0 ) in equation (3).

v _ { max } = \omega A

A graph for variation of velocity with respect to displacement is shown in figure.

__Acceleration in Simple Harmonic Motion__

Expression for velocity in simple harmonic motion is –

v = - \omega A \sin ( \omega t + \phi _ { 0 } ) …….. (2)

Differentiating this equation, we get –

\left ( \frac { d v }{ d t } \right ) = - \omega ^2 A \cos ( \omega t + \phi _ { 0 } ) = - \omega ^ 2 x

From kinematic equations, *acceleration* is given as –

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

Therefore, \quad a = - \omega ^ 2 x …….. (4)

Here, minus sign indicates that acceleration is directed in opposite direction of undergoing displacement.

From equation it is evident that, acceleration will be maximum when displacement is maximum. But maximum possible value of displacement is ( \pm A ) *.*

Therefore, \quad a _ {max} = \omega ^ 2 A

A graph for variation of acceleration verses displacement is shown in figure.

__Time Period__

*Time taken to complete one cycle of an oscillating body is called its time period.*

In one full cycle particle moves through *angular displacement* of ( 2 \pi ) radians with uniform angular speed of ( \omega ) .

Therefore, time period of simple harmonic motion will be –

T = \left ( \frac { 2 \pi }{ \omega } \right ) ……… (5)

Or, \quad T = 2 \pi \left ( \frac { 1 }{ \omega } \right )

From equation (4), we have –

\omega = \sqrt {\left ( \frac {a}{x} \right ) }

Therefore, \quad T = 2 \pi \left ( \frac { x }{ a } \right ) ………. (6)

Or, \quad T = 2 \pi \sqrt { \frac {\text {Displacement}}{\text {Acceleration}}}

For a *Simple harmonic motion* –

( F = m a ) \quad And \quad ( F = - k x )

So, \quad \left ( \frac { x }{ a } \right ) = \left ( \frac { m }{ k } \right )

Therefore, \quad T = 2 \pi \sqrt { \left ( \frac { m }{ k } \right ) } ……. (7)

__Frequency__

*Number of oscillations completer per unit time is called frequency of Simple Harmonic Motion.*

Frequency is the reciprocal of time period.

Therefore frequency \quad \nu = \left ( \frac { 1 }{ T } \right )

From equation (6), we get –

\nu = \left ( \frac { 1 }{ 2 \pi } \right ) \sqrt {\left ( \frac { a }{ x } \right )} ……… (8)

From equation (7), we get –

\nu = \left ( \frac { 1 }{ 2 \pi } \right ) \sqrt {\left ( \frac { k }{ m } \right )} ……… (9)

__Angular Frequency__

*Angular displacement per unit time is called angular frequency or angular velocity of a periodic motion.*

In unit time, angular displacement is –

\nu \times 2 \pi

Therefore, angular frequency is obtained by multiplying the circular frequency ( \nu ) by ( 2 \pi ) .

Hence, angular frequency \quad \omega = 2 \pi \nu = \left ( \frac { 2 \pi }{ T } \right )

See numerical problems based on this article.

__Restoring Force__

*When a particle execute a simple harmonic motion, a force always acts on particle which has a tendency to bring the particle in its mean position. This force is called restoring force. *

For a simple harmonic motion –

F = - k x

*Restoring force is proportional to displacement of the particle from its mean position.*

From acceleration of simple harmonic motion –

a = - \omega ^ 2 x

Therefore, \quad F = m a = - m \omega ^ 2 x

Or, \quad k x = m \omega ^ 2 x

Here, ( k ) is called *force constant.*

Therefore, force constant of simple harmonic equation is –

k = m \omega ^ 2

Also, \quad \omega = \sqrt { \left ( \frac { k }{ m } \right ) }

A graph for variation of restoring force verses displacement is shown in figure.

__Different forms of Simple Harmonic Equations__

Depending upon the initial conditions of particle, general equations for simple harmonic motion can be expressed in different forms. These are illustrated in following figures.

**(1) Case – A ( Initially particle is at extreme right & start moving in negative x direction )**

Consider about the figure in **Case -A**

- Initial position of particle is at extreme right position. This means, when ( t = 0 ) , position of particle is at ( x = + A ) .
- Thus, the particle moves in
**“negative x direction”**i.e. in direction from ( x = + A ) to ( x = 0 )

Then, general form of this simple harmonic equation will be –

x = A \cos ( \omega t + \phi _ { 0 } ) ……… (1A)

**(2) Case – B ( Initially particle is at extreme left & start moving in positive x direction)**

Consider about the figure in **Case -B**

- Initial position of particle is at extreme left position. This means, when ( t = 0 ) , position of particle is at ( x = - A )
- Thus, the particle moves in
**“positive x direction”**i.e. in direction from ( x = - A ) to ( x = 0 )

Then, general form of this simple harmonic equation will be –

x = - A \cos \left ( \omega t + \phi _ { 0 } \right ) …….. (1B)

**(3) Case – C (Initially particle is at mean position & start moving in negative x direction)**

Consider about the figure in **Case -C**

- The initial position of particle is at mean position. This means, when ( t = 0 ) , position of particle is at ( x = 0 )
- Thus, the particle moves in
**“negative x direction”**i.e. in direction from ( x = 0 ) to ( x = - A )

Then, general form of this simple harmonic equation will be –

x = - A \sin ( \omega t + \phi _ { 0 } ) ………. (1C)

**(4) Case – D (Initially particle is at mean position & start moving in positive x direction)**

Consider about the figure in **Case -D**

- The initial position of particle is at mean position. This means, when ( t = 0 ) , position of particle is at ( x = 0 )
- Thus, the particle moves in
**“positive x direction”**i.e. in direction from ( x = 0 ) to ( x = + A )

Then, general form of this simple harmonic equation will be –

x = A \sin ( \omega t + \phi _ { 0 } ) ……… (1D)

See numerical problems based on this article.