Simple Harmonic Motion

If an object is free to oscillate around an equilibrium point and experiences a linear restoring Force, it performs a motion which we refer to as 'Simple Harmonic Motion'. 

In order for an object to display simple harmonic motion (SHM), two key factors must be met.

It must:-

1. Store Potential Energy in some form

2. Have mass in order to possess Kinetic Energy

The most simple example of an object undergoing Simple Harmonic Motion is a mass suspended from a coiled spring :-

In the above diagram, the mass is caused to oscillate by being pulled down and released. When the mass is moved away from the equilibrium point, energy is stored as elastic potential energy within the Spring. 

For a spring following this type of motion :-

There is always an unbalanced Force acting on the mass (except when passing the equilibrium point)

This Force is always in the opposite direction to the motion of the Mass

These two factors apply to all objects undergoing Simple Harmonic Motion. 

We call the opposing Force, the restoring Force. 

The video below gives a summary of Simple Harmonic Motion, whilst giving demonstrations of several experiments

Defining Simple Harmonic Motion 

The following is the standard written definition of Simple Harmonic Motion.

"When an object is displaced from its equilibrium or rest position, and the unbalanced Force is proportional to the Displacement of the object and acts in the opposite direction, the motion is said to be Simple Harmonic. "

Force against Displacement

As seen in the statement above, the unbalanced Force experienced by the Mass undergoing SHM is  proportional to the Displacement. If this Force is measured experimentally, then the following graph can be generated :-

As can been seen above, the graph shows also that the force acts in the opposite direction to the displacement. By introducing a Force constant, we can generate the following :-

where :- 

F = The restoring Force ( N )

k = The Force constant ( Nm-1 )

x = The Displacement ( m )

Note - The negative sign in the formula shows that the Force acts in the opposite direction to the displacement.

The video below gives a brief overview of Hookes Law and how it applies to springs in SHM

Non Examinable Derivation - SHM in terms of Acceleration

By equating the above formula with Newton's Second Law, we can derive a relationship for Acceleration :-

Example 1 - 

A 1.0 kg mass attached to a horizontal spring  ( k = 5.0 Nm-1 ) is performing SHM on a horizontal, frictionless surface. What is the Acceleration of the Mass when its displacement is 4.0 cm ? 

The above calculation is based on a spring ( with a spring constant = k ) obeying Hooke's Law, but in order to apply our understanding of SHM universally, we must convert this into general terms.

In general terms the constant ( k / m ) is related to the period of the motion by :-

giving the general formula for acceleration in SHM as :-

Note - Omega in this case is Angular Frequency , whereas in unit one  (Rotational Dynamics), we used the same symbol for Angular Velocity when working with circular motion. This is because the two concepts are very much related, as will be shown below.

Examinable Derivation - Equations of Motion and SHM

For ease of explanation, we can compare on object undergoing SHM with an object moving in a circle. 

Consider an object Q travelling with a constant linear speed (v) anti-clockwise around the circle above. If we project a line from Q to the vertical axis we can find its vertical displacement on the Y axis (line OP).

The Displacement on the vertical axis is equal to :- 

By differentiation, an equation of the Velocity can be obtained :-

Note - Differentiation of Trig Identities can be found within the CfE Higher Maths course

By further differentiation, the acceleration can also be found :-

By combining the two equations above and by using the trig identity below, the following can be derived :-

Velocity -

The particular equation used for each motion depends on the starting conditions. this means that if :-

y = 0 at t = 0              y =   A Sin wt

y = A at t = 0              y =  A Cos wt

Note - for assessment, you must be able to show by differentiation how both the sin and cos variant can be used to find velocity and acceleration.

Example 1 - 

An object is vibrating with Simple Harmonic Motion of amplitude 0.02 m and a frequency of 5.0 Hz. Assume that at t = 0, y = 0 and that the motion starts in the positive y direction.

1. Calculate the maximum values of the Velocity and Acceleration

2. Calculate the Velocity and Acceleration when Displacement is equal to 0.008 m 

3. Find the Time taken for the object to move from an equilibrium position to a Displacement of 0.012 m

Simple Pendulum as SHM proof

The following diagram shows a simple pendulum undergoing Simple Harmonic Motion as well as the Forces acting upon it

Note - assumptions made

1. Inextensible string of length L

2. Negligible Mass string

In the case of the pendulum above, the restoring Force is equal to the component of the Weight acting towards the centre -   F = - mg Sinq 

If the angle is small ( less than ~ 10o  ) then sinq  = q  in radians and therefore q  = x / L 

Therefore if the above relationship is shown to be true experimentally, the simple pendulum does follow Simple Harmonic Motion.