• Simple harmonic motion (SHM) is defined as follows:

The motion of an object whose acceleration is directly proportional but opposite in direction to the object's displacement from a central equilibrium position

• An object is said to perform?simple harmonic oscillations?when all of the following apply:
  • The oscillations are?isochronous
  • There is a central equilibrium point
  • The object's displacement, velocity and acceleration change continuously
  • There is a?restoring force?always directed towards the equilibrium point
  • The magnitude of the restoring force is proportional to the displacement

• The defining conditions of simple harmonic oscillations are that the?restoring force?and the?acceleration?must always be:
  • Directed towards the equilibrium position, and hence, is always in the?opposite?direction to the?displacement
  • Directly?proportional?to the?displacement

a?∝ ?x

• Where:
  • a?= acceleration (m s^?2)
  • x?= displacement (m)

The Restoring Force

• One of the defining conditions of simple harmonic motion is the existence of a restoring force
• Examples of restoring forces are:
  • The component of the weight of a pendulum's bob that is parallel and opposite to the displacement of the bob
  • The force of a spring, whose magnitude is given by?Hooke's law

For a pendulum, the restoring force is provided by the component of the bob's weight that is perpendicular to the tension in the pendulum's string. For a mass-spring system, the restoring force is provided by the force of the spring.

• For a mass-spring system in simple harmonic motion, the relationship between the restoring force and the displacement of the object can be written as follows:

F = – kx

• Where:
  • F?= restoring force (N)
  • k?= spring constant (N m^–1)
  • x?= displacement from the equilibrium position (m)

Graph of force against displacement for an object oscillating with SHM

• Force and displacement in SHM have a?linear?relationship where the gradient of the graph represents the constant
  • In this case, the spring constant?k
• An object in SHM will also have a restoring force to return it to its equilibrium position
  • This restoring force will be directly proportional, but in the?opposite direction, to the displacement of the object from its equilibrium position

Acceleration & Displacement

• According to?Newton's Second Law, the net force on an object is directly proportional to the object's acceleration,?F?∝?a?for a constant mass

F?=?ma

• Where
  • F?= force (N)
  • m?= mass (kg)
  • a?= acceleration (m s^?2)
• Since?F?=?ma?(Newton's second law), and?F?= ?kx?(Hooke's law), the equations can be set as equal to one another:

ma = – kx

• Rearranging to show the relationship between acceleration and displacement gives:

• This equation shows that
  • There is a?linear?relationship between the?acceleration?of the object moving with simple harmonic motion and its?displacement?from its equilibrium position

• The?minus?sign shows that when the mass on the spring is displaced to the?right
  • The direction of the acceleration is to the?left?and vice versa
  • In other words,?a?and?x?are always in opposite directions to each other

• This equation shows acceleration is?directly proportional?but in the?opposite direction?to displacement for an object in SHM

a?∝ ?x

• Therefore, it can be stated that:

a?= ?kx

• Where
  • a?= acceleration
  • k?= is a constant but in this instance?not?the spring constant
  • x?= displacement

• Note that in physics,?k?is the standard letter used for an?undefined constant

Graph of acceleration against displacement for an object oscillating with SHM

Step 1: List the known quantities

• • Amplitude of the oscillations,?x0?= 4.0 cm = 0.04 m
  • Maximum acceleration,?a?= 2.0 m s^–2
  • Displacement,?x?= 1.0 cm = 0.01 m

Remember to convert the amplitude of the oscillations and the displacement from centimetres (cm) into metres (m)

Step 2: Recall the relationship between the maximum acceleration?a?and the displacement?x

• • The maximum acceleration?a?occurs at the position of maximum displacement?x?=?x₀

a = – kx₀

Step 3: Rearrange the above equation to calculate the constant of proportionality?k

Step 4: Substitute the numbers into the above equation

k?= – 50 s^–2

Step 5: Use this value of?k?to calculate the acceleration?a'?when the displacement is?x?= 0.01 m

a'?= –?kx

a'?= – (– 50) s^–2× 0.01 m

a'?= 0.50 m s^–2