Further Mechanics Tutorial 8 - Energy in SHM

We have seen how we can work out the velocity at any given point by v2 = (2pf )2(A2 – x2).  We also know that kinetic energy, Ek = ˝ mv2, so it doesn’t take a genius to see that:

Ek = ˝ m(2pf )2(A2 – x2)

We can see that the maximum potential energy is found at each end of the swing.  As the pendulum swings away from the rest position, work is done against the restoring force.  If x = 0, the equilibrium position, the restoring force is 0 as well.  At any displacement s we can find the restoring force F.  We can do this by using Newton II, F = ma and a = (2pf )2x.

So the force at any displacement s is given by F = m(2pf )2x.

We know that work done = force ´ distance moved.  However the force is not constant and we need to consider the average force, which is half the maximum force.

Work done = average force ´ displacement = ˝ m(2pf )2x ´ x = ˝ m(2pf )2x2.

So we can work out the potential energy at any point using:

Ep = ˝ m(2pf )2x2

The total energy at any point is simply the sum of the potential and kinetic energy;

Etot = Ep + Ek

          = ˝ m(2pf )2(A2 – x2) + ˝ m(2pf )2x2           

= ˝ m(2pf )2A2