Energy in SHM
We have seen how we can work out the velocity of a simple harmonic oscillator at any given point by:
v2 = (2pf )2(A2 – x2)
We also know that kinetic energy is given by:
so it doesn’t take a genius to see that:
m(2pf )2(A2 – x2)
can see that the maximum potential energy
is found at each end of the swing of a pendulum. As
the pendulum swings away from the rest position, work is done against the
restoring force. If
the equilibrium position, the restoring force is
At any displacement
find the restoring force
We can do this by using Newton II,
a = (2pf )2x.
So the force at any displacement x is given by:
F = m(2pf )2x.
know that work done = force ×
distance moved in the direction of the force. However the force
is not constant and we need to consider the average
force, which is half the
Work done = average force × displacement = ˝ m(2pf )2x ´ x = ˝ m(2pf )2x2.
we can work out the potential energy at any point using:
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
We can also write this as:
A pendulum bob has a mass of 50.0 g, and the length of the very thin string is 1.55 m from its suspension point to the centre of mass of the bob. The bob is displaced by 5.0 cm to the right of the rest position and is released. The bob swings freely.
(a) Show that the period of the swing is about 2.5 s.
(b) Calculate the angular velocity of the swing.
(c) Calculate the displacement after 1.0 s, referring to the rest position.
(d) Calculate the velocity at this point, stating the direction. Give your answer to an appropriate number of significant figures.
(e) What is the kinetic energy at this point?
(f) What is the maximum kinetic energy? When would the bob have the maximum energy?
(g) Calculate the maximum force that occurs on the bob when it's at the left-hand extremity of the swing. State the direction.
Use g = 9.8 m s-2.