During simple harmonic motion, there is a constant exchange of energy between two forms: potential and kinetic. We can see this by considering the mass–spring system shown in Figure 18.21.
When the mass is pulled to one side (to start the oscillations), one spring is compressed and the other is stretched. The springs store elastic potential energy. When the mass is released, it moves back towards the equilibrium position, accelerating as it goes. It has increasing kinetic energy. The potential energy stored in the springs decreases while the kinetic energy of the mass increases by the same amount (as long as there are no heat losses due to frictional forces). Once the mass has passed the equilibrium position, its kinetic energy decreases and the energy is transferred back to the springs. Provided the oscillations are undamped, the total energy in the system remains constant.

Energy graphs

We can represent these energy changes in two ways. Figure 18.22 shows how the kinetic energy and elastic potential energy change with time. Potential energy is maximum when displacement is maximum (positive or negative). Kinetic energy is maximum when displacement is zero. The total energy remains constant throughout. Note that both kinetic energy and potential energy go through two complete cycles during one period of the oscillation. This is because kinetic energy is maximum when the mass is passing through the equilibrium position moving to the left and again moving to the right. The potential energy is maximum at both ends of the oscillation.

Energy graphs

We can represent these energy changes in two ways. Figure 18.22 shows how the kinetic energy and elastic potential energy change with time. Potential energy is maximum when displacement is maximum (positive or negative). Kinetic energy is maximum when displacement is zero. The total energy remains constant throughout. Note that both kinetic energy and potential energy go through two complete cycles during one period of the oscillation. This is because kinetic energy is maximum when the mass is passing through the equilibrium position moving to the left and again moving to the right. The potential energy is maximum at both ends of the oscillation.

A second way to show this is to draw a graph of how potential energy and kinetic energy vary with displacement (Figure 18.23).
The graph shows that:
- kinetic energy is maximum when displacement $x = 0$
- potential energy is maximum when $x =  \pm {x_0}$
- at any point on this graph, the total energy ($k.e. + p.e.$) has the same value.

It follows that if the maximum speed is ν0 then maximum kinetic energy .$energy - \frac{1}{2}m{v_0}^2$ At this point in the cycle, all the energy is in the form of kinetic energy, so the total energy of the system is:

${E_0} = \frac{1}{2}m{v_0}^2$

Since:

${v_0} = \omega {x_0}$

Then:

${E_0} = \frac{1}{2}m{\omega ^2}{x_0}^2$

Questions

 

20) To start a pendulum swinging, you pull it slightly to one side.
a: What kind of energy does this transfer to the mass?
b: Describe the energy changes that occur when the mass is released.

21) Figure 18.23 shows how the different forms of energy change with displacement during s.h.m. Copy the graph, and show how the graph would differ if the oscillating mass were given only half the initial input of energy.

22) Figure 18.24 shows how the velocity ν of a $2.0 kg$ mass was found to vary with time t during an investigation of the s.h.m. of a pendulum. Use the graph to estimate the following for the mass:
a: its maximum velocity
b: its maximum kinetic energy
c: its maximum potential energy
d: its maximum acceleration
e: the maximum restoring force that acted on it.