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12.3Energy in simple harmonic motion
Kinetic and potential energy
Consider again a particle at the end of a horizontal spring. Let the extension of the spring be x at a particular instant of time, and let the velocity of the particle be v at that time. The system has elastic potential energy and kinetic energy. Figure 12.31 shows how these energies vary with displacement. These shapes are parabolic. The total energy ET of the system is then:
ET=Ep+EK
In the absence of frictional and other resistance forces, this total energy is conserved, and so E = constant.
The maximum velocity is achieved when x = 0, i.e. as the mass moves past its equilibrium position. Here there is no extension, so the elastic potential energy is zero. At the extremes of the motion, x = Â± A and v = 0, so the kinetic energy is zero. Thus at x = Â± A the system has elastic potential energy only, and at x = 0 it has kinetic energy only. At intermediate points the system has both forms of energy: elastic potential energy and kinetic energy. During an oscillation, we therefore have transfers from one form of energy into another.

Figure 12.31: Graphs showing the variation with displacement of the potential energy and kinetic energy of a particle on a spring. The total energy is a horizontal straight line. The graphs are parabolas.

CHECK YOURSELF 9
Figure 12.32 shows the variation with displacement of the potential energy of a particle in SHM.

Figure 12.32: For check yourself question 9.1

What is the kinetic energy when the displacement is 1.0 cm?
The variation of the kinetic and potential energies with time is a bit more complicated.
Suppose that we have a SHM where the displacement versus time graph is that of Figure 12.33.

Figure 12.33: Displacement versus time graph.

We see that the speed will be zero at the following times: 0 s, 2 s and 4 s. The speed will be maximum at times: 1 s, 3 s and 5 s. So a graph of kinetic energy versus time will look like the red curve in Figure 12.34. The potential energy is maximum when the displacement is maximum or minimum, i.e. at times 0 s, 2, s and 4 s and zero when the displacement is zero, i.e. at times 1 s, 3 s and 5 s. This gives the blue curve in Figure 12.34.
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Figure 12.34: Variation of kinetic and potential energies with time.

EXAM TIP
The graphs above are specific to the displacement graph of Figure 12.33! You must be able to deduce from such a graph that the period of oscillations is 4 s.
CHECK YOURSELF 10
Figure 12.35 shows the variation with time of the kinetic energy of a particle in SHM.

Figure 12.35: For check yourself question 10.

Draw the corresponding displacement versus time graph. (No numbers are necessary on the displacement axis.)
WORKED EXAMPLE 12.6
The graph in Figure 12.36 shows the variation with displacement of the kinetic energy of a particle of mass 0.40 kg performing SHM at the end of a spring.

Figure 12.36: Graph showing the variation with displacement of the kinetic energy of a particle.

aUse the graph to determine:
ithe total energy of the particle
iithe maximum speed of the particle
iiithe amplitude of the motion
ivthe potential energy when the displacement is 2.0 cm.
bOn a copy of the axes, draw the variation with displacement of the potential energy of the particle.
Answer
aiThe total energy is equal to the maximum kinetic energy, i.e. 80 mJ.
iiThe maximum speed is found from:
12mv2maxv2maxvmax=Emax=2Emaxm=2Ã80Ã10â30.40âââââââââââââ=0.63Â mÂ sâ1
iiiThe amplitude is 4.0 cm.
ivWhen x = 2.0 cm, the kinetic energy is 60 mJ and so the potential energy is 20 mJ.
bThe graph is a parabola like the blue curve in Figure 12.31.
