## Illustration 16.3: **Energy and Simple Harmonic Motion**

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In this Illustration we shall look at energy and simple harmonic motion of both a pendulum and a mass on a spring. We shall consider small-amplitude motion for the pendulum since this will yield simple harmonic motion (see Illustration 16.2 for details). In addition, like Illustration 16.2, we have chosen the mass of the pendulum bob to be 1 kg and the length of the pendulum to be 15 m, while choosing the mass of the ball on the spring to be 2 kg and the spring constant to be 1.30666 N/m (**position is given in meters, angle is given in radians, and time is given in seconds**). Restart. These values tune the motion of the two systems to be the same:

ω_{mass-spring} = (k/m)^{0.5} = ω_{pendulum} = (k_{effective}/m)^{0.5} = (g/L)^{0.5}.

In the following animations we will show graphs of the kinetic and potential energy of the mass-spring system but will not show the kinetic and potential energy of the pendulum. However, the kinetic and potential energy of the pendulum will look the same with *exactly half *the kinetic and potential energy (and therefore half the total energy) of the mass-spring system. Why half? For the mass-spring system, kinetic energy is (1/2 mv^{2}) and the potential energy is (1/2 kx^{2}), and for the pendulum the kinetic energy of the bob is (1/2 mv^{2}) and the potential energy is (1/2 k_{effective}x^{2}). In this Illustration, since the mass on the spring is twice the mass of the pendulum bob, the mass-spring system will always have twice as much kinetic energy as the pendulum bob. Since the spring constant for the mass-spring system is twice the effective spring constant for the pendulum (k_{effective} = m_{pendulum}g/L = 0.6533 N/m), the mass-spring system will always have twice as much potential energy as the pendulum bob.

*When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.*

Consider Animation 1, which shows the graph of kinetic and potential energy vs. position. What can you say about the total energy of the system? It is a constant and about 1.89 J. The energy starts out all potential and at the equilibrium position the energy is all kinetic. At maximum compression the energy is all potential again. Given that the total energy is kinetic plus potential, we have that

E = 0.5 mv^{ 2} + 0.5 k x^{2} = 0.5 k x_{max}^{2} = 0.5 m v_{max}^{2}.

Now consider Animation 2, which shows the graph of kinetic and potential energy vs. time. Notice how the two graphs are different in their functional form.

The graphs in Animation 1 have the form of 0.5 k x^{2} (the potential energy) and the form of A - 0.5 k x^{2} (the kinetic energy), where A is a constant, the total energy. In this animation the total kinetic energy is 1.89 J. The form of the kinetic energy can be understood from the energy function shown above. The potential energy is 0.5 k x^{2}, which is proportional to x^{2}. The kinetic energy can be written in terms of the total energy and the potential energy as E - 0.5 k x^{2}.

The graphs in Animation 2 have the form of cos^{2} (the potential energy) and the form of sin^{2} (the kinetic energy) since both trigonometric functions are a function of time. Why? We know from simple harmonic motion that if the object is initially displaced from equilibrium with no initial velocity that

x = x_{0} cos (ωt) and v = -ω x_{0}sin (ωt).

Given the form of the kinetic energy and the potential energy, we have that

KE(t) = 0.5 k x_{0}^{2} sin^{2}(ωt) and PE(t) = 0.5 k x_{0}^{2} cos^{2}(ωt),

where we used ω^{2} = k/m to simplify the kinetic energy. Therefore the total energy will always add up to 0.5 k x_{0}^{2} = 1.89 J.

Illustration authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.

Script authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.