SHM Displacement

The oscillation of a body is simple harmonic if:

The restoring force is directly proportional to the displacement from the rest position, and acts in the opposite direction to this displacement.

This leads to the following formula:

a = -ω2x.

where a is the acceleration of the body, ω the angular frequency of the oscillation, and x the displacement of the body from the rest position. This has been explained in the page on simple harmonic motion.

Here, we shall look at the solution to this equation. With some mathematics, the equation can be solved to give:

x = x0 sin ωt

where x0 is the amplitude of the oscillation, and t is the time. I shall not go into to mathematics, which requires the use of calculus. Instead, I shall discuss how we can understand the connection between this solution and the above equation physically.

Consider a body moving up and down on a spring in simple harmonic motion. Suppose that there is an identical body next to the spring that is moving at uniform speed around a circle, and that the radius of this circle is equal to the amplitude of the oscillation at the spring. Then the two bodies can actually move up and down together, always keeping the same height. Assuming that this is possible, we can easily get a formula for the displacement x from the rest position at the spring. First, note that both the circular motion and the oscillatory motion would then have the same period T. So the angular velocity of the circular motion is

ω = 2π/T.

Notice that x is also the vertical displacement of the body at the circle, from the centre of the circle. Using trigonometry, x and x0 are related by the sine function:

x = x0 sin θ

Suppose that the the time t starts from the rest position. Then the angle to the horizontal of the radius to the body is

θ = ωt.

Substituting into the previous equation, we get

x = x0 sin ωt,

which is the equation we set out the understand.

You may notice that I have not actually explained why a body moving at uniform speed on the circle can stay at the same height as the body on the spring in simple harmonic motion. I shall only mention here that it is because the centripetal force on the body at the circle has a vertical component that is proportional to the vertical displacement x of the body (because the vertical component is also related to sin θ.) This is the same as the relation between force and displacement on the spring. In other words, the vertical component of the circular motion is also simple harmonic.

Copyright 2010 by Kai Hock. All rights reserved.
Last updated: 27 March 2011.