In post 17.12 we saw how motion in a circle can be described, although some ideas on this subject appeared earlier in post 16.14. We met the simple harmonic oscillator in post 18.6 but some of the equations used were derived more simply in post 18.7. Finally, in post 18.10, we looked at wave motion.
Now I want to look at some of the connections between these ideas. These connections sometimes make it easy to understand ideas and derive equations. This post will be much easier to understand if you’ve read the posts mentioned in the first paragraph.
Let’s think about an object going round in a circle of radius a and whose centre is at O; in the picture above it starts moving from the point P and, a time t later, it is at P’. If it has an angular speed of ω (post 17.12), the angle it sweeps out in this time is ωt, as shown in the picture. Remember that we usually measure ω in rad.s-1 and so measure this angle in radians (post 17.11).
Now we’ll use the line OP to define the x-axis (OX) and the line OY, that is perpendicular to it, to define the y-axis of an orthogonal Cartesian coordinate system (appendix 2 post 16.50). Let’s see how the projections of the line OP’ on these axes depend on time.
The projection on the x-axis is given by
x = acos(ωt) (1).
(For more information on projections, see appendix 3, post 16.50). This equation has the same form as the solution of the differential equation of a simple harmonic oscillator. For a simple pendulum, in which a string of length L supports an object, the angular displacement of the object is given by
θ = cos(g/L)1/2t,
where g is the magnitude of the earth’s gravitational field (see post 18.7). The right-hand side of these equations are identical when ω= (g/L)1/2.
If we differentiate equation 1 twice, with respect to time, we obtain a general differential equation (see post 18.6) that describes the behaviour of a simple harmonic oscillator:
d2x/dt2 = – ω2x (2)
where ω is the angular frequency of the oscillation; the idea of angular frequency is explained in post 17.12 (it is related to frequency, f, by ω = 2πf) and appendix 3 of post 18.6 shows how to obtain equation 2 from equation 1.
How can we use this result? Let’s think about a spring that supports an object of mass m, as shown in the picture above. The weight of the object, mg (post 16.17), will stretch the spring; the centre of gravity of the object (post 17.22) is at O. Now we’ll think what happens when we stretch the spring by an extra distance, x, so that the centre of gravity moves to P. If we release the object, the potential energy of the spring pulls the object upwards and it oscillates, up and down, between P and P’, where the distances OP and OP’ are equal; this is explained in more detail in the appendix. If the viscosity of the air is negligible, the system will oscillate indefinitely – it is a simple harmonic oscillator. Of course, in a real system, there will always be some viscosity opposing motion and the simple harmonic oscillator is a model that helps us to understand how real systems behave – it is not an exact description of a real system (see post 18.6).
The magnitude of the force required to stretch the spring by an extra distance x is given by
F = – kx
where k is the stiffness of the spring (post 16.49). The minus sign shows that F acts in the opposite direction to the displacement x. Since the mass, m, of the object is constant, F is equal to its mass multiplied by its acceleration (post 16.13) so that we can write the previous equation in the form
m(d2x/dt2) = – kx.
Dividing both sides of this equation by m gives
d2x/dt2 = – (k/m)x. (3)
Comparing equations 2 and 3 shows that the object oscillates with an angular frequency of
ω = (k/m)1/2
which is the same as a frequency
f = (1/2π)(k/m)1/2.
The time period (post 16.14) that it takes the object to move from P, through O, through P’, through O again, and back to P, is given by
ΔtΔt
This equation shows that an object with a higher mass moves more slowly and, if the spring is stiffer, it moves more slowly.
Now let’s look at equation 1 again. It’s the same as the final equation for representing a wave, in post 18.10, if we simply replace x by ψ. So, if we plot the projection of a simple harmonic oscillator on to the x-axis, we get a wave.
Suppose that now we think of a second object Q, moving around our circle. When our first object is at P, the second is at Q; when the first is at P’, the second is at Q’. Both move with the same angular speed, ω. If the angle between OP and OQ is θ, the projection of OQ on the x-axis is
x’ = x = acos(ωt + θ) (4).
So equation 1 and 4 represent two waves that are θ radians out of phase (see post 18.10).
Finally, let’s think about the projection of OP on the y-axis. It is
y = asin(ωt) (5).
Equation 5 is an alternative solution to the differential equation for a simple harmonic oscillator, as discussed in post 18.6. Once again, if we plot y against t we get a wave of frequency ω/2π. The phase difference between this wave and the wave of equation 1 is π/2 radians.
In conclusion, motion in a circle, simple harmonic oscillation and wave motion all use the same ideas, and the same equations can be used to describe many of the properties of all three – even though an object going round in a circle, an oscillating pendulum and a wave don’t all look the same.
Related posts
18.10 Waves
18.8 Natural frequency and resonance
18.7 The simple pendulum
18.6 The pendulum: a simple harmonic oscillator
17.13 Centripetal force
17.12 Going round in circles
16.14 Aliasing
Follow-up posts
Appendix
When the spring is stretched, by moving the object from O to P, it gains potential energy
Us = kx2/2
where k is the stiffness of the spring, as explained in post 16.49.
When the object is released from P, it converts this potential energy into kinetic energy
K = – mv2/2 = – (m/2)(dx/dt)2.
The minus sign arises because the velocity, v, acts in the opposite direction to the displacement x; the final step arises from the definition of velocity (post 17.39).
Since the object moves by exchanging potential and kinetic energy then
– (m/2)(dx/dt)2 = kx2/2.
If we multiply both sides of this equation by – 2 and then divide both sides by m, we get the result that
(dx/dt)2 = – (k/m)x2.
This equation has the same form as equation 3 of post 18.6 which has an oscillatory solution.
Thinking about Science with David Hukins 