This post is very mathematical, so you might want to stop reading after the paragraph on “models” – if you decide to do this, it might be worth reading the final two paragraphs.
A pendulum is an object, like that shown in the picture above, that is free to swing around a pivot at P. We can imagine that the object has a horizontal hole bored through it, parallel to the plane of the picture and the pivot consists of a horizontal peg that passes through the hole.
In this post, we will assume that the friction (post 16.19) at the pivot and the drag (post 17.17) acting on the object, when it moves, are both negligible. So, the system conserves mechanical energy when the pendulum swings – it continues to swing by exchanging potential and kinetic energy (post 16.21).
If the centre of gravity, G, of the pendulum is vertically below P, it does not move. This is because the gravitational force is acting vertically downwards and, according to Newton’s third law of motion (post 16.2), is opposed by a reaction force, at the pivot, acting vertically upwards. This is position A of the pendulum its rest position, if it is not oscillating.
Now let’s apply an angular displacement, ϑ (a positive value denotes an anticlockwise displacement – see post 17.11) to the pendulum, so that it is in position B. If the pendulum has mass m, it has gained potential energy
U = mg(L – Lcosϑ) = mgL(1 – cosϑ)
where L is the distance between P and G and g is the magnitude of the earth’s gravitational field (see post 16.21 for information on potential energy and post 16.50 to find out about “cos”). The expression for U is calculated directly from the definition of gravitational potential energy – mg multiplied by height above the rest position – for an extended object this height is measured from the position of the centre of gravity.
If we release the pendulum, it will fall to position A and its potential energy falls to zero because it is all converted to kinetic energy, like any falling object that doesn’t dissipate its mechanical energy (post 17.30). This kinetic energy is then converted to potential energy so that the pendulum moves in the opposite direction to position B but at the same height. Since the system does not dissipate mechanical energy, the pendulum will continue to swing backwards and forwards indefinitely. It is behaving like a ball that bounces by elastic collisions with the horizontal ground, as described in post 17.30.
A system that oscillates indefinitely by exchanging potential and kinetic energy is called a simple harmonic oscillator. There are many examples of simple harmonic oscillators and they are not confined to mechanical systems. For example, the current in an electrical circuit, that contains a charged capacitor and a component called an inductor, oscillates in the same way.
Of course, the simple harmonic oscillator is an idealised model of a real system. There will always be some drag, because of the viscosity of the air, and some friction at the pivot. In post 16.42 we saw that simple models that neglect factors that have little influence on a system, enable us to understand how real systems behave.
When the pendulum is falling, it has kinetic energy
K = – (1/2)Iω2 = – (1/2)I(dϑ/dt)2
as described in posts 17.38 and 17.39. Here I represents the rotational inertia of the pendulum (post 17.38), ω is its angular speed (post 17.12), ϑ is its angular displacement (post 17.12) and dϑ/dt represents differentiation of ϑ with respect to time, t (post 17.4). The minus sign arises because ω acts in a clockwise direction (opposite to the direction of ϑ).
Since the falling pendulum has converting potential energy into kinetic energy (like an object in free fall – see post 16.21), we can write that
(- 1/2)I(dϑ/dt)2 = mgL(1 – cosϑ). (1)
Equation 1 is the most general statement that we can make about how the position of a pendulum depends on time.
Now suppose ϑ is restricted to very small values; then (1 – cosϑ) ≈ ϑ2/2 when ϑ is measured in radians (post 17.11). This result is explained in appendix 1. Equation 1 then becomes
(- 1/2)I(dϑ/dt)2 = mgLϑ2/2. (2)
Equation 1 is a statement that we can make about the position of any pendulum, provided it swings through small angles.
To progress further, we have to decide on the type of pendulum we’re going to investigate. I’ve chosen the simplest – a small object suspended by a string from P. The string is considered to be so stiff that any extension is negligible and to have negligible mass. Then I = mL2 (see post 17.38); substituting this value of I into equation 2, gives the result for the simple pendulum that
(- 1/2)mL2(dϑ/dt)2 = (1/2)mgLϑ2.
If we divide both sides of this equation by mL2/2, and then multiply both sides by -1, we get the result that
(dϑ/dt)2 = – (g/L)ϑ2 (3)
The next stage is to differentiate both sides of equation 3 with respect to time, using the idea explained in appendix 1.2 of post 17.13. The steps involved are complicated and are fully explained in appendix 2. The result is that
2(dϑ/dt)(d2ϑ/dt2) = – (g/L).2ϑ(dϑ/dt).
Dividing both sides of this equation by 2(dϑ/dt) gives
(d2ϑ/dt2) = – (g/L)ϑ. (4)
Equation 4 is a differential equation that we need to solve to find how ϑ depends on t. With a bit of mathematical intuition, it is not very difficult to find a solution to equation 4 by trial-and-error. Two possible trigonometric solutions are
ϑ = sin(g/L)1/2t (5a)
ϑ = cos(g/L)1/2t (5b)
Both equations 5a and 5b are equally valid mathematical solution of equation 4, as shown in appendix 3. However, I will choose 5b because it provides a better description of the physical system we are investigating – a pendulum that starts to oscillate (at t = 0) with a non-zero displacement.
Now let’s compare equation 5b with the cosine wave in post 16.50. The cosine wave repeats itself every 360o or 2π rad (post 17.11). If T is the time it takes our pendulum to return to its original position, then 2π = (g/L)1/2T with the result that the time period of the pendulum (the time to make a complete oscillation) is
T = 2π(L/g)1/2.
Since frequency, f, is related to time period by f = 1/T (post 16.14), the frequency of oscillation of our pendulum is given by
f = (2π)-1(g/L)1/2.
Taking g = 9.8 m.s-2 (post 16.16), the frequency of a simple pendulum of length 1.0 m is
f = (2π)-1(9.81)1/2 = 0.5 Hz.
At this point, I must confess that it’s simpler to derive equation 5b from analysing the forces acting on the pendulum – this is the way I learnt to do it at school.
But the method I have chosen follows directly from the properties of the simple harmonic oscillator model – it doesn’t dissipate energy and continues to oscillate indefinitely by exchanging potential and kinetic energies.
The purpose of this appendix is to how that (1 – cosϑ) ≈ ϑ2/2 when ϑ is very small and measured in radians.
This works because cosϑ can be written as the convergent series (post 17.11)
cosϑ = 1 – ϑ2/2! + ϑ4/4! – ϑ6/6! + ϑ8/8! – ϑ10/10! +…
Here, for example, 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720; so that you can see that each term in the series will be much smaller than its predecessor.
I’m not going to prove this result but I will show that it makes sense. When ϑ equals zero, cosϑ = 1 which we know to be true (post 16.50). We can also show that it works by showing that it gives the correct value for other values of ϑ. For example, you can draw a line of length 10 cm that makes and angle of 60o (π/3 rad) with the horizontal (see picture above). Now measure the length of your line: the result is equal to 10cos60o cm; so your measurement gives you a value of cos60o within the precision of your measurement (post 16.24).
Now put π/3 into the series above and keep adding terms until it converges within the precision of your measurement (post 17.11). You will see that the results are the same.
From the series, we can see that
cosϑ ≈ 1
when ϑ is very small.
Remember that d(cosϑ)/dϑ = -sinϑ (post 17.13, appendix 1.3). Differentiating our series for cosϑ gives
d(cosϑ)/dϑ = 0 – 2ϑ/2! + 4ϑ3/4! – 6ϑ5/6! + 8ϑ7/8! – 10ϑ9/10! +
= – ϑ + ϑ3/3! – ϑ5/5! + ϑ7/7! – ϑ9/9! +…
The first step is true because d(xn)/dx = nxn-1; see the appendices to post 17.4 for two examples. The second step is true because for example 6/6! = 6/(6 × 5 × 4 × 3 × 2 × 1) = 1/5!.
So, from the result of differentiating cosϑ we can see that
sinϑ = ϑ – ϑ3/3! + ϑ5/5! – ϑ7/7! + ϑ9/9! -…
Differentiating the terms of this series will give you the series for cosϑ. Then
sinϑ ≈ ϑ
when ϑ is very small.
The purpose of this appendix is to show that if
(dϑ/dt)2 = – (g/L)ϑ2 (1A)
2(dϑ/dt)(d2ϑ/dt2) = – (g/L).2ϑ(dϑ/dt). (2A)
According to appendix 1.2 of post 17.13, there is good reason to suppose that
dv/dt = (dv/du).(du/dt). (3A)
To differentiate the left-hand side of equation 1, let u = dϑ/dt and v = (dϑ/dt)2 so that v = u2. So
dv/dt = 2u(du/dt).
Substituting the expressions for u and v back into this result gives the result that “the derivative of (dϑ/dt)2, with respect to t is 2(dϑ/dt)(d2ϑ/dt2)” – let’s call this Statement 1.
To differentiate the right-hand side of equation 1A, substitute ϑ for u in equation 3A to give
dv/dt = (dv/dϑ).(dϑ/dt).
If v = – (g/L)ϑ2
dv/dt = – (g/L)(2ϑ)(dϑ/dt).
So “the derivative of – (g/L)ϑ2 with respect to time is = – (g/L)(2ϑ)(dϑ/dt)” – let’s call this Statement 2.
Comparing statements 1 and 2 shows that equation 2A follows from equation 1A by differentiating both sides with respect to t.
The purpose of this appendix is to show that equations 5a and 5b are both solutions of equation 4.
Equation 5a states that
ϑ = sin(g/L)1/2t
So dϑ/dt = (g/L)1/2cos(g/L)1/2t
and d2ϑ/dt2 = – (g/L)sin(g/L)1/2t = – (g/L)ϑ.
Starting with equation 5b gives
ϑ = cos(g/L)1/2t
So dϑ/dt = – (g/L)1/2sin(g/L)1/2t
and d2ϑ/dt2 = – (g/L)cos(g/L)1/2t = – (g/L)ϑ.