The pendulum


We analyzed our prototype example for oscillations, the mass on a spring, using forces and Newton's 2nd law. Oscillations can also be analyzed using mechanical energy. Let's see how this work with another example of an oscillation, the swinging pendulum. (This one has more obvious biological applications and is useful as a starting model for the analysis of gait and the motion of animal limbs.)

Setting up the pendulum

Another nice example of harmonic (or nearly harmonic) motion is the long pendulum. Imagine hanging a small sphere attached to a long thin rod from a vertical support and allowing it to swing back and forth. If the swing is small, the back and forth motion looks a lot like a mass on a spring. It can also provide a nice example of how an energy analysis gives us insight into the motion of an oscillator.

In order to make our first model as simple as possible, we will assume:

  • The sphere is small enough that we can ignore its extent.
  • The rod is light enough (compared to the sphere) that we can ignore its mass.
  • The rod is rigid enough that we can ignore any bending of the rod.
  • Any friction in the pivot is small enough that it can be ignored.
  • The effect of air drag and viscosity on the mass can be ignored.

Of course some of these effects build up, so it might be okay to ignore them for a few dozen swings (or even hundred in a good case).

The energy of the pendulum

Let's look at the free-body diagram for the sphere. Since we're ignoring air drag, the only forces acting on the sphere is the pull of the earth (its weight) and the tension force from the rod. The rod constrains the sphere to move along a circular path, so the sphere's change of position is always perpendicular to the tension force from the rod. This means that the tension force does no work on the sphere — it only acts to change its direction of motion, not its speed. This, together with the fact that we have decided we can ignore any resistive forces, tells us that the only work we need to consider is that done by the force of gravity. So the kinetic energy plus the gravitational potential energy should be conserved.

The result is that the speed of the sphere is related to how high up it is by

$$E_0 = \frac{1}{2}mv^2 + mgh$$

where we have written $E_0$ for the (constant) value of the energy (and to indicate that we mean a particular value, not a variable). If we take the 0 of $h$ to be at the lowest position of the pendulum, then a little trigonometry shows us that

$$h = L - L \cos{\theta} = L(1-\cos{\theta)}$$

so we get
$$E_0 = \frac{1}{2}mv^2 + mgL(1-\cos{\theta)}$$

If we take the starting angle of the pendulum (when $v = 0$) to be $θ_0$ then we get

$$mgL(1-\cos{\theta_0)} = \frac{1}{2}mv^2 + mgL(1-\cos{\theta)}$$

The $mgL$ term on both sides cancels giving

$$mgL (\cos{θ} - \cos{θ_0}) = \frac{1}{2}mv^2$$

or, solving for $v^2$,

$$v^2 = 2 gL (\cos{θ} - \cos{θ_0}).$$

In principle, this solves everything. We could take the square root of both sides, write $v$ as the derivative of position along the arc($Lθ$ with θ in radians) and then integrate the RHS over time giving the angle as a function of time. This is a fair mess with the result being something called an "elliptic integral" — something rarely seen outside of the advanced math Tripos exams at Cambridge University (UK). But if we're willing to accept a useful and important result from our small angle approximations, we can make the connection with the harmonic oscillator.

The pendulum as SHM

Since the cosine function can be given by the power series in $\theta$,

$$\cos{\theta} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!}...$$

if we only let our pendulum do small angle oscillations, since $\theta^4 << \theta^2$, we can approximate the cosine in the energy by $1 - θ^2/2$. In our energy expression, the 1 cancels giving

$$E_0 = \frac{1}{2}mv^2+ mgL (1 - \cos{θ}) \approx ½mv2 + \frac{1}{2}mgLθ^2$$

but since by our small angle approximation,  $Lθ \approx L \sin{θ} = x$, we can write our energy as

$$E_0 = \frac{1}{2}mv^2 + \frac{1}{2}\bigg(\frac{mg}{L}\bigg)x^2$$

If we write $k = mg/L$ for our combination of constants, then the energy becomes just the same as the energy for a mass on a spring

$$E_0 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$$

If the relation between the velocity and the position is the same for two systems they have to move in the same way. So our pendulum will oscillate back and forth just like a harmonic oscillator — as long as the angle doesn't get too big.

We can even figure out the period by using our analogy. We know for SHM the period, $\tau$ (we've already used the symbol $T$ in this problem to mean tension, so let's call it "tau") is $2π/ω_0$ or

$$\tau = \frac{2\pi}{\omega_0} =  2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{mL}{mg}} = 2\pi\sqrt{\frac{L}{g}}$$

Note that the mass has dropped out! The period of an (ideal) pendulum is independent of the mass. We probably shouldn't be surprised since we know that when gravity is the only force, the mass cancels out and all objects accelerate the same way. Here, only the gravitational force is changing the sphere's speed so we get the same kind of result as for free fall.

Note that the dimensional analysis suggests this result. We are working with the parameters [$m$] = M, [$L$] = L, and [$g$] = L/T2. The only way to get a time is to take [$1/g$] = T2/L and multiply by $L$ to get the result [$L/g$] = T2. Taking the square root gives the correct result up to a factor.

The pendulum has some valuable applications. It was important in the design of the first clocks and it plays a role in the analysis of the gaits of animals. The way the animals arms and legs swing naturally have a lot to do with how efficient motion is patterned.

Workout: The pendulum

Joe Redish 3/15/12

Article 665
Last Modified: May 22, 2019