Do you know what a pendulum is? It's just a small object at the end of a string that oscillates back and forth due to gravity. As you could probably guess, this is also a sort of simple harmonic motion, though only for certain cases. The oscillatory motion of a pendulum is very similar to rotation, so we will be using techniques of torque rather than just force analysis.

Our first kind of pendulum is a simple pendulum, consisting of a small point object hung by a massless string of some length that is set to oscillate. As the name suggests, it is the most simple kind of pendulum we'll be talking about here.

The simple pendulum can be analyzed using forces. At each point in the pendulum's motion, the string will make some angle $\theta$ with the vertical. Gravity is the only force that causes a torque about the pivot point in the ceiling and causes the bob to rotate, so we can figure out the torque due to gravity in terms of the bob's mass $m$, the angle $\theta$ and the string length $L$. Only the perpendicular component causes a torque.

$$ \tau = - mg \sin \theta L $$ $$ mL^2 \alpha = - mg \sin \theta L $$ $$ \alpha = - \dfrac{g}{L} \sin \theta $$
We have angular acceleration related to angular position, but it's not in the exact form $\alpha = - \omega ^2 \theta $ needed for simple harmonic motion. We next want to use an approximation technique, where we can approximate $\sin \theta \approx \theta $ for small angles, typically angles less than $15 \degree$. If you've noticed, we are actually just taking the first term in the Taylor expansion for $\sin \theta$ as an approximation.

What this means is that the pendulum is actually not perfectly simple harmonic in nature, but rather can deviate quite a bit from traditional simple harmonic motion if the angles are large! Most of the time, we will deal with small angles, so the motion is very close to simple harmonic in nature and is just approximated as simple harmonic. After we do the approximation, we are left with the new equation:

$$ \alpha = - \dfrac{g}{L} \theta $$
From this, we can conclude that:

$$ \omega = \sqrt{\dfrac{g}{L} } $$
This result holds for a simple pendulum set to oscilalte at small angles. Of course, this lets us write the period and frequency as well.

$$ T = 2\pi \sqrt{\dfrac{L}{g} }$$ $$ f = \dfrac{1}{2\pi} \sqrt{\dfrac{g}{L} } $$
Notice that mass is nowhere to be seen. The mass of the object does not affect the oscillation period! The only things you can do to change it are to change the gravity (which is usually unfeasible) or to change the length of the string (which is much more feasible). The math for the pendulum is a bit more complex than for the spring-block oscillator, so I will not go through the entire derivation here. It uses rotational dynamics methods to relate angular acceleration and angular position, then determines the frequency of oscillation from there. The only things in consideration are the gravitational acceleration and the length of the string holding the pendulum bob.

$$ \omega = \sqrt{\dfrac{g}{L} } $$
Once we have that, we can write the period and frequency relatively easily, by folliwng the formulas we were given before.

$$ T = 2\pi \sqrt{\dfrac{L}{g} }$$ $$ f = \dfrac{1}{2\pi} \sqrt{\dfrac{g}{L} } $$
I had stated that this was a type of simple harmonic motion, but that's sort of a lie. See, this kind of motion is only very close to simple harmonic and is approximated as simple harmonic. The approximation is very good if the maximum angle that the pendulum swings up to is less than $15 \degree$, and starts to break down at larger angles. We won't be worrying about that here, though.

Let's see how well you understand pendulums now!


As that last question stated, what is actually oscillating sinusoidally for our simple pendulum is not the position of the bob, but rather the angle. Indeed, the position of the bob changes simultaneously in two dimensions at once, since it moves both left to right as well as up and down. This shouldn't be too surprising, considering our derivation for the simple harmonic nature of this system related angular acceleration and angular position. The "amplitude" for this kind of system is a maximum angle, called $\theta_0$.

I won't write the exact equations for the pendulum because they are not too important and follow the same form as the equations I presented previously in the spring-block oscillator lesson.

With that, let's try and apply our skills to an actual pendulum problem.

How long does the arm on a granfather clock have to be in order to function? (Hint: The clock records time in seconds.)

Okay, so the key here is to realize that the clock records time whenever the bob reaches a maximum amplitude and starts to move the other way. (At least, I think that's how it works. I don't own one of these.) This means that in one full oscillation cycle, it "ticks" twice. That's what we're going with here. Don't question me! I probably got the mechanism wrong, but what matters is the physics, okay?

If the clock records time twice during an oscillation period, this means that the period of oscillation must be $T = 2.0 \textrm{s}$. We can then list out our formula for the period of oscillation of a simple pendulum:

$$ T = 2\pi \sqrt {\dfrac{L}{g} }$$
We know what $g$ is because we're (presumably) on Earth, so now all that's left is to solve for $L$ given $T$.

$$ L = g \left( \dfrac{T}{2\pi} \right)^2 = \bbox[3px, border: 0.5px solid white]{0.994 \textrm{m} }$$
Google says the pendulum length for a granfather clock is around one meter, so my guess of how the clock worked was right!

What happens if we have entire objects acting as a pendulum instead of just having small bobs? Well, that sounds an awful lot like rigid bodies rotating, doesn't it? That's exactly what the next kind of pendulum is: the physical pendulum. Try out this demo to see what happens, paying special attention to the moments of inertia of the objects here.

Physical Pendulum Demo

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Gravity: 10

As you might have seen, the moment of inertia plays a role in the period of oscillation! The circle had a much greater moment of inertia than the rod, so it had a longer period of rotation. The other factor that influenced it, as you may have seen, is the strength of the gravity on the objects.

We can derive the equation for the angular velocity of such a system. We can say an object of mass $m$ has some arbitrary moment of inertia $I$ about an axis that is a distance $d$ from the center of mass. As always, the gravitational acceleration is $g$. The gravity acts at the center of mass, so for an arbitrary rotation angle $\theta$ we can do some torque analysis.

$$ - mg \sin \theta d = I \alpha $$ $$ \alpha = - \dfrac{mgd}{I} \sin \theta $$
Once again, we can use the small angle approximation for $\theta$.

$$ \alpha = - \dfrac{mgd}{I} \theta$$
Now, we can easily get the angular frequency for this kind of oscillator. You see that the derivation process was very similar to that for the pendulum bob, just generalized to rigid bodies.

$$ \omega = \dfrac{mgd}{I} $$
The derivation for the physical pendulum requires some rotational dynamics and a bit of trigonometry, so I'll skip it again. It uses much of the same reasoning and methodology as the derivation for the simple pendulum, and if you're interested in seeing those two you can bump the level up to algebra-based. In short, the angular frequency for a physical pendulum depends on its mass $m$, the distance $d$ the pivot is from the center of mass, and the moment of inertia $I$ through that pivot.

$$ \omega = \dfrac{mgd}{I} $$
Of course, once we know the angular frequency we can't forget about our good friends, the period and frequency of oscillation.

$$ T = 2\pi \sqrt{\dfrac{I}{mgd} }$$ $$ f = \dfrac{1}{2\pi} \sqrt{\dfrac{mgd}{I} } $$
Do recall that $I$ here is measured about the pivot point, not the center of mass. If you want to find this moment of inertia for one of our six common objects, simply use the parallel axis theorem. You remember that, right?

With that, let's try and apply our skills to a simple physical pendulum problem.

We still need to talk about the energy behind the pendulum, don't we? The pendulum is a system that has gravitational potential energy and kinetic energy, so it can be analyzed using conservation of energy. The pendulum bob can be said to have zero potential energy at the lowest point of its swing, which means it will have maximum kinetic energy there. At the highest points of the motion, all energy is potential.

We are able to write the potential energy as a function of the angle that the pendulum makes with the vertical. It's a bit more complex than just using a single trig function, though. We need to find the change in height of the pendulum with respect to the lowest point.

We want to look for the yellow line's length. This involves subtracting $R\cos\theta$ from the entire length $R$ of the string, which gives us that the height at any angle $\theta$ is $h = R (1- \cos \theta)$. This means that the potential energy at any angular position is equal to:

$$ U = mgR(1-\cos \theta) $$
The total energy is equal to the potential energy at the maximum angular position $\theta_0$, and at any other point there is some kinetic energy and some potential energy. This can be easily dealt with using familiar energy conservation techniques, so I won't reiterate too much of what you've heard a million times here. If you need a refresher, look back at our energy unit.

A very similar method can be used for physical pendulums. You want to recall that the energy of rigid objects is measured to their center of mass, so instead of $R$ we have $d$, the distance between the pivot and the center of mass. Moreover, we use the formula for rotational kinetic energy rather than translational kinetic energy. Those are the only changes; the rest of the analysis is very similar.

This concludes the pendulums lesson. Technically, there are more kinds of pendulums, but the ones that can be easily dealt with at our level are the simple and physical pendulums we analyzed. Next, we'll go over a few more kinds of oscillators that don't fit clearly into any category. If you're ready, let's move on!