# Video: Oscillating Pendulums

In this video we see how pendulums oscillating at small angles follow simple harmonic motion, and learn that a pendulum’s period depends only on its length and gravity.

11:30

### Video Transcript

In this video, we’re going to learn about oscillating pendulums. We’ll see how pendulums can act as simple harmonic oscillators. And we’ll also see what factors do and do not affect a pendulum’s period.

To get started, imagine that you are the keeper of a clock tower in a medieval town. The clock, which everyone in town uses to keep time, has recently been repaired. And you’re keeping careful watch on it to see that it keeps accurate time. Unfortunately, as you track the clock throughout the days and weeks, you realize that it’s running a bit fast. Knowing that the clock keeps time thanks to a gigantic pendulum that swings back and forth inside the tower, what alteration can you make so that the clock slows down a bit to keep proper time?

To figure out, it will be helpful to know something about oscillating pendulums. The type of pendulum we’ll focus on is called a simple pendulum, simple because it’s represented by a mass which is exactly at the end of a string or a solid arm. And of course, when this mass is moved off of its equilibrium line and then released, the pendulum swings back and forth in a regular way.

There’s an interesting fact about simple pendulums. And that is, when the angle between the pendulum arm and the vertical is small, that is, if the pendulum mass is not very far displaced from its equilibrium position, then the pendulum moves back and forth in simple harmonic motion. We’ve recall that simple harmonic motion, or SHM for short, happens when an object is restored to its equilibrium position with a force that’s proportional to its displacement from that position.

We found earlier that masses on the ends of springs, which move up and down according to Hooke’s law, exhibit simple harmonic motion, and so these simple pendulums when the angle that they make with the vertical is small. And when it came to those masses on springs oscillating up and down or left and right, we found that the position of that mass as a function of time was equal to the amplitude of its oscillations times the sine of the angular frequency times 𝑡.

This equation describes not just a mass on a spring, but any simple harmonic oscillator position. That said, there is a difference between this equation for mass on a spring and for a simple pendulum. That difference has to do with the angular frequency 𝜔. We can recall that, for a mass on a spring, the angular frequency 𝜔 is equal to the square root of the spring constant divided by the mass 𝑚. This angular frequency applied to an object which was subjected to a force magnitude according to Hooke’s law which is equal to the spring constant 𝑘 times the displacement of the mass from its equilibrium 𝑥.

And we can rearrange this expression to say 𝑘, the spring constant, is equal to the force acting on the mass divided by its displacement from equilibrium. When we consider our simple pendulum, we know we’ll get a different result for the angular frequency 𝜔, if for no other reason than a pendulum doesn’t have a spring constant 𝑘.

To find out what 𝜔 would be for a simple pendulum, let’s draw the forces that are acting on our mass as part of the pendulum. The two forces that act on our mass at all times are the weight force, 𝑚 times 𝑔, and the tension force mediated by the string or our solid bar. If we focus exclusively on horizontal motion, motion in what we’ve called the 𝑥-direction, then we can see the only force that acts in that direction is the horizontal component of the tension 𝑡.

Referring to Newton’s second law of motion, we can say that the net force in the 𝑥-direction on our pendulum is equal to the mass of the pendulum times its acceleration in that dimension. And looking carefully at the right triangle we’ve drawn with the hypotenuse of the tension 𝑡 and the legs of the 𝑥- and 𝑦-components of that tension, we see that the horizontal component, the 𝑥-component, is equal to 𝑡 times the sin of 𝜃. So we have that the mass of the pendulum times its acceleration in the 𝑥-direction is equal to 𝑡 sin 𝜃.

Now here is where the fact that 𝜃 is a small angle comes in. When 𝜃 is a very small angle, much much less than one radian, then we can approximate the sin of 𝜃 as the angle 𝜃 itself. That means that, down at our Newton’s second law expression, we can replace 𝑡 sin 𝜃 with 𝑡 times 𝜃. And again, we’re able to do this because of the small angle approximation that we’re making. This wouldn’t work if the amplitude of our swinging pendulum was very great so that 𝜃 would approach a large angle.

So far, we’ve considered the 𝑥-direction or horizontal component of the tension force. But now let’s consider the vertical component for a moment. When our pendulum is at its maximum displacement from equilibrium, when our angle is equal to 𝜃, at that moment, the vertical forces that act on our pendulum bob cancel out. Once again, using the fact that 𝜃 is small, this means that 𝑚 times 𝑔 is approximately equal to the tension 𝜃. This means we can replace the 𝑡 in our current expression for Newton’s second law with 𝑚 times 𝑔.

And now we can take advantage of a geometric fact of the right triangle that’s formed by a vertical line, the horizontal displacement of our pendulum, and the pendulum arm length. For any angle 𝜃, we can say that the sin of that angle is equal to 𝑥, the horizontal displacement, divided by 𝑙, the length of the pendulum. And since 𝜃 is small, we can say that this is approximately equal to 𝜃 itself. So we’ll now take 𝑥 over 𝑙, the pendulum length, and substitute that in for 𝜃.

We now have an expression for the force acting on our pendulum in the 𝑥-direction, which is working under the assumption of a small angle 𝜃. We can now take this force in the 𝑥-direction and substitute it in for the force 𝐹 in our Hooke’s law equation. And remember, our objective in all this is to come up with an expression for the angular frequency 𝜔 that’s appropriate for a simple pendulum compared to the expression for 𝜔 for an oscillating mass on a spring.

With this substitution for 𝐹, we see that 𝑘 is equal to 𝑚 times 𝑔 times 𝑥 all over 𝑙 times 𝑥. So the displacements 𝑥 cancel out. If we now divide both sides of our equation by the mass 𝑚, then that value cancels out on the right, leaving us with our result. 𝑘 over 𝑚 is equal to 𝑔 over 𝑙. What we’ve done is establish a correspondence between the factors affecting angular frequency for a mass oscillating on a spring and the factors affecting that angular frequency for a simple pendulum.

So for a simple pendulum, the angular frequency 𝜔 is equal to the square root of 𝑔 over the length of the pendulum 𝑙. This means we can now write a complete expression for the simple harmonic motion of a simple pendulum. Knowing that 𝜔 is equal to the square root of 𝑔 over 𝑙, that means that the period of a simple pendulum is equal to two 𝜋 over 𝜔 or two 𝜋 times the square root of 𝑙 over 𝑔.

Consider what this result means in terms of the motion of a pendulum swinging back and forth. This expression for period doesn’t say anything about the amplitude of our pendulum. So if we had two different pendulums of the same length and started one at a much lower amplitude than the other, this equation says that they would actually have the same period as they swung back and forth. And what’s more, this equation for period says nothing about the mass of the pendulum. So on our second pendulum, we could create a much larger mass. And yet the equation tells us this would have no effect on its period of oscillation.

This expression tells us that the only physical parameter of a pendulum that affects its period is the length of its pendulum arm. So if we made our pendulum arm shorter, that will make our period shorter, meaning the pendulum would swing back and forth in less time than before. And if we were to make the arm longer, that would increase the period by a factor of the square root of that change in length. All this means that if we go back to our clock tower, the way we could change the time being kept would be to change the length of the clock pendulum arm.

Keeping these relationships for oscillating pendulums in mind, let’s get some practice with these ideas through an example.

A pendulum of length 𝑙 one has a period 𝑇 one. 𝑙 one is decreased by 9.00 percent. The pendulum’s new period is 𝑇 two. What percent of 𝑇 one is 𝑇 two?

To start on our solution, we can recall that the period of a pendulum is equal to two 𝜋 times the square root of the pendulum’s length divided by the acceleration due to gravity 𝑔. For this exercise, to find the percent of 𝑇 one that 𝑇 two is, we wanna take the ratio 𝑇 two to 𝑇 one and then multiply it by 100. That will give us that value as a percent.

First, we write out 𝑇 two in terms of 𝑙 two and 𝑇 one in terms of 𝑙 one based on the relationship for pendulum period. Then we’ll divide these equations one by another. And when we do that, we see that the factors of two 𝜋 cancel out, as well as one over the square root of 𝑔. The fraction simplifies to the square root of 𝑙 two over 𝑙 one.

We’re told in the problem statement that 𝑙 two, the shorter pendulum arm length, is equal to 𝑙 one minus nine percent of 𝑙 one. Mathematically, this is 𝑙 one times the quantity one minus 0.09. Or 𝑙 two is equal to 0.91 𝑙 one, that is, 91 percent of 𝑙 one’s length. When we substitute that expression in for 𝑙 two in our square root, the factors of 𝑙 one cancel out. And we’re left with the square root of 0.91. To three significant figures, that’s 0.954.

To get this result as a percent, we multiply it by 100. And that gives us 95.4 percent. That’s the percent of 𝑇 one that 𝑇 two is.

Let’s summarize what we’ve learned about oscillating pendulums. We’ve seen that when a simple pendulum swings at small angles, its motion follows SHM, simple harmonic motion. Written as an equation, we can say that the pendulum’s horizontal position from equilibrium as a function of time is equal to its amplitude, that is, its maximum displacement from equilibrium, times the sine of its angular frequency of oscillation times 𝑡.

We’ve also seen that the angular frequency for a simple pendulum is 𝜔 equals the square root of 𝑔 over 𝑙, the pendulum’s arm length. We noted that, for an oscillating spring, 𝜔 is equal to the square root of 𝑘 over 𝑚. And finally, we saw that pendulum period 𝑇 does not depend on the pendulum’s mass or on its amplitude of oscillation, but only on the length of its arm. 𝑇 is equal to two 𝜋 times the square root of 𝑙 over 𝑔.