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### Course: AP®︎/College Physics 1 > Unit 5

Lesson 1: Period of simple harmonic oscillators# Period of a Pendulum

A pendulum behaves as a simple harmonic oscillator. Learn about the period of a pendulum, how it can be adjusted, and how it compares to a mass on a spring. Created by David SantoPietro.

## Want to join the conversation?

- At6:34, David says that only 1 L shows up in the equation for the pendulum's period because torque is proportional to length. What does torque have to do with the period of a pendulum?(2 votes)
- Torque produced by gravity acts as the restoring force for the pendulum. Thus, the time period would be inversely proportional to the restoring force or the square root of gravitational acceleration or, the higher the torque on the pendulum, the lesser the time period would be and vice versa.(2 votes)

- So I've been studying bout large angles in a pendulum, and I know that the formula is

T = sqrt(l/g).[1+∅²/16]

Where ∅ is the angle of pendulum.

Any way to derive this one?(2 votes) - what about a pendulum motion that has got a spring instead of an ordinary string(1 vote)
- That would actually make the problem quite a bit more difficult!

You actually can solve this problem but it involves the usage of Lagrangian mechanics and multivariable calculus. If you want to learn more, try Googling it. If you need more help with understanding Lagrangian mechanics, feel free to ask here and I'll try my best to answer!(3 votes)

- he said "If you wanna get technical, rotational inertia is proportional to length squared, but the torque would only be proportional to length. That's why only one L shows up here." what did he meant?(2 votes)
- Learn calculus or you can't understand(0 votes)

- 8:08, So wait, the Period of the Pendulum doesn't matter if the angle is 180 degrees or 10 degrees? Will it still be 2 seconds even though it has more distance to travel?(1 vote)
- the period is based only on the gravitational constant and the length of the string so any other variables will not matter(1 vote)

- Does pendulum with large angular displacement obey Hook's law?(1 vote)
- How did he get the answer as 2(1 vote)

## Video transcript

- [Instructor] So, a simple pendulum is just a mass hanging from a string. And if you are to pull this mass, sometimes it's called a pendulum bob, if you are to pull it
back and then let go, gravity would act as a restoring force and this mass would swing
back and forth over and over. And because this simple pendulum is a simple harmonic
oscillator, it's motion. Its angle as a function of time would be accurately described
by a sine or a cosine graph. So, in other words, if you are to pull this back
initially by 15 degrees, you might get a graph that
looks something like this. Now, every simple harmonic oscillator has a characteristic period of motion. Now, the period of motion
is the time it takes to complete one full cycle. So, the time it takes to swing from here all the way to there, all the way back to here, would be one full period. And in this graph, I've
written it as .5 seconds, but the period of every simple pendulum is not gonna be .5 seconds. The period of a simple pendulum depends on the characteristics
of that pendulum and the environment that it's in. So, to derive this formula
for the period of a pendulum, you would need calculus. So, I'm just gonna write it down and give you a quick tour and compare it with the period formula
for a mass on a spring, to the period of a pendulum is gonna be equal to two
pi times the square root of the ratio of the
length of the pendulum, L, so the length of that
string here, the length L, divided by g, the
gravitational acceleration of the planet that the
pendulum is being used on. Now, if you look at this and you've been paying
attention in physics, you might be like, "Wait,
that looks really similar to the formula for the period
of a mass on a spring." So, if you are to take a mass on a spring, displace it 15 centimeters, you'd get a similar graph and this would also have a
characteristic period of motion. And if you wrote down the
period for a mass on a spring, it looks like this, its
period for a mass on a spring is also two pi, so that's identical. And then, it's also the
square root of a ratio, but instead of L over g
for the mass on a spring, it's m, the mass of the block
connected to the spring, divided by k, the spring
constant of the spring. So, one obvious similarity
between these two formulas is just their format. They're both a two pi times
a square root of a ratio, but another important
similarity between these that might not be evident is that neither of these formulas depend on the amplitude of the motion. So, the period of a pendulum does not depend on the
amplitude of the motion and the period of a mass on a spring also does not depend on the
amplitude of the motion. What I mean by that is if you are to pull this pendulum back, instead of 15 degrees, pull it back 20 degrees and let go, it would have farther to swing. So, its motion might
look something like this, and we get down farther
and it would get back up, but it would take the
exact same amount of time. The period would not change if you pull this amplitude back farther. Same goes with this mass on a spring. Instead of pulling it 15 centimeters, let's say you pulled it
20 centimeters, again, this would start up higher. It would get down lower. But the time it takes to
complete one full cycle would not vary as you vary this amplitude. And that might seem weird. You're like, "Wait, don't these objects have farther to go now that you've pulled it back
to a larger amplitude?" That's true, they'll both
have farther to travel, but they'll be going faster now. And faster motion over a bigger
distance is gonna offset. And the amplitude does
not affect the period of the motion of a pendulum
or of a mass on a spring. As for the differences, well, the denominator here
for the mass on a spring depends on k, and that's
the spring constant k, and that should make sense. More spring constant means
you get more restoring force, the spring force is the
restoring force here. More restoring force means this mass is gonna be moving faster. That means it's gonna take less time to go through a full period. So, dividing by a bigger
number, bigger spring k, gives you less period. That's also true here but it's not k. The force that's the
restoring force for a pendulum isn't a spring, it's the force of gravity. So, mg depends on the g. So, a bigger g would give
you a bigger restoring force over here for the pendulum. That means the pendulum
could be moving faster. Moving faster means it's
gonna take less time to complete a full period. So, dividing by a bigger g,
if you took this pendulum to the surface of Jupiter or something, where the g is bigger, it would swing back and forth faster. It would take less time to complete it because that restoring force is bigger. So, even though these denominators
are different letters, they're arriving from the same source. They're both arriving
because the restoring force is larger when you increase
these denominators, which increases the speed of the object. Now, maybe the biggest difference here is that the numerator here for the mass on a spring depends on mass, but nowhere is mass to be found in this period of a pendulum. The period of the pendulum
does not depend on the mass. Now, this is kind of interesting. So, you know, really big,
heavy person gets on a swing, swings back and forth, very light child gets on the same swing, they will take the same amount of time to complete a full cycle. Their mass does not factor in here to the period of pendulum, but it does for the mass
on a spring, why is that? Well, bigger mass on a spring gives you more inertia in the system. If you have more inertia in the system, it's more sluggish to movement. It's gonna go slower. That means it's gonna take more time to complete a full cycle. Now, you might think, "Wait a minute. don't that hold true up here? Look, if we have a
bigger mass pendulum bob, that should increase
the rotational inertia. And so, that should make this take longer. It should be more sluggish to movement. It should take longer to
go through a full period." But look at the restoring
force is also, for a pendulum, proportional to mass. So, if you increase the
mass of this pendulum, you do get more inertia, but you're getting more restoring force because the restoring force is gravity. Those completely offset mass does not end up showing
up into this pendulum formula, even though it does down here. So, the spring force is
not proportional to mass. Spring force is kx. Increasing the mass of
this block on the end does not increase the
spring restoring force. So, this mass stays in the numerator here, but it does not affect
the period of a pendulum. So, why does this L show up then? Why is it length of the
pendulum in the numerator? Well, the rotational
inertia does get increased when you increase the
length of the pendulum. But increasing that length that does not increase
the force of gravity. If you wanna get technical, rotational inertia is
proportional to length squared, but the torque would only
be proportional to length. That's why only one L shows up here. Long story short, if you increase
the length of a pendulum, you're gonna increase the
inertia of that pendulum. That's gonna make it take longer to go through a full cycle. This is while I go into the park and finding the long swings. The longer the swing, the
longer it actually takes, the more time it actually
takes to swing back and forth. I think that's more fun
than the little short swings that go back and forth really quick. So, let's try a sample problem to see how this period formula works. Let's say you went to the park. You're 60 kilograms. You're swinging on a swing and your friend pulls you back 20 degrees on a swing that's one meter in length. Let's find the period of the motion. So, in other words, the time it takes to go all the way to here and then all the way back to there. We use the period formula for a pendulum. It's two pi, root L over g. And so, we would do two
pi times the square root, the length here is the
length of the string here. So, one meter, technically, it'd be to wherever
your center of mass is, but we're gonna assume our center mass is right here at the end, divided by g, well, we're
on earth, I'm assuming, so, 9.8 meters per second squared. If you solve all of that, you get a period of about
two seconds exactly. So, this swing would have a
period of about two seconds. Now, notice we did not
use this 20 degrees. That's the amplitude. This period does not
depend on the amplitude. We also did not use the fact
that I was 60 kilograms. The period of a pendulum
also does not depend on the mass of the bob at
the end of the pendulum. We only use the length
and if you're on earth, the denominator here
is always gonna be 9.8. Now, one thing I should be
clear about is that a pendulum is technically not a perfect
simple harmonic oscillator. It's only approximately a
simple harmonic oscillator. So, this formula here is
only approximately correct, but for small angles, it's almost perfect. So, at 20 degrees, this formula is only gonna be off. The error's only going to be about 1%, which isn't bad. If you get this back to like 70 degrees, even then, the error's only around 10%. So, this is a really,
really good approximation if you're in small angles,
like around 20 to 30 degrees. The bigger the angle gets though, the worst the approximation gets. For small angles, most
physicists just treat a pendulum as if it's a perfect
simple harmonic oscillator. As you get to those higher angles though, you gotta be careful. This can start deviating
more significantly from the actual value. So, to recap, the period of a pendulum depends on the length of the pendulum and the surface gravity of
the planet that you're on. It does not depend on the amplitude or the mass of the pendulum. And the form it takes is very similar to the period of a mass on a spring where the numerator increases due to increased inertia and the denominator increases due to increased restoring force.