00:01
We're given the linearized equation for a pendulum.
00:05
So we assume that sine of theta is small, or theta is small, so that sine of theta is approximately theta.
00:12
So we get theta double dot plus g over l theta equals zero.
00:15
And we're given some initial conditions, so that this pendulum is at initial angle of 0 .2 radiance, and it has an initial velocity of one rating per second squared.
00:27
So you can say here that it's like here, and we give it some of velocity, so it's going to come back up and then it's going to swing.
00:34
And there's no dissipation in here, so it's going to, you know, in theory, swing forever.
00:43
So this is just a second order differential equation, and with a solution that looks like this, c1 and c2 we could find from initial conditions.
00:54
So c1 is 0 .2, and c2 is square root of l over g.
01:00
And i guess at some point, let's see here, i guess i didn't write that down here.
01:06
They told us that l was one meter.
01:10
I didn't use that until the very end or until, you know, getting solutions.
01:16
So our solution is given by this, right? and so if we plug in, you know, if we plug in l is one and we know g is 9 .8.
01:30
One meters per second squared.
01:34
Here's our solution.
01:37
Now, they ask us for the period, so how long it takes to go through one cycle.
01:45
And so that's 2 pi divided by the natural frequency.
01:48
And the natural frequency is the square root of this, square root of g over l.
01:54
And so that winds up being 2 pi times the square root of l over g.
01:58
And that turns out to be very roughly, very close to two seconds.
02:02
So it takes basically two seconds to go.
02:05
It's pretty slow...