00:01
In this problem, we have a clock pendulum oscillating at a frequency of 2 .5 hertz, and it begins at rest at a starting position of 15 degrees.
00:10
However, we want our answers to be in radians.
00:15
So we can look here, here, to see how to convert degrees to radians, and we simply multiply by 2 pi over 360 degrees.
00:30
There are two pi radians for every 360 degrees, so that's how we convert that, and we get 0 .2618 radians, which i wrote over in our givens section.
00:43
We also want to figure out the position of the pendulum at a certain time.
00:53
And fortunately, the equation for that is quite easy.
00:58
We have what our distance will be in radians the angle theta equals theta initial times cosine of omega t omega is the angular frequency and t is just time so for part a we have t is equal to 0 .35 seconds but we don't know what omega is so we know that for simple harmonic motion omega is equal to 2 pi times the frequency, which would give us 5 pi radians per second.
01:43
Because we have 2 pi there, and frequency is 2 .5 hertz.
01:47
You get 5 pi.
01:48
So if we just plug into our equation above, we will get 0 .2618 radiance times cosine.
02:09
Of 5 pi times t, which is 0 .35 seconds.
02:21
I'm not going to include radiance per second there because all of this, all the units inside our trig function will just cancel out.
02:29
We only really care about the units outside the trig function.
02:34
So for that, we get roughly 0 .19 radians.
02:48
And i am rounding this to two significant figures like the problem asks.
02:54
So then we have part b, and we would not like to know the position of the pendulum at 2 .8 seconds.
03:14
So again, we already have omega, we have t.
03:17
We're just plugging in again...