00:01
Right, hello, in this question we're told that we have a spring with a mass of one kilogram on the end and the spring constant k is 200 newtons per meter and then we're told that we're going to stretch this back and cause it to oscillate with an amplitude equal to five centimeters and we're asked various things.
00:17
The first part, a, how much time elapses before the mass first returns to its equilibrium position? so in order to answer that we're going to need to write an equation for the position of this object.
00:30
So i know that for simple harmonic motion my position at some time t is going to my amplitude times the cosine of omega t or it's going to be the amplitude times the sine of omega t.
00:44
In this case we know that at time equals zero we're going to be at our amplitude so if we plug in zero into here cosine becomes one so we're going to discredit that one in this case.
00:53
Though if we started at the equilibrium position that would be our function there.
00:58
So we know that the amplitude is going to be 5 .0 centimeters and that's going to be times the cosine of omega.
01:04
Well i know omega is equal to 2 pi over the period and i know that the period of a spring is going to be 2 pi root m over k.
01:14
So if i can find those two i'm going to find that omega is going to be root k over m and i have k and i have m.
01:22
K is 200, m is one so if i square root of that i'm going to get that omega is 14 .1 per second times t.
01:31
So this is going to be my equation for my position.
01:34
I'm going to go ahead and drop the units just noting that x is in centimeters here just because it makes it a little easier to work with.
01:41
And then i want to find the time when it first returns to its equilibrium position.
01:44
So i want to set this equal to zero and solve for that first time.
01:49
Well i know this expression is going to equal zero when cosine is equal to pi over two and so i want 14 .1 t to equal pi over two.
02:00
So time is going to be pi over 28 .2 which is going to be 0 .11 seconds.
02:07
That will be the time it first returns to equilibrium.
02:10
Part b asks how much time elapses before the mass reaches its maximum displacement in the direction opposite.
02:17
So in this case it starts over here and then we're going to release it.
02:20
It's going to go through equilibrium and then it's going to end up over here and we want to figure out that time.
02:25
Well in that case its position x of t is going to be negative times its amplitude which we know is 5 .0 centimeters.
02:34
So we want that to equal 5 .0 cosine of 14 .1 t...