00:01
So here the graph makes it clear that from 4 part a, the period t is equaling 0 .20 seconds.
00:07
So we can then, for part b, we can use equation 1513, and this is giving us that the period of a simple harmonic oscillator, t, would be equaling 2 pi times the square root of m over k.
00:26
Therefore we can find the mass the mass would then be equal to t squared divided by four pi squared multiplied by k and so this would be equal to 0 .20 seconds divided by 4 pi squared and then multiplied by the spring constant of 200 newtons per meter and we find that the mass is giving us approximately 0 .203 kilograms.
01:05
For part c now, the graph indicates that the speed is momentarily zero at t equals zero.
01:13
So we can say that the speed equals zero at t equals zero.
01:20
This would be your answer for part b.
01:23
So we can then say that x initial would be equal to plus or minus the amplitude.
01:30
From the graph we also note that the slope of the velocity curve and hence the slope of the velocity curve is equaling the acceleration.
01:38
This is positive at t equals zero.
01:41
So the acceleration is positive at t equals zero.
01:45
And we can then say that the value of x is negative.
01:51
And this is simply from applying newton's second law.
01:54
The mass times the acceleration would be equal to negative kx...