00:01
So for part a, from the graph, we can see that the period is equaling 3 .0 seconds.
00:05
And so we have that then the period t is equaling 2 pi multiplied by the square roots of the mass divided by the spring constant k.
00:12
Solving for the mass, this would be equalling to the period squared, multiplied by the spring constant, divided by 4 pi squared.
00:21
And this is equaling then 3 .0 seconds quantity squared multiplied by 240 newtons per meter.
00:29
This is divided by 4 pi squared and we find that then the mass m is equaling to 55 kilograms approximately.
00:39
For part b, oscillations occur about an equilibrium position of one meter.
00:47
So here we can say that the equilibrium position, we can say x of bq, x equal 1 .0 meter.
00:58
So the amplitude will always be plus, rather it'll range from 1 .0.
01:02
Meter plus the amplitude 1 .0 meter minus the amplitude and the amplitude we find to be equal equaling one half multiplied by 0 .80 meters and this is giving us 0 .40 meters we know that then the phase constant or rather from the graph the face constant is zero radiance so we can find that then the angular frequency equaling 2 pi times the time period t this would be 2 pi divided by 3 0 .0 seconds.
01:35
This is giving us then 2 .1 radians per second.
01:41
The x position as a function of time would then equal a cosine of omega -t plus 1 .0 meters...