Simple Harmonic Motion and amplitude help

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ntox101
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Homework Statement



A 0.50 kg particle executes linear harmonic motion about the point x = 0.0. At t= 1.0, it has a displacement x = 0.50m and a speed of 5.0 m/s to the right. The frequency of the motion is 2Hz. Determine (a) the spring constant, (b) the amplitude of the motion, (c) the displacement as a function of time t...blah blah blah...

The rest of the questions wants you to form equations and find displacements at other times which is a piece of cake.



Homework Equations



Acos(wt+ phase)
-wAsin(wt+phase)


The Attempt at a Solution



Spring constant is easy - 78.87 N/m.

Ok, when it is asking for amplitude I am little confused. Maybe someone can help clear it up. It says it executes linear harmonic motion at x = 0.0. Does this mean that there isn't a phase angle since it is initially at rest at t = 0s and x = 0? If there isn't a phase angle that makes solving for amplitude so much simpler since I can just set it to 0 and calculate the amplitude using the conditions at t = 1.0s.

Anybody?
 
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At x = 0m, the velocity is at a maximum (if it was at rest, it wouldn't oscillate). The frequency is f = 2Hz, which would mean the period is T = 0.5s. At t = 1.0s, it made two full oscillations, which would mean that at t = 0.0s, x = 0.50m. You can calculate the phase angle from this.
 
How so? The way I learned to find phase angles is take the displacement equation x(t) and the velocity equation v(t). If you divide these 2, you are left with x(t) / v(t) = tan(wt + phase angle).
 
I overlooked the amplitude...sorry about that. Since you know the mass, position, velocity, and spring constant, you can use the conservation of energy to find the amplitude.
 
Hey there, I got the problem solved.

What I did actually was took v(t) / x(t) . If you write out the equations you will see that the amplitude cancels out and you are left with tangent of some number. I then used guess and check to find the the phase angle which took a while. It was a very painstaking procedure that took up a huge mess of paper.

I checked it using the conservation of energy and I got the same answer.

Thanks for the help.