Simple Harmonic Motion and equilibrium problem

[robbondo]

Homework Statement

A partridge of mass 4.94 kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.17 s. When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it?

Homework Equations

The Attempt at a Solution

Well, I've already solved for the k of the spring to be 11.21, and that the velocity at the equivalence point is .151m/s and the acceleration when the mass is .05m above the equilibrium point is-.114 m/s^2. I'm assuming that I'll need to do some type of integral from -.5 to .5, but I have no idea which equation to use, or how to approach this problem.

 

[hage567]

You don't need to integrate. Set up a wave equation for the position of the simple harmonic motion. You can solve this for the time at each position you're interested in.

 

[robbondo]

What is the equation for that graph?

is it $$X= Acos(\omega t + \phi)$$

So, if this is right and I solve for x=.05, A=.1, $$\omega=\sqrt{k/m}=\sqrt{11.21/4.94}$$ and $$\phi=\pi$$

When I solve this equation for t, i get a negative number. So obviously I've done something very wrong. Should I solve for the arclength between the two points? If so, then I'd have to take the derivative, is the equation I have correct to do that?

 

[hage567]

Your equation isn't quite right.

If the mass is pulled down (so negative) to 0.1m and released, what does that tell you about the starting point of your curve? Think of the graph of a cosine function. You must make it suit your conditions. I don't think you will have any phase angle if you use a cosine function (so phi=0).

You do not have to take derivatives or anything like that.

Note: you were given the period of the motion (4.17s), you can use that to get w as well.

 

[robbondo]

cool. I think that I was just messing up with that whole phase angle thing. I was thinking that since it started at negative a that it the phase angle was pi. Which still kinda makes sense to me, but it worked ok without the phase angle, so I guess you were right. Thanks for all your help.

 

[hage567]

If you had used a sine function, then you would have had to worry about a phase angle (try it!).

Glad it worked out!

phi=pi should have worked, by the way (depending on how you set it up). I think I misunderstood what you were saying when you were talking about it. I'm not sure why it didn't work for you.