Determine the particle's equation of motion

1. Apr 25, 2008

Sheneron

1. The problem statement, all variables and given/known data
A particle with a mass of 0.500 kg is attatched to a spring with a force constant of 50.0 N/m. At time t = 0 the particle has its maximum speed of 20.0 m/s and is moving to the left. (Use t as necessary.)
(a) Determine the particle's equation of motion, specifying its position as a function of t, time.
(b) Where in the motion is the potential energy three times the kinetic energy?
(c) Find the length of a simple pendulum with the same period.
(d) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m.

3. The attempt at a solution
I only need help with the first part for now, but here is what I tried.

$$\omega = \sqrt{\frac{k}{m}}$$
$$\omega = 10$$

$$v_{max} = A\omega$$
$$A = 2$$
So my equation of motion looked like this.
$$x(t) = 2cos(10t)$$

but that was wrong and I am not entirely sure why

2. Apr 25, 2008

Staff: Mentor

You are off by a phase factor. Note that at t = 0 the particle is at max speed--so what's its position with respect to equilibrium?

3. Apr 25, 2008

Sheneron

It's position, with respect to equilibrium, would be 0, which means the phase factor needs to be pi?

4. Apr 25, 2008

Staff: Mentor

Right.
Write the general expression and see if that works.

5. Apr 25, 2008

Sheneron

Ok that worked and I am on part D) now.

x(t)= 2sin(10t + pi)

I set x to 1 and solved for t but that didn't work. Can you not do that?

6. Apr 25, 2008

Staff: Mentor

That should work fine. What did you get? (Remember that you're working in radians.)

7. Apr 25, 2008

Sheneron

$$\frac{sin^{-1}(.5)-\pi}{10}$$

that gives me a negative .261

perhaps I am not doing my algebra right?

8. Apr 25, 2008

Staff: Mentor

Two things you need to do:
(1) Choose a value of $sin^{-1}(.5)$ that gives you a positive time.
(2) Figure out the times when x = 0. (t = 0 is one such time, but it's not the one you want. At t = 0 the particle is moving left; you want the time when it's moving right.)

9. Apr 25, 2008

Sheneron

but my time = all that stuff, and sin of .5 - pi always comes out a negative number..

10. Apr 25, 2008

Staff: Mentor

Realize that there's more than one angle that satisfies $\sin\theta = .5$.

11. Apr 25, 2008

Sheneron

For arcsin of 0.5 I get a value of 0.52359, so you are saying there is another value in addition to this?

12. Apr 25, 2008

Sheneron

Can anyone that is still awake help me solve part D?

13. Apr 26, 2008

Staff: Mentor

I think you've figured it out by now, but sure:
$$\sin\theta = \sin(\theta + 2\pi)$$

That's just one example (but it's the one you need). Look at a graph of a sine function and see all the places where it equals .5.

14. Apr 26, 2008

Sheneron

I figured it out, but I still didn't understand it, so I am glad you posted again.