Determine the particle's equation of motion

Homework Statement

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.

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

Related Introductory Physics Homework Help News on Phys.org
Doc Al
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?

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

Doc Al
Mentor
It's position, with respect to equilibrium, would be 0,
Right.
which means the phase factor needs to be pi?
Write the general expression and see if that works.

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?

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

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

that gives me a negative .261

perhaps I am not doing my algebra right?

Doc Al
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.)

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

Doc Al
Mentor
but my time = all that stuff, and sin of .5 - pi always comes out a negative number..
Realize that there's more than one angle that satisfies $\sin\theta = .5$.

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

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

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

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