Determine the particle's equation of motion

[Sheneron]

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

 

[Doc Al]

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?

 

[Sheneron]

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

 

[Doc Al]

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.

 

[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?

 

[Doc Al]

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

 

[Sheneron]

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


that gives me a negative .261

perhaps I am not doing my algebra right?

 

[Doc Al]

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.)

 

[Sheneron]

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

 

[Doc Al]

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 $\sin\theta = .5$.

 

[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?

 

[Sheneron]

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

 

[Doc Al]

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.

 

[Sheneron]

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