# Homework Help: Simple Harmonic Motion and time equation

1. Oct 29, 2005

### erik-the-red

Question:

An object is undergoing simple harmonic motion with period 0.305 s and amplitude 6.05 cm. At $$t=0$$ the object is instantaneously at rest at $$x=6.05$$ cm.

A.
Calculate the time it takes the object to go from $$x=6.05$$ cm to x = -1.49 cm.

I'm thinking $$\omega = \frac{2\pi}{T} = 20.6$$ rad/s. If we assume left to be the negative direction, then the $$\Delta x = -0.0754$$ m.

So, I can use the equation $$x = A\cos(\omega t + \phi)$$. $$\phi = 0$$.

$$-0.0754 = .0605cos(20.6t)$$.

But, wait a minute here, this can't possibly be right. If I divide both sides by .0605, then I have cos equaling a magnitude greater than one.

I'm stuck.

Last edited: Oct 29, 2005
2. Oct 29, 2005

### Tom Mattson

Staff Emeritus
Your problem is that you set $\Delta x$ equal to $x$. $\Delta x$ is the displacement between the initial and final positions, and $x$ is the $x$-coordinate of the particle, which they tell you is $-1.49cm$.

3. Oct 29, 2005

### erik-the-red

Thanks. I guess the problem wasn't as hard as I thought it was.