Simple Harmonic Motion and time equation

[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.

 

[quantumdude]

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.

 

[erik-the-red]

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