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heyyalliee · Nov 10, 2010, 04:20 PM

The position of a mass oscillating on a spring is given by x = ( 3.2cm)Cos[2(pi)(t)/ .74seconds]
What is the first time the mass is at the position x = 0?

I was able to calculate the period of motion which is .74 seconds.

I am having the hardest time figuring out the time when the position is 0. Maybe its my algebra?

Thanks for the help :)

Unknown008 · Nov 10, 2010, 11:35 PM

When is $\cos\theta = 0$ ?

Then, equate theta to what you have in brackets.

heyyalliee · Nov 11, 2010, 07:31 AM

So cos=0 when theta is 90. So then I set up what's in the brackets =to 90?
Sorry I'm a little confused.

Unknown008 · Nov 11, 2010, 10:39 AM

Yes, that's it :)

But since you have pi in your expression in brackets, it means you have to work in radians.

That is 90 degrees in radians?

heyyalliee · Nov 11, 2010, 11:28 AM

Okay so 90 degrees in radians is 1.5708. I set 2(pi)(t)/ .74 equal to 1.5708.
I got t =1.8258. Is this right?

Unknown008 · Nov 11, 2010, 11:39 AM

I'm getting this:

$90^o = \frac{\pi}{2}$

You might want to keep pi so that you don't have the problem of rounding off.

$\frac{2\pi t}{0.74} = \frac{\pi}{2}$

Pi cancel out, giving:

$\frac{2 t}{0.74} = \frac{1}{2}$

$\frac{t}{0.74} = \frac{1}{4}$

$t = \frac{0.74}{4} = 0.185$

heyyalliee · Nov 11, 2010, 12:12 PM

Thank you so much for help:)

Unknown008 · Nov 11, 2010, 12:15 PM

You're welcome :)