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 :)

When is \cos\theta = 0?

Then, equate theta to what you have in brackets.

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.

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?

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?

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

Thank you so much for help:)

You're welcome :)