Simple Harmonic Motion Acceleration Calculation and Equations

[Winegar12]

Homework Statement

I have been stuck on this question for quite some time. I'm trying to study for the test and actually have the answer, but I can't figure it out. The answer is 1.3X10^-6cm/s²
The following graph represents an object oscillating in simple harmonic motion. What is the
acceleration of the object at t = 10.0 s?
You can find the graph here http://rwdacad01.slcc.edu/academics/dept/physics/tvanausdal/2210/exams/sampleexam5.pdf and scroll down to number 9.
I'm assuming that T=.5 and A=20cm

Homework Equations

These are the equations I've been trying to use

a=$$\omega$$²x=-($$\frac{k}{m}$$)x
a=-($$\frac{2pi}{T}$$)²(X_o)sin($$\frac{2pi*t}{T}$$)

The Attempt at a Solution

I have tried to plug in numbers and read somewhere that X_o is amplitude, but that didn't work either. I don't know mass, so I don't know how to use the first equation, and I don't know how to find k. Anyways, I hope someone can give me a few pointers on what I am doing wrong and if the equation I am using is wrong or what equation I should use.

 

[americanforest]

First write down the position function.

$$x(t)=x_{0}sin(\omega t)$$

We use sine since, at t=0, the position is zero. Then differentiate twice with respect to time to get the acceleration.

The first equation, involving $$k/m$$, is not relevant here.

 

[Winegar12]

So was I right to put the amplitude in for x_o and then for $$\omega$$ what would I use to plug in for that, is it 2pi/T?

 

[americanforest]

It seems to me that the answer should be zero, unless there is some discrepancy and the period is not actually half of a second. If the period was half a second then the argument of the sine function would be

$$\omega t=\frac{2 \pi}{T}t=4 \pi t= 40 \pi$$

the sine of which is zero. Since the second derivative of the position function given in my above post will be proportional to the sine function, acceleration should be zero.