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

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.

[tex]x(t)=x_{0}sin(\omega t)[/tex]

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 [tex]k/m[/tex], is not relevant here.

 

[Winegar12]

So was I right to put the amplitude in for x_o and then for [tex]\omega[/tex] 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

[tex]\omega t=\frac{2 \pi}{T}t=4 \pi t= 40 \pi [/tex]

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.

 

[rwisz]

I understand your frustration with trying to solve this problem. It can be difficult to figure out the correct equations and values to use without proper guidance. Here are a few pointers to help you with this question:

1. First, let's clarify the given information. The graph shows an object oscillating with an amplitude of 20 cm and a period of 0.5 seconds. This means that the object is moving back and forth between 20 cm and -20 cm every 0.5 seconds.

2. To find the acceleration of the object at t = 10.0 s, we can use the equation a = -\omega^2x, where \omega is the angular frequency and x is the displacement of the object from its equilibrium position.

3. To find \omega, we can use the equation \omega = \frac{2\pi}{T}, where T is the period of the oscillation. In this case, T = 0.5 seconds, so \omega = \frac{2\pi}{0.5} = 4\pi.

4. Now, we need to find the displacement of the object at t = 10.0 s. Since the object is oscillating with a period of 0.5 seconds, at t = 10.0 s, the object will have completed 20 oscillations. This means that the displacement will be 20 times the amplitude, or 20(20 cm) = 400 cm.

5. Plugging in the values we have found, we get a = - (4\pi)^2(400 cm) = -1600\pi^2 cm/s^2.

6. Finally, we can convert this to the desired units of cm/s^2 by dividing by 100 and multiplying by 1 second^2, giving us a = - 16\pi^2 cm/s^2. This can also be written as a = - 1.6\times 10^2 \pi^2 cm/s^2 or approximately -1.3\times 10^{-6} cm/s^2.

I hope this explanation helps you understand how to approach this problem. Remember to always check your units and use the correct equations for the given situation. Good luck on your test!