# Homework Help: Finding Oscillation Amplitude?

1. Apr 21, 2010

### erok81

1. The problem statement, all variables and given/known data

I have a problem where I am given the following values:
Angular Oscillation Frequency which I have assigned to omega
Spring Constant, which is k
The system's kinetic energy in Joules.
Phi is assumed to be 0

I am asked to find the oscillation amplitude at a certain time, t.

2. Relevant equations

I am using two different formulas.

$$KE=\frac{1}{2}(kA^{2})$$ I am using this one to solve for A.

$$x(t)=Acos(\omega t+\varphi)$$ Then this one to solve for the oscillation amplitude at time "t"

3. The attempt at a solution

I am stuck on the oscillation amplitude. I know this is also referred to as A (which is the max amplitude of the object) in my formula, but since it is asking for the amplitude at a certain time, would this be the same thing as x(t), displacement? I've never seen it worded this way, so I am not sure if they are the same thing.

I can provide the values if that will help and I can show my final answer to see if it is correct.

2. Apr 22, 2010

### rock.freak667

Unless your oscillation is damped, 'A' would be constant. I think they want you to find the displacement at a certain time.

Total energy = 1/2 k A2 = 1/2 kx2+1/2mv2

3. Apr 23, 2010

### erok81

Yeah, you are right. I read the question wrong. It just wants the amplitude from an energy at time t. But your answer helps because that clears up A.

So with those givens I posted, is it possible to just use KE=1/2kA^2, and then solve for for A? Or do I need anything extra?

4. Apr 23, 2010

### rock.freak667

Depending on the time t, you will need to use KE+PE = constant which is the equation which I posted and find 'A' from there:

5. Apr 23, 2010

### erok81

Here is the exact question.

A mass on a spring has an angular oscillation frequency of 2.56 rad/s. The spring constant is 27.2 N/m, and the system's kinetic energy is 4.47J when t = 1.56 s. What is the oscillation amplitude? Assume that φ = 0.00.

At first I some how missed the period and thought the time was referring to the oscillation amplitude. But now that I read it again, I think it's just asking for A. Since I don't know the mass, is it possible to solve it just using the KE=1/2kA^2?

I ended up with 0.5 something meters. (I don't have my paper with me right now, so I can't remember my exact answer)

6. Apr 23, 2010

### ehild

The total energy of the oscillator is KE+PE = 1/2 kA^2. The potential energy at a given position x is PE=1/2 kx^2. KE at a given time is

KE(t)=1/2kA^2-1/2 k x(t)^2

Supposing that x=A cos (wt), you know everything to get A.

ehild

7. Apr 23, 2010

### erok81

Hmm...in order to get x, I have to find A first, which is the exact question. I guess that is where I confused. With what I am given, the only thing that makes sense is KE=1/2kA^2.

I don't see why one would need to solve anything related to x=A cos (wt). Since in order to use that I have to know either A or the position, neither of which I have.

***SEE POST FIVE FOR CORRECT QUESTION. ORIGINAL PROBLEM WORDED INCORRECTLY.***

Last edited: Apr 23, 2010
8. Apr 23, 2010

### erok81

Oooh....I see I think.

So if I use the total energy) KE(t)=1/2kA^2-1/2 k x(t)^2, I have two unknowns. But I can sub in cos(wt) for x(t). Solving for A I get....

$$A=\sqrt{\frac{2KE+k[cos(\omega t)]^2}{k}}$$

Which comes out to be 0.87m for A.

Does that look right?

Last edited: Apr 23, 2010
9. Apr 23, 2010

### erok81

Except this is where I am confused. Your equation says

Total energy = 1/2 k A^2 = 1/2 kx^2+1/2mv^2

If the total energy is KE=1/2kA^2. I should be able solve from there??

10. Apr 23, 2010

### ehild

No. You have to sub in Acos(wt) for x(t).

ehild

11. Apr 23, 2010

### erok81

Oh yeah, duh.

$$A=\sqrt{\frac{KE}{0.5k-0.5k[cos(\omega t)]^2}$$

That gives me 0.762m.

The part I still don't get is why can't I just solve for the total energy KE=1/2kA^2 since that represents the total energy of the system?

Last edited: Apr 23, 2010
12. Apr 23, 2010

### ehild

"the system's kinetic energy is 4.47J when t = 1.56 s". 1/2 kA^2 is the maximum potential energy which is the same as the maximum kinetic energy, but different from the kinetic energy at t=1.56 s.

ehild

13. Apr 23, 2010

### erok81

I finally get it. Thanks for including that explanation at the end. That helps me out a ton because now I actually understand why the way I was doing it didn't work.

Thanks again.

14. Apr 23, 2010

### |<ings

Also note that if energy is conserved then 1/2 kA2 is the total energy of the system (K + U). So erok81 the original equation you supplied for KE is actually, as the other posters have pointed out, the equation for total energy of the system, . I think that's where your confusion arised/arose?

Last edited: Apr 23, 2010