# 1D oscillator solving for Amplitude

1. Mar 1, 2012

### aaj92

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

You are told that, at the known positions x$_{1}$ and x$_{2}$, an oscillating mass m has speed v$_{1}$ and v$_{2}$. What are the amplitude and angular frequency of the oscillations?

2. Relevant equations

x(t) = Acos(wt - $\delta$)
v(t) = -Awsin(wt -$\delta$)

w = $\sqrt{\frac{k}{m}}$

probably others?

3. The attempt at a solution

I need help solving this. I know what the answer should be but I'm not sure if these are the equations I should be using. And if they are, I'm not really sure how to start it. I wrote out all the equations and have just sort of been staring at them. Can someone just help get me started?

2. Mar 1, 2012

### genericusrnme

How would you usually go about solving two equations for two unknowns?

3. Mar 1, 2012

### schliere

I don't think you have to worry about the angular displacement term $\delta$ since it's just one harmonic oscillator without any reference displacement.

4. Mar 1, 2012

### aaj92

yeah the thing is is I got 4 equations? One for each position and speed so like,

x$_{1}$ = Acos(wt-$\delta$)
v$_{1}$ = -Awsin(wt-$\delta$)

and then the same thing for x$_{2}$ and v$_{2}$. Do you only need the ones for x(t) then to solve for the amplitude?

edit: but I do need the equations for velocity because the answer is

A = $\sqrt{\frac{x^{2}_{2}v^{2}_{1}-x^{2}_{1}v^{2}_{2}}{v^{2}_{1}-v^{2}_{2}}}$

:/

5. Mar 1, 2012

### aaj92

yeeeeah there's other equations involved... I somehow managed to solve for w using the fact that A =$\sqrt{x^{2}+\frac{v^{2}}{w^{2}}}$

but I can't really find anything that would help me find A? any ideas?

6. Mar 1, 2012

### Staff: Mentor

Small hints:

1. Forget the offset angle. As was mentioned by schleire above, it's totally arbitrary and meaningless if no reference positions or angles are given at the outset.

2. Rather than using cos() for the position, use sin(). Why? Because you can use either if you don't know where the starting reference position is and, more importantly, the derivative of sin() is cos() so you don't have to deal with the negative sign that comes about when you take the derivative of cos()

3. Take advantage of the fact that $cos(\theta) = \sqrt{1 - sin(\theta)^2}$.

7. Mar 1, 2012

### aaj92

Thanks for the help... apparently there was a really simple way of solving for it using energy

E = T+U

I just completely forgot about the equation :/ but thanks again :)