# 2D isotropic mass-spring system, need to find maximum distance from origin

1. Feb 6, 2010

### naele

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

I have a 2D isotropic mass-spring system. The mass is pulled a distance A=1m, and then given an upwards kick with a velocity v_0. The k=1, m=1kg.

I need to find the furthest distance from the origin the mass will travel in its orbit.

2. Relevant equations
$$x(t)=A_x\cos(\omega t)$$
$$y(t)=A_y\cos(\omega t-\delta)$$

3. The attempt at a solution
The first thing I'm unsure of is how to handle the relative phase. I initially just solved for it using the initial conditions so when t=0
$$0=A_y\cos(-\delta)$$
$$\delta=-\cos^{-1}(0)$$
$$A_y=\frac{v_0}{\sin(\delta)\omega}$$
I'm not 100% sure if this is acceptable.

Next, my initial approach was to set a function r(t)=sqrt(x(t)^2+y(t)^2), take the derivative set to zero and solve for t that way. In principle this would have given me the times at which the distance is at a max or a min. I only got one solution which I'm not sure makes sense.

So I have

$$r(t)=\sqrt{(\cos(\omega t))^2+(A_y\cos(\omega t-\delta))^2}[\tex] where for simplicity I substituted $A_x=1$. Taking the derivative gives a really messy result [tex]r'(t)=\frac{-2\cos(\omega t)\sin(\omega t)\omega+2A_y^2\cos(-\omega t+\delta)\sin(-\omega t+\delta)\omega}{2\sqrt{(\cos(\omega t))^2+(A_y\cos(\omega t-\delta))^2}}$$

Setting equal to 0 then

$$2\cos(\omega t)\sin(\omega t)=2A_y^2\sin(-\omega t+\delta)\sin(-\omega t+\delta)$$

Using a trig identity

$$\sin(2\omega t)=A_y^2\sin(2(-\omega t+\delta))$$

I should mention I used Maple to find the derivative, and then I used Maple again to solve for t. This is the expression it gave me

$$t=\frac{\arctan(\frac{A_y^2\sin(2\delta)}{1+A_y^2\cos(2\delta)})}{2\omega}$$

This is the point where I am. I'm not at all sure about my work.

edit: it seems like I broke the latex interpreter. The equation it's not displaying is Ay=v_0/(sin(delta)w)

Last edited: Feb 6, 2010
2. Feb 6, 2010

### diazona

Well for starters, those aren't the right equations. You're probably thinking of circular motion, which is described by the equations
$$x(t) = A\cos(\omega t + \delta)$$
$$y(t) = A\sin(\omega t + \delta)$$
But then again, this isn't circular motion. So there's no reason you'd expect those equations to work. (Note that when using the proper equations in the appropriate situation, you would solve for the phase using the initial conditions.)

That would be the painful way to do it Think about it this way: what conservation laws apply to the system?

3. Feb 6, 2010

### naele

Hmm, I would say conservation of energy applies. In which case the maximum distance is achieved all energy = potential energy, or $$U=\frac{1}{2}k(x+y)^2$$.

I'm just not sure how I can use that to solve for the maximum distance.

Last edited: Feb 6, 2010