# Find the period of the motion and the farthest distance from equilibrium

## Homework Statement

an object of mass m experiences a force of the form
Fx = -kx
Fy = -ky

Its initial speed is v and at that time it is at (x,y) = (0,0)

Find the period of the motion and the farthest distance from equilibrium

## Homework Equations

U = -$$\int F(r) dr$$
T = 2$$\pi$$$$\sqrt{m/k}$$

## The Attempt at a Solution

I think this is a superposition of SHM, so the period would just be 2pi root(m/k), yes?

Furthermore since the force acting on the particle is conservative it has a potential energy given by .5kr^2 and its initial energy is .5mv^2, so when its speed equals zero the potential energy is maxed so .5kr^2 = .5mv^2? v is again the initial speed

What happens if the k's in the force equation aren't equal though? In terms of the period i mean. Is the motion just characterized by two separate periods? One in x and one in y?

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collinsmark
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I think this is a superposition of SHM, so the period would just be 2pi root(m/k), yes?
'Sounds reasonable to me.
Furthermore since the force acting on the particle is conservative it has a potential energy given by .5kr^2 and its initial energy is .5mv^2, so when its speed equals zero the potential energy is maxed so .5kr^2 = .5mv^2? v is again the initial speed
That also sounds reasonable.

But using the above relationship, the exact maximum distance from equilibrium is contingent on the fact that at some point in the oscillation, it passes through point (0,0). That is specified in the problem statement. If that tidbit of information wasn't given, there wouldn't be enough information to solve the problem (without giving at least some sort of initial condition -- and even then, it might not be solved in exactly the same way). For example, the mass might just be going around in a circle (with a constant speed and a constant distance).
What happens if the k's in the force equation aren't equal though? In terms of the period i mean. Is the motion just characterized by two separate periods? One in x and one in y?
Yes, that's right. And you have to be a little more careful too about how you define your U and T equations, because there are now two ks to be concerned with, instead of just one. So you won't be able to factor out the k.