# SHM Quick Conceptual Question

Say a mass is suspended vertically from a spring and is in equilibrium.

What is its potential energy? I think its zero because while it has gravitational potential that would allow it to do work downwards, it also has potential energy from the spring upwards. So do these cancel out? It seems like they have to for SHM equations to work out.

What is its potnetial energy when it is pulled down m meters? (1/2)km^2

This isn't homework just a a concept I'm confused on.

vsage
the m in your equation is relative to the equilibrium position so I think you're probably right. Someone else may want to weigh in on this.

Doc Al
Mentor
With a mass hanging from a spring, it's simpler to measure displacement from the equilibrium point. (Thus, if the mass is at the equilibrium point, the potential energy is zero.) If you do that, then when the mass is moved a distance x from that equilibrium point, the change in potential energy is given by $1/2 k x^2$. Note that this includes gravitational PE.

If you wanted to keep measuring things from the original unstretched position of the spring, then you'd have to add the gravitational PE term. But you'd get the same net $\Delta {PE}$ when you displace the mass from equilibrium. (Convince yourself of this by doing the calculation.)

It's of course much easier to use the equilibrium position as the reference for analyzing the resulting SHM.

Ok, I think I understand just tell me if this is right....

You only need gravitational PE if you consider equilibrium to be mass on the unstretched string. If you consider equilibrium to be mass on the stretched string(not moving) then gains in gravitational PE are included in the .5kx^2 term. This is probably why I've never seen a problem that tells you the height at which the mass is oscillating. Sounds pretty much like I reiterated what Doc Al said but it helped typing it. I'd do the math but I have a policy of no homework the day before the test 