Simple Harmonic Motion for a spring hanging from the ceiling

[jgalt42]

So we know that at equilibrium point the restoring force is 0, potential energy is 0, and KE is at max and so is the velocity of a mass, if we are assuming the everything is on a horizontal plane.

What about a mass that's attached to a spring and hanging from the ceiling? Wouldn't there be a gravitational force at all times? Therefore at equilibrium point, is the restoring force still 0? Is PE still 0? And is KE and V still at max?

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[chiddler]

What you wrote is true for an ideal spring. If you attach a mass as you've described, then you've damped the spring and that changes the game.

 

[milski]

The 0 for PE is a matter of convention, you can set it anywhere you want. The equilibrium with a weight hanging is just as good as any other position. With a bit of work it can be proven that even with the weight attached the spring will continue to have a simple harmonic motion with a changed equilibrium point. What you know about max/min KE stays the same.

 

[milski]

What you wrote is true for an ideal spring. If you attach a mass as you've described, then you've damped the spring and that changes the game.

Not really. You need something that changes the F=k.Δx to dampen the spring. Friction or heat losses would be a way to do that. Attaching a weight will change the equilibrium point but the motion stays simple harmonic.

 

[chiddler]

Not really. You need something that changes the F=k.Δx to dampen the spring. Friction or heat losses would be a way to do that. Attaching a weight will change the equilibrium point but the motion stays simple harmonic.

oh yes... you're right. energy can be conserved if no friction or heat in such a case.

 

[milski]

oh yes... you're right. energy can be conserved if no friction or heat in such a case.

Here's a quick proof by the way:
For a spring flat table: F=k.Δx
For a spring with a weight: F=k.Δx+mg=k.Δx+k.mg/k=k(Δx+mg/k)=k(x-x0+mg/k)=k(x-x0')=kΔx'

 

[chiddler]

Here's a quick proof by the way:
For a spring flat table: F=k.Δx
For a spring with a weight: F=k.Δx+mg=k.Δx+k.mg/k=k(Δx+mg/k)=k(x-x0+mg/k)=k(x-x0')=kΔx'

how did you do

k(x-x0+mg/k)=k(x-x0')

where is my mg/k

 

[milski]

how did you do

k(x-x0+mg/k)=k(x-x0')

where is my mg/k

It's a constant - x0+mg/k turned into a new constant x0'. The only thing that changed is from what point you measure Δx. k stays the same, which means that frequency and amplitude for the same initial displacement will stay the same as well.

 

[chiddler]

OHH because X(final) is just distance from x0! not some absolute distance

that helped a lot, thanks.