Vertical spring, the relationship between potential energies

[al33]

I'm stuck with a solution to the vertical spring with a body suspended. The solution says that when the body's gravitational potential energy is decreasing, the spring's potential energy must be increasing. How is this correct?

I understand the analysis leading to the result that the vertical SHM doesn't differ from horizontal SHM, but I can't reach to that statement above. When the body moves up or down from the equilibrium position, the spring are both stretching, but the gravitational potential energy is either increasing or decreasing, right? Can someone help?

 

[jrmichler]

In the world of physics, a spring has zero potential energy when it is in its equilibrium position. A physics spring, unlike a real world spring, can be either lengthened or shortened from its equilibrium position. Moving a spring from its equilibrium position increases the potential energy of that spring regardless of the direction of that movement.

 

[al33]

Yes.
So,
(1) moving up from the equilibrium, potential energy of spring increases, gravitational potential energy increases (less negative)
(2) moving down from the equilibrium, potential energy of spring increases, gravitational potential energy decrease (more negative)
Is that correct?

 

[jrmichler]

Yes.
Keep in mind that you have two separate equilibrium positions:
1) No weight on spring. Your case #2 starts from this position.
2) With weight on spring. SHM starts from this position.

 

[Dale]

The solution says that when the body's gravitational potential energy is decreasing, the spring's potential energy must be increasing. How is this correct?

Hmm, I have a problem with that too. The spring’s PE has a minimum, and increases in either direction from that minimum. The gravitational PE does not have a minimum and only increases in one direction.

 

[al33]

Soon after I posted that question, I figured it out. For small oscillation, the spring with mass is always moving with a total length longer than its natural length, which means no matter what position it is at now, its potential energy is never zero. And this reaches its maximum at the bottom and minimum at the top, which is just opposit to the case of gravitational potential energy with zero at the bottom. This is just the condition we set as to the derivation of the fact that vertical SHM has no difference with the horizontal one.

 

[256bits]

Yes.
So,
(1) moving up from the equilibrium, potential energy of spring increases, gravitational potential energy increases (less negative)
(2) moving down from the equilibrium, potential energy of spring increases, gravitational potential energy decrease (more negative)
Is that correct?

You have a difficulty with the meaning of equilibrium position.
It seems that you are making the interpretation as if the spring is in the horizontal position rather than in the vertical.

For a horizontal spring, the equilibrium position is the un-stretched length of the spring.
For a vertical spring, the equilibrium position is that length at which the mass would hang motionless if not disturbed.

In both cases, the net force on the mass is zero. .
For the vertical spring that would be when the weight of the mass is balanced by the spring force,

So (1) is incorrect for a mass hanging from a spring.

The problem is making an assumption that the spring does not go onto compression during oscillations of the mass.

 

[al33]

So (1) is incorrect for a mass hanging from a spring.

The problem is making an assumption that the spring does not go onto compression during oscillations of the mass.

Right! I was confused by taking the compression case into the consideration.

 

[256bits]

Right! I was confused by taking the compression case into the consideration.

You do understand the correct meaning of equilibrium position for a vertical spring?

 

[al33]

I think so, before challenged. It is the position when kx equals mg. And the kinetic energy gained then can be thought of just as the potential energy lost with respect to the new equilibrium position only. For the spring stretched down from that equilibrium position and let go, we don't need to worry about potential energy with respect to the natural length of the spring.

 

[256bits]

Looks good from my perspective.