How do energy conversions in a spring work during one complete oscillation?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 4K views
xJJx
Messages
56
Reaction score
2
Hi, I'm having trouble understanding the energy conversions in a spring. I know that whilst a spring is being deformed, it gains elastic potential energy and at maximum deformation it has max elastic potential energy. But, does a spring have maximum kinetic energy at its un-deformed state? if so, how? it will have zero velocity at its un-deformed state so how can it have max kinetic energy?
 
Physics news on Phys.org
xJJx said:
Hi, I'm having trouble understanding the energy conversions in a spring. I know that whilst a spring is being deformed, it gains elastic potential energy and at maximum deformation it has max elastic potential energy. But, does a spring have maximum kinetic energy at its un-deformed state? if so, how? it will have zero velocity at its un-deformed state so how can it have max kinetic energy?
Welcome to the PF.

Say you have a mass at the end of a spring. When you compress that system to some Δx, you invest energy in that compression, and that is the potential energy that you store in the compressed mass+spring system. When you release that compressed spring+mass, it extends and gains KE at the expense of PE. The KE is maximum as the mass passes the zero-displacement position, and the stored PE is zero at that position.

The system oscillates about that position indefinitely barring loss in the system. Does that help?
 
berkeman said:
Welcome to the PF.

Say you have a mass at the end of a spring. When you compress that system to some Δx, you invest energy in that compression, and that is the potential energy that you store in the compressed mass+spring system. When you release that compressed spring+mass, it extends and gains KE at the expense of PE. The KE is maximum as the mass passes the zero-displacement position, and the stored PE is zero at that position.

The system oscillates about that position indefinitely barring loss in the system. Does that help?
So you're saying the spring has max Ke at that very point of zero displacement? but it then a
 
xJJx said:
So you're saying the spring has max Ke at that very point of zero displacement? but it then a
Your post got cut off a bit...

But yes, when you pull the mass back and let go, the spring & mass undergo an oscillation. If there is little damping, it just rings like a bell. If you plot the KE and PE as functions of position, you will see that PE is max and KE is zero at the ends of the mass' travel, and KE is max and PE is zero in the middle (the place where the non-moving mass was settled before you pulled it back).

Does that make sense? It's probably easy to find such a plot with a Google Images search...
 
Here are nice plots of the KE and PE of a harmonic oscillator (like your mass + spring) as functions of time and position:

http://www.kshitij-iitjee.com/Study/Physics/Part1/Chapter13/41.jpg
41.jpg
 
berkeman said:
Your post got cut off a bit...
Thank you so much, I understand it a lot better now, especially with the help of the graphs! So do the energy transfer stages go like this for one complete oscillation? (assuming there is no damping in the system):

One complete oscillation of a spring: The spring starts off stationary, meaning it has no kinetic energy and no EPE, it only has GPE. As the spring is being deformed, it is gaining EPE and KE. The spring then reaches its maximum possible deformation; at this point, the spring has maximum EPE and zero KE.

Once the deforming forces stop acting on the spring, it eventually returns back to its original shape; the spring oscillates towards its equilibrium position whilst all of its EPE is getting transferred into KE. At the equilibrium position, all of the springs EPE has now been transferred into KE, so the spring has maximum KE and zero EPE.

The spring then oscillates towards its maximum possible deformation (the type of deformation is the opposite to its first type of deformation) whilst all of its KE is getting transferred into negative EPE. At the maximum possible deformation, all of the springs KE has now been transferred into negative EPE, so the spring has maximum negative EPE and zero KE.

The spring then oscillates back towards its equilibrium position whilst all of its negative EPE is getting transferred into KE. At the equilibrium position, all of the springs EPE has now been transferred into KE, so the spring has maximum KE and zero EPE. The spring has now returned back to its original shape.

(Sorry its so long and detailed, I have an assignment where I have to describe it in a lottt of detail haha)
 
Last edited by a moderator: