What is the maximum potential energy of an oscillating mass?

AI Thread Summary
The discussion centers on calculating the maximum potential energy of a 2-kg mass oscillating on a spring in simple harmonic motion. At the equilibrium position, the kinetic energy (KE) is determined to be 25 J, leading to the conclusion that the total mechanical energy (KE + potential energy, PE) remains constant. Since potential energy is zero at equilibrium, all energy is kinetic at that point. At the maximum displacement, the kinetic energy drops to zero, and the potential energy reaches its maximum value of 25 J. The key takeaway is that the maximum potential energy of the oscillating mass is 25 J.
lussi
Messages
14
Reaction score
0
1. Homework Statement
A 2-kg mass attached to a spring oscillates in simple harmonic motion and has a speed of 5 m/s at the equilibrium point. What is the maximum potential energy of this oscillating mass?

2. Homework Equations
I know that the potential energy is: Ep = 1/2 kx2
k = mω2
 
Physics news on Phys.org
In SHM,

KE+PE = Constant.

At the equilibrium position, the PE is what value? Once you have that then at the max position of the spring, the KE is what value?
 
So, at the equilibrium position KE = PE, therefore KE = 1/2 mv2 = 25 J. And PE = 25 J. If that's correct. But I don't know what will be the value at the max position
 
Not that KE= PE. It is that KE+PE= a constant. So, you can (for example), find all the different energies at the equilibrium and then find all the energies at your maximum potential. And then you know that they must equal the same constant (in the end you should only have max PE which you don't know).
 
I really don't know how to do it, and I have an exam tomorrow. If you tell me the solution, I might find the logic by looking at it.
 
lussi said:
I really don't know how to do it, and I have an exam tomorrow. If you tell me the solution, I might find the logic by looking at it.

KE +PE = Constant

At the equilibrium position you found that PE=0 so that KE = Constant = 25 J

Therefore

KE+PE = 25

At the maximum positionm what would be the KE? (when the spring is oscillating after it reaches its max position does it keep going or does its velocity change?)
 
Since KE + PE = const, and at the equilibrium position PE = 0, therefore KE = Constant = 25 J, as you rock.freak667 said. And since "they must equal the same constant", as Jufro said, that means that at the maximum position KE = 0 and PE = const = 25 J. Is that correct, or am I mistaken again?
 
lussi said:
Since KE + PE = const, and at the equilibrium position PE = 0, therefore KE = Constant = 25 J, as you rock.freak667 said. And since "they must equal the same constant", as Jufro said, that means that at the maximum position KE = 0 and PE = const = 25 J. Is that correct, or am I mistaken again?

Yes that is correct. So PE=25 J at the max position.

For SHM KE+PE = constant at any point in the motion
 
Thank you very much :)
 
Back
Top