Solving Spring Energy: Half Amplitude & Kinetic Energy %

AI Thread Summary
At half amplitude in an ideal spring system, the total mechanical energy is conserved, with potential energy and kinetic energy being interconvertible. At maximum amplitude, potential energy is at its peak while kinetic energy is zero. Conversely, at the equilibrium point, kinetic energy is maximized and potential energy is zero. To find the percentage of kinetic energy at half amplitude, one can calculate the potential energy at that position and subtract it from the total energy. The remaining energy at half amplitude will represent the kinetic energy, allowing for the determination of its percentage.
go2cnavy
Messages
19
Reaction score
0
For an Ideal spring...At half Amplitude, what % of the energy is kinetic?

I know at A the kinetic energy it instantly zero which makes max potential energy (.5kx^2)

I also know that after released, at the instant passising through the equilibrium point kinetic energy is max and potential energy is zero.

From the energy conservation approach Et = Ei so, .5mv^2 = .5kA^2


I have no clue where to start this problem. Maybe I am making it too difficult
 
Physics news on Phys.org
go2cnavy said:
For an Ideal spring...At half Amplitude, what % of the energy is kinetic?

I know at A the kinetic energy it instantly zero which makes max potential energy (.5kx^2)

I also know that after released, at the instant passising through the equilibrium point kinetic energy is max and potential energy is zero.

From the energy conservation approach Et = Ei so, .5mv^2 = .5kA^2


I have no clue where to start this problem. Maybe I am making it too difficult
You know how to calculate the potential energy at full amplitude when it is the total energy. You should also be able to calculate it at half amplitude when it is not the total energy. Any energy that is not potentail energy at half amplitude (or anywhere else) is kinetic energy.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top