Spring Potential Energy

  • Thread starter forty
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  • #1
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(a) Consider a particle of mass m moving along the x axis under the influence of a spring with spring constant k. The equilibrium point is at x = 0, and the amplitude of the motion is A.

(i) At what point x is the kinetic energy of the particle equal to its potential energy?

For this do i equate KE = PE

.5mv^2 = .5kx^2

*** EDIT ***

I worked this out to be:

.5kx^2 = .25kA^2 (at what value x does the potential energy = half of its maximum)

and worked x out to be A/root(2)

(ii) When the particle reaches point x = A/2, what fraction of its total energy is kinetic energy, and what fraction is potential energy?

For this i don't know how to relate x = A/2 to kinetic energy

*** EDIT ***

if the max energy of the system is .5kA^2 then at x = A/2 the PE is .5k(A/2)^2 which is one quarter of the total PE so the rest is KE?

so does this mean that .25 is PE and .75 is KE?

This has become quite frustrating so any ideas would be great!
 
Last edited:

Answers and Replies

  • #2
Doc Al
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Don't panic! You are correct! :approve:

Energy is conserved, so at any point: KE + PE = .5kA^2
 

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