Spring Potential Energy

  • Thread starter forty
  • Start date
  • #1
forty
135
0
(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
Mentor
45,462
1,950
Don't panic! You are correct! :approve:

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

Suggested for: Spring Potential Energy

  • Last Post
Replies
6
Views
430
  • Last Post
Replies
12
Views
931
  • Last Post
Replies
17
Views
407
Replies
3
Views
440
  • Last Post
Replies
13
Views
434
Replies
12
Views
230
  • Last Post
Replies
2
Views
628
Replies
20
Views
332
  • Last Post
Replies
1
Views
346
Top