How Is Kinetic and Potential Energy Balanced in Harmonic Motion?

[forty]

(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!

 

[Doc Al]

Don't panic! You are correct!

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

 

[Moetasim]

Your approach to the first part is correct. Equating the kinetic energy to the potential energy gives you the point where they are equal. This point is x = A/√2, as you have correctly calculated.

For the second part, you are on the right track. At x = A/2, the potential energy is equal to half of its maximum value, which is .5kA^2. This means that the remaining half of the maximum energy must be kinetic energy. So, at x = A/2, the fraction of kinetic energy is .5 and the fraction of potential energy is .5. In other words, half of the total energy is kinetic and half is potential.

To summarize, at x = A/√2, the kinetic energy is equal to the potential energy. At x = A/2, the kinetic energy is half of the total energy and the potential energy is also half of the total energy. I hope this helps clarify things for you.