Energy of simple harmonic oscillator

[clemente]

Homework Statement

A particle moves along x-axis subject to a force toward the origin proportional to -kx. Find kinetic (K) and potential (P) energy as functions of time t, and show that total energy is contant.

Homework Equations

K = (1/2)m*v^2
P = (1/2)k*x^2
E = K+P

x = Asin(wt + \tau)

v = dx/dt = wA(cos(wt + \tau)

The Attempt at a Solution

K = (1/2)m*v^2 = (1/2)m*w^2A^2(cos^2(wt + \tau)

P = (1/2)k*x^2 = (1/2)k*A^2(sin^2(wt + \tau))

But when I add these to get the total energy, the terms with t do not cancel, and so the total energy is not constant. I can only imagine then that I've done something wrong in the above, very basic steps. Any suggestions would be appreciated. Thanks.

 

[joeyar]

Another equation that comes in handy is w^2 = k/m. Rearranging this you get k = m*w^2 which can be substituted into the first equation, to get K = (1/2)m*v^2 = (1/2)k*A^2(cos^2(wt + \tau )).

When adding K+P, the (1/2)k*A^2 (which is present in both terms) can now be taken as a common factor and using the trigonometric identity cos^2(x) + sin^2(x) = 1 shows the energy is constant.

 

[clemente]

Of course, that seems so obvious now! Thank you very much!