# Energy of simple harmonic oscillator

1. Nov 19, 2008

### clemente

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 + $$\tau$$)

3. 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.

2. Nov 19, 2008

### 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.

3. Nov 19, 2008

### clemente

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