# Energy in simple Harmonic motion

We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible.

The kinetic energy of the body is K = 1 /2 mv^{2} and the potential energy of the spring is U = 1 /2 kx^{2} , just as in Section 7.2. There are no nonconservative forces that do work, so the total mechanical energy E = K + U is conserved:

(Since the motion is one-dimensional, v2 = vx 2.) The total mechanical energy E is also directly related to the amplitude A of the motion. When the body reaches the point x = A, its maximum displacement from equilibrium, it momentarily stops as it turns back toward the equilibrium position. That is, when x = A (or -A), vx = 0. At this point the energy is entirely potential, and E = 1 2 kA2 . Because E is constant, it is equal to 1/ 2 kA^{2} at any other point. Combining this expression with Eq. (14.20), we get

We can verify this equation by substituting x and vx from Eqs. (14.13) and (14.15) and using w^{2} = k/m from Eq. (14.9):

Hence our expressions for displacement and velocity in SHM are consistent with energy conservation, as they must be.

We can use Eq. (14.21) to solve for the velocity vx of the body at a given displacement x:

The ± sign means that at a given value of x the body can be moving in either direction. For example, when x = ±A/2,

Equation (14.22) also shows that the maximum speed vmax occurs at x = 0. Using Eq. (14.10), v = (k)^{1/2}/m, we find that

This agrees with Eq. (14.15): wx oscillates between -wA and +wA

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