Energy is conserved for SHM; the sum of the potential and kinetic energies is constant.

The potential energy is:  U = ½ k x² = ½ k A² cos² (ω t)

The kinetic energy is:     K = ½ m v² = ½ m ω² A² sin² (ω t)

The total energy E = K + U is:  

½ k A² cos² (ω t) + ½ m ω² A² sin² (ω t)

Using k = m ω², we find

E = ½ k A² cos² (ω t) + ½ k A² sin² (ω t) = ½ k A²

Notice that E also equals the maximum potential energy:

U_max = ½ kA²

E also equals the maximum kinetic energy:

K_max = ½ m v_max² = ½ mA² ω² = ½ k A²

The potential and kinetic energies go through two cycles for each amplitude cycle.

The first set of graphs is for ω = 1 rad/s. The second set is for ω = 0.8 rad/s. This change of ω is accomplished either by decreasing the spring constant or increasing the mass. Which change was made?

1. Decreased the spring constant
2. Increased the mass
3. Either, you can't tell the difference