Energy in oscillations

Let's figure out what the kinetic and potential energy is during different phases of an oscillation. The kinetic energy is ${1 \over 2}mv^2$, and we calculated $v$ in 1.7. Using that, and also that $\omega^2 ~=~ k/m$,

$$K ~=~ {1 \over 2}m A^2\omega^2\sin^2(\omega t + \delta) ~=~ {1 \over 2}k A^2 \sin^2(\omega t + \delta)$$ (1.36)

The potential energy is ${1 \over 2}kx^2$ so

$$U ~=~ {1 \over 2}k A^2\cos^2(\omega t + \delta)$$ (1.37)

If we calculate $E ~=~ K +U$ we get

$$E ~=~ {1 \over 2}k A^2 \sin^2(\omega t + \delta) + {1 \over 2}k A^2\cos^2(\omega t + \delta) ~=~$$ (1.38)

$${1 \over 2}k A^2 (\sin^2(\omega t + \delta) + \cos^2(\omega t + \delta) ~=~ {1 \over 2}k A^2$$ (1.39)

Which is constant, and equal to the potential energy when the spring is stretched to its maximum value. This seems correct.

Here we plot the both the position as a function of time and underneath that, kinetic and potential energy.

See how the energy shifts form, constantly transforming between kinetic and potential. For example, Initially all the energy is potential, as the mass is at it's maximum and the velocity is zero. A quarter of the way through a cycle, all that potential energy has gone to kinetic as the mass passes through $x=0$. Then it starts slowing down as it heads over to negative $x$. Eventually at half a cycle, the spring is maximally compressed and the energy is all potential again.