Consider
[link] , which shows the energy at specific points on the periodic motion. While staying constant, the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring:
The motion of the block on a spring in SHM is defined by the position
$x\left(t\right)=A\text{cos}\left(\omega t+\varphi \right)$ with a velocity of
$v\left(t\right)=\text{\u2212}A\omega \text{sin}\left(\omega t+\varphi \right)$ . Using these equations, the trigonometric identity
${\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta =1$ and
$\omega =\sqrt{\frac{k}{m}}$ , we can find the total energy of the system:
The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude
${E}_{\text{Total}}=(1\text{/}2)k{A}^{2}.$ The total energy of the system is constant.
A closer look at the energy of the system shows that the kinetic energy oscillates like a sine-squared function, while the potential energy oscillates like a cosine-squared function. However, the total energy for the system is constant and is proportional to the amplitude squared.
[link] shows a plot of the potential, kinetic, and total energies of the block and spring system as a function of time. Also plotted are the position and velocity as a function of time. Before time
$t=0.0\phantom{\rule{0.2em}{0ex}}\text{s,}$ the block is attached to the spring and placed at the equilibrium position. Work is done on the block by applying an external force, pulling it out to a position of
$x=+A$ . The system now has potential energy stored in the spring. At time
$t=0.00\phantom{\rule{0.2em}{0ex}}\text{s,}$ the position of the block is equal to the amplitude, the potential energy stored in the spring is equal to
$U=\frac{1}{2}k{A}^{2}$ , and the force on the block is maximum and points in the negative
x -direction
$\left({F}_{S}=\text{\u2212}kA\right)$ . The velocity and kinetic energy of the block are zero at time
$t=0.00\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$ At time
$t=0.00\phantom{\rule{0.2em}{0ex}}\text{s,}$ the block is released from rest.
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