Determine the total energy of the oscillator and the speed of the block when it’s $4.0\ \mathrm{cm}$ from equilibrium.

$v=1.1\ \mathrm{\displaystyle \frac{m}{s}}$

The total energy of the oscillator is the sum of its kinetic and potential energies. By Conservation of Mechanical Energy, the sum $K+U_S$ is a constant, so if we can determine what this sum is at some point in the oscillation region, we’ll know the sum at every point. When the block is at its amplitude position, $x=8\ \mathrm{cm}$, its speed is zero; so at this position, $E$ is easy to figure out:

$$E=K+U_S$$

$$E=0+\frac{1}{2}kA^2$$

$$E=0+\frac{1}{2}\left(500\ \mathrm{\frac{N}{m}}\right)(0.08\ \mathrm{m})^2$$

$$E=1.6\ \mathrm{J}$$

This gives the total energy of the oscillator at every position. At any position $x$, we have

$$\frac{1}{2}mv^2+\frac{1}{2}kx^2=E$$

$$v=\sqrt{\frac{E-\displaystyle \frac{1}{2}kx^2}{\displaystyle \frac{1}{2}m}}$$

so when we substitute in the numbers, we get

$$v=\sqrt{\frac{(1.6\ \mathrm{J})-\displaystyle \frac{1}{2}\left(500\ \mathrm{\displaystyle \frac{N}{m}}\right)(0.04\ \mathrm{m})^2}{\displaystyle \frac{1}{2}(2.0\ \mathrm{kg})}}$$

$$v=1.1\ \mathrm{\frac{m}{s}}$$