Calculate the maximum speed of the block.

$v_{\mathrm{max}}=4\ \mathrm{\displaystyle \frac{m}{s}}$

First, let’s get an expression for the maximum elastic potential energy of the system:

$$U_S=\frac{1}{2}kx^2$$

$$U_{S,\ \mathrm{max}}=\frac{1}{2}kx^2_{\mathrm{max}}$$

$$U_{S,\ \mathrm{max}}=\frac{1}{2}kA^2$$

When all this energy has been transformed into kinetic energy - which, as we discussed earlier, occurs just as the block is passing through equilibrium - the block will have a maximum kinetic energy and maximum speed of

$$U_{S,\ \mathrm{max}}\to K_{\mathrm{max}}$$

$$\frac{1}{2}kA^2=\frac{1}{2}mv^2_{\mathrm{max}}$$

$$v_{\mathrm{max}}=\sqrt{\frac{kA^2}{m}}$$

$$v_{\mathrm{max}}=\sqrt{\frac{\left(500\ \mathrm{\displaystyle \frac{N}{m}}\right)(0.04\ \mathrm{m})^2}{0.05\ \mathrm{kg}}}$$

$$v_{\mathrm{max}}=4\ \mathrm{\frac{m}{s}}$$