A spring has natural length 40 cm and spring constant 500 N/m. A block of mass 1 kg is attached at one end of the spring and other end of the spring is attached to a ceiling. The block is released from the position, where the spring has length 45 cm

(1) $k{x}_0=mg$

$\therefore x_0=\frac{mg}{k}=\frac{1\times 10}{500}$

=0.02 m=2 cm

So, equilibrium is obtained after an extension of 2 cm of at a length of 42 cm. But it is released from a length of 45 cm.

$\therefore A=3 \mathrm{cm}=0.03 \mathrm{m}$

(2) $A=3 \mathrm{cm}=6A=\sqrt{\frac{k}{m}}A$

$$\begin{gathered} =\left(\sqrt{\frac{500}{1}}\right)(0.03)=0.3\sqrt{5} \mathrm{m}/\mathrm{s} \\ =30\sqrt{5} \mathrm{cm}/\mathrm{s} \end{gathered}$$

(3) $a_{\max }={\omega }^{2}A$

$=\left(\frac{k}{m}\right)\left(A\right)=\left(\frac{500}{1}\right)(0.03)=15 \mathrm{m}/{\mathrm{s}}^2$

(4) Mean position is at 42 cm length and amplitude is 3 cm. Hence, block oscillates between 45 cm length and 39 cm. Natural length 40 cm lies in between these two, where elastic potential energy =0.