91Ó°ÊÓ

• Mass of the block is,$M=4.00\mathrm{kg}$.
• Displacement of the spring is,role="math" localid="1657262737853" $x=16.0\mathrm{cm}\mathrm{or}0.16\mathrm{m}$.
• Mass of the body is, $m=500\mathrm{kg}$.

Using Hooke’s law, we can find the value of the spring constant. Then using the formula for the period of oscillation for S.H.M we can find the period of oscillation of spring.

Formulae:

The force of a body usingHooke’s law,$F=kx$ (i)

The period of oscillation, $T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ (ii)

Using equation (i) to the given system, we get the spring constant of an oscillation as:

$$\begin{gathered} k=\frac{mg}{x}(âˆ\mu F=kx=Mg) \\ =\frac{4\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{0.16\mathrm{m}} \\ =245\mathrm{N}/\mathrm{m} \end{gathered}$$

Therefore, the value of the spring constant is$245\mathrm{N}/\mathrm{m}$

Using equation (ii), the period of oscillations of the system is given as:

$$\begin{gathered} T=2(3.142)\sqrt{\frac{0.5\mathrm{kg}}{245\mathrm{N}·\mathrm{m}}} \\ =0.284\mathrm{s} \end{gathered}$$

Therefore, the period of oscillation of spring is$0.284\mathrm{s}$