Question

A $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$block hangs from a spring, extending it 16.0 cmfrom its unstretched position.

1. What is the spring constant?
2. The block is removed, and a$\mathbf{0}\mathbf{.}\mathbf{500}\mathit{k}\mathit{g}$body is hung from the same spring. If the spring is then stretched and released, what is its period of oscillation?

Short answer

1. The value of the spring constant is $245\mathrm{N}/\mathrm{m}$
2. The period of oscillation of spring is$0.284\mathrm{s}$.

Step-by-step solution

The given data

• 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, $m=500\mathrm{kg}$.

Understanding the concept of Hooke’s law and the period of oscillations

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, $T=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ (ii)

a) Calculation for the spring constant

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

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

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

b) Calculation of period of oscillations

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$0.284\mathrm{s}$