Question

A spring suspended vertically is $\mathbf{18}\mathbf{}\mathit{c}\mathit{m}$long. When you suspend a $\mathbf{30}\mathbf{}\mathit{g}$ weight from the spring, at rest, the spring is $\mathbf{22}\mathbf{}\mathit{c}\mathit{m}$ long. Next you pull down on the weight so the spring is $\mathbf{23}\mathbf{}\mathit{c}\mathit{m}$ long and you release the weight from rest. What is the period of oscillation?

Short answer

The period of oscillation is $T=0.4s$.

Step-by-step solution

Identification of given data

Given data can be listed below,

• Mass, $m=30g$

• Stretching of spring length $L_f=22cm$

Evaluating spring stiffness

The weight is suspended vertically so that the change of length of the spring is

$$\begin{gathered} ∆L=L_f-L_i \\ =22-18 \\ =4cm \end{gathered}$$

Now the spring stretched due to weight vertically suspended its counterbalance or equilibrium the by spring force

$$\begin{gathered} mg=k∆L \\ \left(30g\right)\left(9.8m/s^2\right)=k\left(4cm\right) \\ k=7.35N/m \end{gathered}$$

The period of oscillation in spring

The period of oscillation of spring is given as

$$\begin{gathered} T=2\sqrt{\frac{m}{k}} \\ =2\pi \sqrt{\frac{\left(30g\right)\left(0.001kg\right)}{\left(7.35N/m\right)\left(1g\right)}} \\ =0.4s \end{gathered}$$

Thus, the period of oscillation of spring is $0.4s$.