91Ó°ÊÓ

• The vertical axis scale is set by, $K_s=10.0\mathrm{mJ}\mathrm{or}0.010\mathrm{J}$.
• Mass of pendulum bob, m = 0.200 kg.
• Graph of $Kvs\mathrm{θ}$.

A particle in simple harmonic motion has, at any time, kinetic energy

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

We can interpret the maximum kinetic energyK from the graph, from which, we can find the maximum speed${\mathit{V}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$. Finally, using the relation betweenand, we can find the length of the pendulum.

Formula:

The maximum kinetic energy of the wave,$K=\frac{1}{2}m{v_{max}}^2$ (i)

The angular frequency of simple pendulum,$\mathrm{Ó\neg }=\sqrt{\frac{g}{L}}$ (ii)

The maximum speed of the pendulum,$V_{max}=L\mathrm{Ó\neg }{\mathrm{θ}}_{max}$ (iii)

From the graph, we can find that maximum kinetic energy is given as;

$$\begin{gathered} K=15\mathrm{mJ} \\ =0.015\mathrm{J} \end{gathered}$$

But from equation (i), we can get the maximum velocity of the wave as:

$$\begin{gathered} 0.015J=\frac{1}{2}×0.20kg×{v_{max}}^2 \\ {v_{max}}^2=\frac{0.015}{0.100} \\ =\sqrt{\frac{0.015}{0.100}} \\ =0.3873\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (ii) and ${\mathrm{θ}}_{max}=100\mathrm{mrad}$in equation (iii), we get the length of the pendulum as:

$$\begin{gathered} v_{max}=L×\sqrt{\frac{g}{L}}×{\mathrm{θ}}_{max} \\ {v_{max}}^2=L^2×\frac{g}{L}×{{\mathrm{θ}}_{max}}^2 \\ =\frac{{V_{max}}^2}{g×{{\mathrm{θ}}_{max}}^2} \\ =\frac{{(0.3873\mathrm{m}/\mathrm{s})}^2}{9.8\mathrm{m}/{\mathrm{s}}^2×{(0.100\mathrm{rad})}^2} \\ =1.53\mathrm{m} \end{gathered}$$

Therefore, the length of simple pendulum is 1.53 m