91Ó°ÊÓ

The pendulum is pulled to a $10°$angle on the left side and released. So, when we calculating the time period we have to sum up the two time periods of the two ends of pendulum of length $0.12m$.

The time period of the pendulum written as $T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

$g=$$9.81m/s^2$

Where,

L is the length of the pendulum and

g is the acceleration due to gravity.

Given $L=0.12m$

The total time period is equal to the sum of periods of two ends of the pendulum

Therefore,

$$\begin{gathered} T=\mathrm{π}\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}+\mathrm{π}\sqrt{\frac{2\mathrm{L}}{\mathrm{g}}} \\ =\mathrm{π}\sqrt{\frac{0.12}{9.81}}+\mathrm{π}\sqrt{\frac{0.12×2}{9.81}} \\ =0.838\mathrm{s} \\ =0.84\mathrm{s} \end{gathered}$$

The pendulum pulled to a $10°$angle on the left side. We have to find the pendulum maximum angle at the right side.$\mathrm{θ}=10°$

For that we have to use the formula of conservation of energy of simple pendulum. Where, as a pendulum swings, its potential energy changes to kinetic energy, then back to potential energy, then back to kinetic energy.

Therefore, the equation of conservation of energy

$mg\frac{l}{2}(1-\cos (\mathrm{θ}\text{'}\left)\right)=mgl(1-\cos (\mathrm{θ}\left(max\right))$

where $l$is the length of the pendulum

$$\begin{gathered} \mathrm{θ}\text{'}=minimumangle \\ \mathrm{θ}\left(max\right)=maximumangle \end{gathered}$$

$$\begin{gathered} mg\frac{l}{2}(1-\cos (10\left)\right)=mgl(1-\cos \mathrm{θ}(max\left)\right) \\ \frac{l}{2}(0.016)=1-\cos \mathrm{θ}\left(max\right) \\ 0.008=1-\cos \mathrm{θ} \\ \cos \mathrm{θ}=0.992 \\ \mathrm{θ}={\cos }^{-1}(0.992) \\ =7.25° \end{gathered}$$