A simple pendulum of length ' $l$ ' and a bob of mass ' $\mathrm{m}$ ' is executing S.H.M. of small…
- $2 \mathrm{mg}$
- $\operatorname{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)^2\right]$
- $\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]^2$
- $\operatorname{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]$
Solution
Tension in the string is given as,
$\begin{aligned}
& \mathrm{T}^{\prime}=\mathrm{mg} \cos \theta+\frac{\mathrm{mv}^2}{l} \\
& \mathrm{~T}_{\max }^{\prime}=\mathrm{mg}+\frac{\mathrm{mv}^2}{l}
\end{aligned}$
Now, time period $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$
$\omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{l}}$
$\omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{l}}$
Maximum Velocity is given as $\mathrm{v}_{\max }=\mathrm{A} \omega$ Substituting the values
$\begin{aligned}
\mathrm{T}_{\max }^{\prime} & =\mathrm{mg}+\frac{\mathrm{m}(\mathrm{A} \omega)^2}{l} \\
& =\mathrm{mg}+\frac{\frac{\mathrm{mA}}{l}}{l} \\
& =\mathrm{mg}+\frac{\mathrm{mA}^2 \mathrm{~g}}{l^2} \\
& =\mathrm{mg}\left(1+\left(\frac{\mathrm{A}}{l}\right)^2\right)
\end{aligned}$
.Asked in: MHT CET 2023 (14 May Shift 1)