A simple pendulum of length ' \(l\) ' and a bob of mass ' \(\mathrm{m}\) ' is executing S.H.M. of small amplitude ' \(\mathrm{A}\) '. The maximum tension in the string will be ( \(\mathrm{g}=\) acceleration due to gravity)

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
\(
\begin{aligned}
\mathrm{T}_{\max }^{\prime} & =\mathrm{mg}+\frac{\mathrm{m}(\mathrm{A} \omega)^2}{l} \\
& =\mathrm{mg}+\frac{\frac{\mathrm{mA}}{l}}{l} \\
& =\mathrm{mg}+\frac{\mathrm{mA}^2 \mathrm{~g}}{l^2} \\
& =\mathrm{mg}\left(1+\left(\frac{\mathrm{A}}{l}\right)^2\right)
\end{aligned}
\)