Physics - SHM Question with Solution | TestHub

Tension in the string is given as, $$\begin{array}{rl}& {\mathrm{T}}^{\prime }=\mathrm{mg}\cos \theta +\frac{{\mathrm{mv}}^2}{l}\\ & {\mathrm{T}}_{\max }^{\prime }=\mathrm{mg}+\frac{{\mathrm{mv}}^2}{l}\end{array}$$ 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 $$\begin{array}{rl}{\mathrm{T}}_{\max }^{\prime }& =\mathrm{mg}+\frac{\mathrm{m}(\mathrm{A}\omega {)}^2}{l}\\ & =\mathrm{mg}+\frac{\frac{\mathrm{mA}}{l}}{l}\\ & =\mathrm{mg}+\frac{{\mathrm{mA}}^2\mathrm{g}}{l^2}\\ & =\mathrm{mg}\left(1+{\left(\frac{\mathrm{A}}{l}\right)}^2\right)\end{array}$$