Physics - SHM Question with Solution | TestHub

(B) tension in the string is maximum when the bob passes through the mean position. ${\mathrm{T}}_{\max }=\mathrm{mg}+\frac{{\mathrm{mV}}^2}{\mathrm{L}}$...(1) In S.H.M. velocity at the mean position is given by $\mathrm{V}=\mathrm{a}\omega$ For simple pendulum $\mathrm{T}=2\pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ $\therefore \omega =\frac{2\pi }{T}=\sqrt{\frac{g}{L}}$ $\therefore V=a\sqrt{\frac{g}{L}}$ or $V^2=a^2\frac{g}{L}$ Putting this value of ${\mathrm{V}}^2$ in $\mathrm{Eq}.(1)$ we get ${\mathrm{T}}_{\max }=\mathrm{mg}\left[1+{\left(\frac{\mathrm{a}}{\mathrm{L}}\right)}^2\right]$