A simple pendulum of length ' $\ell$ ' has a bob of mass 'm'. It executes S.H.M. of small amplitude 'A'. The…

1. 2mg
2. mg
3. $\operatorname{mg}\left(\frac{\mathrm{A}}{\ell}+1\right)$
4. $\operatorname{mg}\left(\frac{\mathrm{A}^{2}}{\ell^{2}}+1\right)$

Solution

$\begin{array}{l} \mathrm{y}=\mathrm{A} \sin \omega \mathrm{t} \\ \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\ \frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\ell}}=\omega \end{array}$ $\begin{aligned} \therefore \text { Tension } &=m g \cos \theta+\frac{m v^{2}}{L} \\ T_{\max } &=m g+\frac{m v^{2}}{L} \quad \cos \theta=1 \\ &=m g\left(1+\frac{v^{2}}{g L}\right) \end{aligned}$ Now, $y=A \sin \omega t$ $\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t} \\ &\left.\frac{\mathrm{dy}}{\mathrm{dt}}\right|_{\max }=\mathrm{A} \omega=\mathrm{A} \sqrt{\frac{\mathrm{g}}{\ell}}=\mathrm{V}_{\max } \\ \therefore & \mathrm{V}_{\max }^{2}=\mathrm{A}^{2} \frac{\mathrm{g}}{\mathrm{L}} \\ \therefore & \mathrm{T}_{\max }=\mathrm{mg}\left(1+\frac{\mathrm{A}^{2} \mathrm{~g}}{\mathrm{~L}^{2} \mathrm{~g}}\right)=\mathrm{mg}\left(1+\frac{\mathrm{A}^{2}}{\mathrm{~L}^{2}}\right) \end{aligned}$ ~

Asked in: MHT CET 2020 (16 Oct Shift 2)

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