A bob of a simple pendulum has mass m and is | vedclass

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(A) The tension in the string is maximum when the bob passes through the mean position.At the mean position,the forces acting on the bob are the tensi

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$mg\left[1+\left(\frac{a}{L}\right)^{2}\right]$

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(A) The tension in the string is maximum when the bob passes through the mean position.At the mean position,the forces acting on the bob are the tension $T$ upwards and weight $mg$ downwards. The net centripetal force is provided by the difference between tension and weight:$T_{\max} - mg = \frac{mV^{2}}{L} \implies T_{\max} = mg + \frac{mV^{2}}{L} \dots (1)$In Simple Harmonic Motion $(SHM)$,the velocity at the mean position is given by $V = a\omega$.For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{L}}$.Thus,$V = a\sqrt{\frac{g}{L}}$,which implies $V^{2} = a^{2}\frac{g}{L}$.Substituting this value of $V^{2}$ into $Eq. (1)$:$T_{\max} = mg + \frac{m}{L} \left(a^{2}\frac{g}{L}\right) = mg + \frac{mga^{2}}{L^{2}} = mg \left[1 + \left(\frac{a}{L}\right)^{2}\right]$.

Column-$I$	Column-$II$
$(a)$ $T^2 \to l$	$(i)$ Linear
$(b)$ $T^2 \to g$	$(ii)$ Parabolic
$(c)$ $T \to l$	$(iii)$ Hyperbolic

$A$ bob of a simple pendulum has mass $m$ and is oscillating with an amplitude $a$. If the length of the pendulum is $L$,then the maximum tension in the string is (given $\cos 0^{\circ}=1, g=$ acceleration due to gravity):

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The time period of a simple pendulum is $T$. When the length is increased by $10 \ cm$, its period is $T_1$. When the length is decreased by $10 \ cm$, its period is $T_2$. Then, the relation between $T, T_1$, and $T_2$ is

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Column-$I$ Column-$II$
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$(c)$ $T \to l$ $(iii)$ Hyperbolic

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