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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$.Since $\omega = \frac{2 \pi}{T}$,we have $\omega = \sqrt{\frac{

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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$.Since $\omega = \frac{2 \pi}{T}$,we have $\omega = \sqrt{\frac{g}{\ell}}$,which implies $\omega \propto \frac{1}{\sqrt{\ell}}$.Given the new length $\ell_2 = \frac{\ell_1}{3}$,the new angular frequency $\omega_2$ is related to the initial frequency $\omega_1$ by $\frac{\omega_2}{\omega_1} = \sqrt{\frac{\ell_1}{\ell_2}} = \sqrt{\frac{\ell_1}{\ell_1/3}} = \sqrt{3}$.The total energy of a simple harmonic oscillator is given by $E = \frac{1}{2} m \omega^2 A^2$.Since the mass $(m)$ and amplitude $(A)$ remain constant,$E \propto \omega^2$.Therefore,$\frac{E_2}{E_1} = \left( \frac{\omega_2}{\omega_1} \right)^2 = (\sqrt{3})^2 = 3$.Thus,the new total energy is $E_2 = 3 E_1$.

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

If the length of an oscillating simple pendulum is made $\frac{1}{3}$ times at a place while keeping the amplitude the same,then its total energy $(E)$ will be: (in $E$)

Similar Questions

In the following table, the relation of the graph in column-$I$ and the shape of the graph in column-$II$ is shown. Match them appropriately.
Column-$I$ Column-$II$
$(a)$ $T^2 \to l$ $(i)$ Linear
$(b)$ $T^2 \to g$ $(ii)$ Parabolic
$(c)$ $T \to l$ $(iii)$ Hyperbolic

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