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If length of oscillating simple pendulum is made $\frac{1}{3}$ times at a place keeping
amplitude same, then its total energy (E) will be
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amplitude same, then its total energy (E) will be
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Verified Answer
The correct answer is:
$4 \mathrm{E}$
$T=2 \pi \sqrt{\frac{\ell}{g}}$
$\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\ell}{g}}$
$\omega=\sqrt{\frac{\mathrm{g}}{\ell}} \quad \therefore \omega \propto \frac{1}{\sqrt{\ell}}$
$\therefore \frac{\omega_{2}}{\omega_{1}}=\sqrt{\frac{\ell_{1}}{\ell_{2}}}=\sqrt{\frac{\ell_{1}}{\ell_{1} / 3}}=\sqrt{3}$
$\therefore \omega_{2}=\sqrt{3} \omega_{1}$
Now $E \propto \omega^{2}$
$\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\omega_{2}^{2}}{\omega_{1}^{2}}=3$
$\therefore E_{2}=3 E_{1}$
$\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\ell}{g}}$
$\omega=\sqrt{\frac{\mathrm{g}}{\ell}} \quad \therefore \omega \propto \frac{1}{\sqrt{\ell}}$
$\therefore \frac{\omega_{2}}{\omega_{1}}=\sqrt{\frac{\ell_{1}}{\ell_{2}}}=\sqrt{\frac{\ell_{1}}{\ell_{1} / 3}}=\sqrt{3}$
$\therefore \omega_{2}=\sqrt{3} \omega_{1}$
Now $E \propto \omega^{2}$
$\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\omega_{2}^{2}}{\omega_{1}^{2}}=3$
$\therefore E_{2}=3 E_{1}$
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