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A light balloon filled with helium of density $\rho_{\mathrm{He}}$ is tied to a long light string of length $\ell$ and the string is attached to the ground. If the balloon is displaced slightly in the horizontal direction from the equilibrium and released then.
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The correct answer is:
The ballon undergoes simple harmonic motion with period $2 \pi \sqrt{\left(\frac{\rho_{\mathrm{He}}}{\rho_{\text {air }}-\rho_{\mathrm{He}}}\right) \frac{\ell}{\mathrm{g}}}$
$$
\tau_{0}=\mathrm{V}\left(\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}\right) \mathrm{g} \ell \sin \theta
$$
For small angular displacement $(\theta)$
$$
\begin{array}{l}
\tau_{0}=\mathrm{V}\left(\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}\right) \mathrm{g} \ell \theta \\
\mathrm{I} \alpha=\mathrm{V}\left[\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}\right] \mathrm{g} \ell \theta \\
\rho_{\mathrm{He}} \mathrm{V} \ell^{2} \alpha=\mathrm{V}\left[\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}\right] \ell \theta \mathrm{g} \\
\alpha=\left[\frac{\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}}{\rho_{\mathrm{He}}}\right] \frac{\mathrm{g}}{\ell} \theta \\
\omega=\sqrt{\left(\frac{\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}}{\rho_{\mathrm{He}}}\right) \frac{\mathrm{g}}{\ell}} \\
=2 \pi \sqrt{\frac{\ell}{\mathrm{g}} \frac{\rho_{\mathrm{He}}}{\left(\rho_{\mathrm{Air}}-\rho_{\mathrm{He}}\right)}}
\end{array}
$$
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