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A pendulum performs S.H.M. with period $\sqrt{3}$ second in a stationery lift. If lift moves up with acceleration $\frac{\mathrm{g}}{3}$, the period of pendulum is $[\mathrm{g}=$ acceleration due to gravity $]$
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$1 \cdot 5$ second
$\mathrm{T}=\sqrt{3} \quad \mathrm{a}=\frac{\mathrm{g}}{3}$
$\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}+\frac{\mathrm{g}}{3}}}$
$\therefore \frac{T^{\prime}}{T}=\sqrt{\frac{g}{4 g / 3}}=\frac{\sqrt{3}}{2}$
$\mathrm{T}^{\prime}=\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}=1.5 \mathrm{~sec}$
$\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}+\frac{\mathrm{g}}{3}}}$
$\therefore \frac{T^{\prime}}{T}=\sqrt{\frac{g}{4 g / 3}}=\frac{\sqrt{3}}{2}$
$\mathrm{T}^{\prime}=\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}=1.5 \mathrm{~sec}$
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