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A block of mass ' $\mathrm{m}$ ' attached to one end of the vertical spring produces extension $\mathrm{x}^{\prime}$. If the block is pulled and released, the periodic time of oscillation is
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The correct answer is:
$2 \pi\sqrt{\frac{\mathrm{x}}{\mathrm{g}}}$
(A)
$\mathrm{mg}=-\mathrm{kx} \quad \therefore \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$
$\omega^{2}=\frac{k}{m}$
$\left(\frac{2 \pi}{T}\right)^{2}=\frac{k}{m}$
$\frac{4 \pi^{2}}{T^{2}}=\frac{k}{m}$
$\therefore T^{2}=\frac{4 \pi^{2} m}{k}=\frac{4 \pi^{2} m x}{m g} \Rightarrow T=2 \pi \sqrt{\frac{x}{g}}$
$\mathrm{mg}=-\mathrm{kx} \quad \therefore \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}$
$\omega^{2}=\frac{k}{m}$
$\left(\frac{2 \pi}{T}\right)^{2}=\frac{k}{m}$
$\frac{4 \pi^{2}}{T^{2}}=\frac{k}{m}$
$\therefore T^{2}=\frac{4 \pi^{2} m}{k}=\frac{4 \pi^{2} m x}{m g} \Rightarrow T=2 \pi \sqrt{\frac{x}{g}}$
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