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If a simple harmonic motion is represented by $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}+\alpha \mathrm{x}=0$, its time period is
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Verified Answer
The correct answer is:
$\frac{2 \pi}{\sqrt{\alpha}}$
$\frac{2 \pi}{\sqrt{\alpha}}$
$\omega^2=\alpha$
$\omega=\sqrt{\alpha}$
$T=\frac{2 \pi}{\sqrt{\alpha}}$
$\omega=\sqrt{\alpha}$
$T=\frac{2 \pi}{\sqrt{\alpha}}$
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