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A body of mass $0.3 \mathrm{~kg}$ hangs by a spring with a force constant of $50 \mathrm{~N} / \mathrm{m}$. The amplitude of oscillations is damped and reaches $\frac{1}{e}$ of its original value in about 100 oscillations. If $\omega$ and $\omega^{\prime}$ are the angular frequencies of undamped and damped oscillations respectively, then percentage of $\left(\frac{\omega-\omega}{\omega}\right)$ is
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The correct answer is:
$\left(\frac{1}{800 \pi^2}\right)$
Hint The displacement of a spring mass oscillator, $X=A e^{-b t / 2 m} \cos \left(\omega^{\prime} t+\phi\right)$ where, $\omega^{\prime}=$ angular frequency of damped oscillation and $\omega^{\prime}=\omega_0 \sqrt{1-\frac{b^2}{4 m^2 \omega_0^2}}$
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