Q132 / 160 · TS EAMCET 2019 (04 May Shift 1) · Physics
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
- A$\left(\frac{1}{800 \pi}\right)$
- B$\left(\frac{\pi^2}{600}\right)$
- Correct answerC$\left(\frac{1}{800 \pi^2}\right)$
- D$\left(\frac{\pi}{400}\right)$
Solution
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 $\omega^{\prime}=\omega_0 \sqrt{1-\frac{b^2}{4 m^2 \omega_0^2}}$
Hence, the correct answer is option: 'C'.