Q 11-13-152JEE MainJEE Main 2017 (9 Apr)Medium
A block of mass $0.1$ kg is connected to an elastic spring of spring constant $640\ \text{N m}^{-1}$ and oscillates in a damping medium of damping constant $10^{-2}\ \text{kg s}^{-1}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:
Answer: (D) $7$ s
For weak damping the amplitude decays as $e^{-bt/2m}$, so energy decays as $e^{-bt/m}$:
$$\frac12 = e^{-bt/m} \;\Rightarrow\; t = \frac{m\ln2}{b} = \frac{0.1\times0.693}{10^{-2}} \approx 6.9\ \text{s}$$
Closest is $7$ s.
Solution by Sreeraj P, M.Sc Physics