Q 11-13-133JEE MainJEE Main 2019 (10 Apr, Shift 1)Easy
The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1t}\cos(10\pi t + \varphi)$. Here $t$ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
Answer: (D) $7\ \text{s}$
The amplitude is $e^{-0.1t}$. Setting it to $\tfrac12$:
$$0.1t = \ln2 \;\Rightarrow\; t = 10\ln2 \approx 6.9\ \text{s} \approx 7\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics