Rankers Physics

miscellaneous: Practice Problem & Solution

The amplitude of a S.H.M. reduces to $1/3$ in first $20 \text{ secs}$, then in first $40 \text{ sec.}$ its amplitude becomes: (1999)
$\frac{1}{3}$
$\frac{1}{9}$
$\frac{1}{27}$
$\frac{1}{\sqrt{3}}$

Solution Explained:

To solve this problem, we apply the core principles of miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:

In damped S.H.M., amplitude at time $t$ is $A(t) = A_0 e^{-bt}$. At $t = 20 \text{ s}$, $A(20) = A_0 e^{-20b} = \frac{A_0}{3}$. At $t = 40 \text{ s}$, $A(40) = A_0 e^{-40b} = A_0 (e^{-20b})^2 = A_0 (\frac{1}{3})^2 = \frac{A_0}{9}$.

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