Question

A block of mass 100 g is connected to an elastic spring of spring constant 450 Nm$^{-1}$ oscillates vertically. The block-spring system is in viscous surrounding medium with a damping constant 69.3 g s$^{-1}$. The time in which the amplitude of oscillations drop to half of its initial value. [take $\ln 2 = 0.693$]

• A. 6.93 s
• B. 2 s
• C. 20 s
• D. 69.3 s

Hint:

Amplitude decay in damped oscillations: $A(t) = A_0 e^{-bt/(2m)}$.
The time for amplitude to reduce by a factor $k$ (i.e., $A(t) = A_0/k$) is $t = \frac{2m \ln k}{b}$.
Ensure all units are consistent (e.g., convert grams to kilograms).

Answer 1

The Correct Option is B

For damped oscillations, the amplitude as a function of time is given by: $A(t) = A_0 e^{-bt/(2m)}$ where $A_0$ is the initial amplitude, $b$ is the damping constant, $m$ is the mass, and $t$ is time. Given: Mass $m = 100 \text{ g} = 0.1 \text{ kg}$. Damping constant $b = 69.3 \text{ g s}^{-1} = 69.3 \times 10^{-3} \text{ kg s}^{-1} = 0.0693 \text{ kg s}^{-1}$. We want to find the time $t$ when the amplitude $A(t)$ drops to half of its initial value, i.e., $A(t) = A_0/2$. So, $\frac{A_0}{2} = A_0 e^{-bt/(2m)}$. Dividing by $A_0$: $\frac{1}{2} = e^{-bt/(2m)}$. Taking the reciprocal of both sides: $2 = e^{bt/(2m)}$. Taking the natural logarithm of both sides: $\ln 2 = \frac{bt}{2m}$. Solving for $t$: $t = \frac{2m \ln 2}{b}$. Substitute the given values: $\ln 2 = 0.693$. $m = 0.1 \text{ kg}$. $b = 0.0693 \text{ kg s}^{-1}$. $t = \frac{2 \times (0.1 \text{ kg}) \times 0.693}{0.0693 \text{ kg s}^{-1}}$. $t = \frac{0.2 \times 0.693}{0.0693} \text{ s}$. Since $0.693 / 0.0693 = 10$: $t = 0.2 \times 10 \text{ s} = 2 \text{ s}$. The spring constant ($450 \text{ Nm}^{-1}$) is not needed for this calculation. \[ \boxed{2 \text{ s}} \]