Question

A block of mass $2\text{kg}$ connected to a spring of spring constant $8{\text{Nm}}^{–1}$ is allowed to oscillate on a rough horizontal surface. If the system experiences a damping force $=0.230\times$ velocity, then the time required for the amplitude of resulting oscillation to fall to half of its initial value is

• A. $0.693\text{s}$
• B. $12\text{s}$
• C. $0.8\text{s}$
• D. $14.3\text{s}$

Answer 1

The correct option is B $12\text{s}$

When damping force $F=bv$ where $v$ is velocity
then the amplitude at time $t$ is given by
$A=A_0{e}^{-\frac{bt}{2m}}$
Here $b=0.23$
Thus, $t=\frac{2m}{b}\ln \left(\frac{A_0}{A}\right)$
$=\frac{2\times 2}{0.230}\ln \left(2\right)=12.05\text{s}$