Question

The displacement of a damped harmonic oscillator is given by $x\left(t\right)=e^{-0.1t}\cos (10\pi t+\phi ).$ Here $t$ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to

• A. $13\text{s}$
• B. $7\text{s}$
• C. $4\text{s}$
• D. $27\text{s}$

Answer 1

The correct option is B $7\text{s}$

Given:
$x\left(t\right)=e^{-0.1t}\cos (10\pi t+\phi ).$
$A_2=A/2$

Amplitude of vibration at time $t=0$ is given by

$A=A_0{e}^{-0.1\times 0}=1\times A_0=A_0$

also at time $\left(t\right)=t$,

Amplitude $A=\frac{A_0}{2}$

$\Rightarrow \frac{A_0}{2}=A_0{e}^{-0.1t}$

$\therefore t=10\ln 2\simeq 7\text{s}$

Hence option $\left(B\right)$ is correct.