Question

The displacement of a damped harmonic oscillator is given by $x\left(t\right)=e^{-0.1t}\cos (10\pi t+\varphi )$, where $t$ is in $\text{seconds}$. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
$[\ln 2=0.7]$

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

Answer 1

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

Let us assume, at $t=0$, the oscillator is at extreme position.
Then, amplitude of the oscillator is, $A=e^{-0.1t}$
So, at $t=0$,
$A_1=e^{-0.1\times 0}=1$

Let at $t=t_0,A_2=\frac{1}{2}$
$\therefore \frac{1}{2}=e^{-0.1t_0}$
$\Rightarrow e^{0.1t_0}=2$
$\Rightarrow 0.1t_0\ln e=\ln 2$
$\Rightarrow 0.1t_0=0.7$
$\Rightarrow t_0=7\text{s}$

Why this question? Concept involved - In damped oscillation, amplitude of the oscillation varies (decays) exponentially.