Question

The amplitude of a damped oscillator becomes one-fourth in one minute. The amplitude after 16 minutes will be ${\frac{1}{x}}^{th}$ of its original amplitude. Find the value of $x.$

• A. $2^4$
• B. $2$
• C. $2^{16}$
• D. $2^{32}$

Answer 1

The correct option is D $2^{32}$

From the equation of damped oscillations, we know that,
Amplitude of damped oscillations $\left(A\right)=A_0{e}^{-\gamma t}$
As $A=\frac{A_0}{4}$ at $t=1\text{min}$, we get
$\frac{A_0}{4}=A_0{e}^{-\gamma }$ or $e^{\gamma }=4.......\left(1\right)$
After $16\text{min}$, the amplitude will be $\frac{A_0}{x}$
$\Rightarrow \frac{A_0}{x}=A_0{e}^{-16\gamma }$
$\Rightarrow x=e^{16\gamma }=(e^{\gamma }{)}^{16}$
Using $\left(1\right)$ in the above equation, we get
$x=(4{)}^{16}\Rightarrow x=2^{32}$
Thus, option (d) is the correct answer.