Physics - SHM Question with Solution | TestHub

The amplitude of a damped oscillator at time $t$ is given by $A=A_0{e}^{-\lambda t}$.

Given that at $t=2$ s, the amplitude becomes $\frac{1}{3}$ of the original amplitude:

$\frac{A_0}{3}=A_0{e}^{-2\lambda }$

$e^{-2\lambda }=\frac{1}{3}$

At $t=6$ s, the amplitude is $A^{\prime }=A_0{e}^{-6\lambda }$.

This can be written as:

$A^{\prime }=A_0(e^{-2\lambda }{)}^3$

Substituting the value of $e^{-2\lambda }$:

$A^{\prime }=A_0{\left(\frac{1}{3}\right)}^3=\frac{A_0}{27}$

Given that the amplitude after 6 seconds is $\frac{A_0}{n}$, comparing the two expressions gives:

$n=27$

Answer: 27