The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of original amplitude in 2 seconds. If its amplitude after 6 second become $\left(\frac{1}{n}\right)$ times the original amplitude, the value of n is ( n is non zero integer)

The amplitude of a damped oscillator follows $A(t) = A_0 e^{-bt}$, where $A_0$ is the initial amplitude and $b$ is the damping constant.

Given that $A(2) = \frac{A_0}{3}$, substitution yields $\frac{A_0}{3} = A_0 e^{-2b}$, which simplifies to $\frac{1}{3} = e^{-2b}$.

After 6 seconds, $A(6) = \frac{A_0}{n} = A_0 e^{-6b}$, giving $\frac{1}{n} = e^{-6b}$.

Since $e^{-6b} = (e^{-2b})^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$, it follows that $\frac{1}{n} = \frac{1}{27}$.

Thus, $n = 27$.