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(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.Given that at $t = 2 \ s$,$A(2) = \frac{1}{3} A_0$.So,$\frac{

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$27$

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(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.Given that at $t = 2 \ s$,$A(2) = \frac{1}{3} A_0$.So,$\frac{1}{3} A_0 = A_0 e^{-b(2)/2m} \implies e^{-b/m} = \frac{1}{3}$.We need to find the amplitude at $t = 6 \ s$,which is $A(6) = A_0 e^{-b(6)/2m} = A_0 (e^{-b/m})^3$.Substituting the value of $e^{-b/m}$,we get $A(6) = A_0 \left(\frac{1}{3}\right)^3 = A_0 \left(\frac{1}{27}\right)$.Comparing this with $A(6) = \frac{1}{n} A_0$,we find $n = 27$.

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

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