The amplitude of a damped harmonic oscillator | vedclass

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Question

(C) The amplitude of a damped harmonic oscillator is given by $A(t) = A_0 e^{-bt}$.Given that the amplitude becomes half in $3 \ s$:$A(3) = A_0 e^{-3b

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$2^3$

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(C) The amplitude of a damped harmonic oscillator is given by $A(t) = A_0 e^{-bt}$.Given that the amplitude becomes half in $3 \ s$:$A(3) = A_0 e^{-3b} = \frac{A_0}{2}$$e^{-3b} = \frac{1}{2} \quad \dots (i)$In the next $6 \ s$,the total time elapsed is $3 \ s + 6 \ s = 9 \ s$. The amplitude at $t = 9 \ s$ is:$A(9) = A_0 e^{-9b} = A_0 (e^{-3b})^3$Substituting equation $(i)$ into the expression:$A(9) = A_0 \left(\frac{1}{2}\right)^3 = A_0 \left(\frac{1}{8}\right)$Since the amplitude becomes $1/x$ of the initial amplitude,we have $1/x = 1/8$,which means $x = 8 = 2^3$.Therefore,the correct option is $C$.

The amplitude of a damped harmonic oscillator becomes half in $3 \ s$ and will become $1/x$ of the initial amplitude in the next $6 \ s$,where $x$ is:

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