The time taken for the amplitude of vibration | vedclass

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(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.Given that $A(t) = 0.25 A_0 = \frac{A_0}{4}$ at time $t$,we have:

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$\frac{t}{4}$

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(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.Given that $A(t) = 0.25 A_0 = \frac{A_0}{4}$ at time $t$,we have:$\frac{A_0}{4} = A_0 e^{-\frac{bt}{2m}} \implies \frac{1}{4} = e^{-\frac{bt}{2m}} \implies 4 = e^{\frac{bt}{2m}}$.Taking the natural logarithm on both sides: $\ln(4) = \frac{bt}{2m} \implies 2\ln(2) = \frac{bt}{2m} \implies t = \frac{4m \ln(2)}{b} \quad ... (1)$The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-\frac{bt}{m}}$.Given that $E(t') = 0.5 E_0 = \frac{E_0}{2}$ at time $t'$,we have:$\frac{E_0}{2} = E_0 e^{-\frac{bt'}{m}} \implies \frac{1}{2} = e^{-\frac{bt'}{m}} \implies 2 = e^{\frac{bt'}{m}}$.Taking the natural logarithm on both sides: $\ln(2) = \frac{bt'}{m} \implies t' = \frac{m \ln(2)}{b} \quad ... (2)$Dividing equation $(2)$ by $(1)$:$\frac{t'}{t} = \frac{m \ln(2) / b}{4m \ln(2) / b} = \frac{1}{4} \implies t' = \frac{t}{4}$.

The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial value is $t$. The time taken for its mechanical energy to drop to $50 \%$ of its initial mechanical energy is

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