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(D) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.Given that at $t = 10 \ s$, $A(t) = A_0/2$.So, $A_0/

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$5$

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(D) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.Given that at $t = 10 \ s$, $A(t) = A_0/2$.So, $A_0/2 = A_0 e^{-b(10)/2m} \implies 1/2 = e^{-5b/m} \implies \ln(2) = 5b/m$.The mechanical energy of the oscillator is $E(t) = \frac{1}{2} k A(t)^2 = \frac{1}{2} k A_0^2 e^{-bt/m} = E_0 e^{-bt/m}$.We want to find the time $t'$ such that $E(t') = E_0/2$.So, $E_0/2 = E_0 e^{-bt'/m} \implies 1/2 = e^{-bt'/m} \implies \ln(2) = bt'/m$.Equating the two expressions for $\ln(2)$: $5b/m = bt'/m \implies t' = 5 \ s$.

If the amplitude of a damped harmonic oscillator becomes half of its initial amplitude in a time of $10 \ s$, then the time taken for the mechanical energy of the oscillator to become half of its initial mechanical energy is (in $s$)

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