The mechanical energy of a damped oscillator | vedclass

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(B) The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-bt/m}$.At $t = 4 \ s$,$E = E_0/2$.So,$E_0/2 = E_0 e^{-4b/m} \Rightarrow e

Vedclass · Accepted Answer

$8$

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(B) The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-bt/m}$.At $t = 4 \ s$,$E = E_0/2$.So,$E_0/2 = E_0 e^{-4b/m} \Rightarrow e^{-4b/m} = 1/2 \Rightarrow e^{4b/m} = 2$.Taking the natural logarithm on both sides,$4b/m = \ln 2$,so $b/m = (\ln 2)/4$.At time $T = 4 + t$,the energy is $12.5 \% E_0 = (1/8) E_0$.So,$E_0/8 = E_0 e^{-b(4+t)/m} \Rightarrow e^{-b(4+t)/m} = 1/8 \Rightarrow e^{b(4+t)/m} = 8 = 2^3$.Taking the natural logarithm,$b(4+t)/m = 3 \ln 2$.Substituting $b/m = (\ln 2)/4$,we get $[(\ln 2)/4] 	imes (4+t) = 3 \ln 2$.Dividing by $\ln 2$,we get $(4+t)/4 = 3 \Rightarrow 4+t = 12 \Rightarrow t = 8 \ s$.

The mechanical energy of a damped oscillator becomes half of its initial energy in $4 \ s$. In another $t \ s$,its mechanical energy becomes $12.5 \%$ of its initial mechanical energy. Then $t=$ (in $s$)

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