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(D) The displacement of a damped harmonic oscillator is given by $x(t) = A(t) \cos(\omega t + \varphi)$,where $A(t) = A_0 e^{-bt/2m}$.Comparing this w

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$7$

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(D) The displacement of a damped harmonic oscillator is given by $x(t) = A(t) \cos(\omega t + \varphi)$,where $A(t) = A_0 e^{-bt/2m}$.Comparing this with the given equation $x(t) = e^{-0.1t} \cos(10\pi t + \varphi)$,we identify the amplitude as $A(t) = A_0 e^{-0.1t}$,where $A_0 = 1$.We need to find the time $t$ when the amplitude $A(t)$ becomes half of its initial value $A_0$.Set $A(t) = \frac{A_0}{2}$,which gives $A_0 e^{-0.1t} = \frac{A_0}{2}$.Dividing both sides by $A_0$,we get $e^{-0.1t} = \frac{1}{2}$,or $e^{0.1t} = 2$.Taking the natural logarithm on both sides: $0.1t = \ln(2)$.Using the value $\ln(2) \approx 0.693$,we get $0.1t = 0.693$.Therefore,$t = \frac{0.693}{0.1} = 6.93 \, s$.Rounding to the nearest integer,we get $t \approx 7 \, s$.

The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1t} \cos(10\pi t + \varphi)$. The time taken for its amplitude of vibration to drop to half of its initial value is close to .... $s$

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