The displacement of a damped harmonic oscilla | vedclass

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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 $x(t) = e^{-0.1 t} \cos(10 \pi t + \varphi)$,the amplitude is $A(t) = A_0 e^{-0.1 t}$,where $A_0 = 1$.We need to find the time $t$ when the amplitude becomes half of its initial value,i.e.,$A(t) = \frac{A_0}{2}$.Substituting this into the amplitude equation: $\frac{A_0}{2} = A_0 e^{-0.1 t}$.Dividing both sides by $A_0$,we get: $\frac{1}{2} = e^{-0.1 t}$.Taking the natural logarithm on both sides: $\ln(0.5) = -0.1 t$.Since $\ln(0.5) = -\ln(2) \approx -0.693$,we have: $-0.693 = -0.1 t$.Solving for $t$: $t = \frac{0.693}{0.1} = 6.93 \ s$.Rounding to the nearest integer,we get $t \approx 7 \ s$.

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

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