A block of mass 200 , g is executing SHM unde | vedclass

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(A) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-\left(\frac{b}{2m}\right)t}$.Given: $m = 200 \, g = 0.2 \, k

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$7$

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(A) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-\left(\frac{b}{2m}ight)t}$.Given: $m = 200 \, g = 0.2 \, kg$,$b = 40 \, g \, s^{-1} = 0.04 \, kg \, s^{-1}$.We want to find $t$ when $A(t) = \frac{A_0}{2}$.Substituting the values: $\frac{A_0}{2} = A_0 e^{-\left(\frac{0.04}{2 	imes 0.2}ight)t}$.$\frac{1}{2} = e^{-\left(\frac{0.04}{0.4}ight)t} = e^{-0.1t}$.Taking the natural logarithm on both sides: $\ln(0.5) = -0.1t$.$-0.693 = -0.1t$.$t = \frac{0.693}{0.1} = 6.93 \, s \approx 7 \, s$.

$A$ block of mass $200 \, g$ is executing $SHM$ under the influence of a spring with spring constant $K = 90 \, N \, m^{-1}$ and a damping constant $b = 40 \, g \, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $\ln \frac{1}{2} = -0.693$).

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