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A particle of mass 2 kg is executing simple harmonic motion with amplitude $A_{0} .$ At $t = 0,$ a small resistive force also starts acting on the particle which is given by $\vec{F} = \frac{- \vec{v} k \ln 2}{10} .$ The value of amplitude of the particle after 20 second will be: (where k = 1 kg/s)

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$A_{0} / 2$

$A_{0} / 4$

$A_{0}$

$A_{0} / \sqrt{2}$

answer is D.

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Detailed Solution

$A = A_{0} e^{\frac{b t}{2 m}} A = A_{0} e^{\frac{\ln 2}{10 \times 2} \times \frac{20}{2}} = A_{0} e^{- \ln \sqrt{2}} = \frac{A_{0}}{\sqrt{2}}$

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