Oscillations

JEE Main level, single correct, medium

138 of 194

Question 138 of 194

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbondioxide will be close to (ln 5 = 1.601, ln 2 = 0.693).

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Answer: (D) 161 s

A=A0e−bt2mA={A}_{0}{e}^{\dfrac{-bt}{2m}}

bisb isrelated to viscosity

8=10e−ba402m8=10{e}^{\dfrac{-{b}_{a}40}{2m}}

5=10e−bCo2t2m5=10{e}^{\dfrac{-{b}_{C{o}_{2}}t}{2m}}

ln⁡54ln⁡2=40t×1.3\dfrac{\ln \dfrac{5}{4}}{\ln 2}=\dfrac{40}{t}\times 1.3

t=40×(ln⁡2)×1.3ln⁡5−2ln⁡2t=\dfrac{40\times (\ln 2)\times 1.3}{\ln 5-2\ln 2}

=40×0.693×1.31.601−2×0.693=\dfrac{40\times 0.693\times 1.3}{1.601-2\times 0.693}

≃161sec⁡≃161\sec

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