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JEE Main 2014 (9 Apr, Shift 1) · Damped, Forced Oscillations & Superposition

Damped, Forced Oscillations & Superposition question from JEE Main 2014 (9 Apr, Shift 1)

Q20JEE Main 2014 · 9 APR · SHIFT 1Difficult

Damped, Forced Oscillations & Superposition

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 carbon dioxide will be close to (ln 5 = 1.609, ln 2 = 0.693).

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Answer: (c)

  1. A=A0e−bt/2mA = A_0 e^{-bt/2m} with b∝ηb \propto \eta (Stokes law).
  2. Air: ln⁡108=ln⁡5−2ln⁡2=0.223=bair2m(40)⇒bair2m=5.575×10−3 s−1\ln\frac{10}{8} = \ln 5 - 2\ln 2 = 0.223 = \frac{b_{air}}{2m}(40) \Rightarrow \frac{b_{air}}{2m} = 5.575 \times 10^{-3}\ \text{s}^{-1}.
  3. CO2_2: bCO22m=5.575×10−31.3=4.29×10−3 s−1\frac{b_{CO_2}}{2m} = \frac{5.575 \times 10^{-3}}{1.3} = 4.29 \times 10^{-3}\ \text{s}^{-1}.
  4. 10 cm to 5 cm: t=ln⁡24.29×10−3≈161 st = \frac{\ln 2}{4.29 \times 10^{-3}} \approx 161\ \text{s}.
  5. So the answer is (c) 161 s.
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