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JEE Main 2015 (11 Apr, Shift 1) · Damped, Forced Oscillations & Superposition

Damped, Forced Oscillations & Superposition question from JEE Main 2015 (11 Apr, Shift 1)

Q17JEE Main 2015 · 11 APR · SHIFT 1Difficult

Damped, Forced Oscillations & Superposition

A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in s−1{}^{-1}) is:

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Answer: (a)

  1. A=A0e−btA = A_0 e^{-bt}, so E∝A2=E0e−2btE \propto A^2 = E_0 e^{-2bt}.
  2. 15 oscillations of 1 s: t=15 st = 15\ \text{s}; 1545=e−2b(15)\frac{15}{45} = e^{-2b(15)}.
  3. 30b=ln⁡3⇒b=130ln⁡3 s−130b = \ln 3 \Rightarrow b = \frac{1}{30}\ln 3\ \text{s}^{-1}.
  4. So the answer is (a) 130ln⁡3\frac{1}{30}\ln 3.
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