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JEE Main 2019 (10 Apr, Shift 1) · Damped, Forced Oscillations & Superposition

Damped, Forced Oscillations & Superposition question from JEE Main 2019 (10 Apr, Shift 1)

Q10JEE Main 2019 · 10 APR · SHIFT 1Medium

Damped, Forced Oscillations & Superposition

The displacement of a damped harmonic oscillator is given by

x(t) = e–0.1t cos (10π\pit + f).

Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

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Answer: (c)

  1. Amplitude =e−0.1t= e^{-0.1t} (initial value 1).
  2. e−0.1t=12⇒0.1t=ln⁡2=0.693e^{-0.1t} = \frac{1}{2} \Rightarrow 0.1t = \ln 2 = 0.693.
  3. t=6.93 s≈7 st = 6.93\ \text{s} \approx 7\ \text{s}.
  4. So the answer is (c) 7 s.
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