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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Oscillations

JEE Main 2012 · Offline · Physics · Oscillations

If a simple pendulum has significant amplitude (up to a factor of 1...

If a simple pendulum has significant amplitude (up to a factor of 1/e1 / e of original) only in the period between t=0st=0s to t=τst=\tau s, then τ\tau may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'bb' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:If a simple pendulum has significant amplitude (up to a factor of 1 / e of original) only in the period between t=0s to

Options

  1. A0.693b\frac{0.693}{b}
  2. Bbb
  3. C1b\frac{1}{b}
  4. D2b\frac{2}{b}

Correct answer

D. 2b\frac{2}{b}

Step-by-step solution

If a simple pendulum has significant amplitude (up to a factor of 1 / e of original) only in the period between t=0s to
As retardation =bv=\mathrm{bv} ∴\therefore retarding force =mbv=\mathrm{mbv} ∴\therefore net restoring torque when angular displacement is θ\theta is given by =−mg⁡ℓsin⁡θ+mbv⁡ℓ=-\operatorname{mg} \ell \sin \theta+\operatorname{mbv} \ell ∴\therefore \quad\mathrm{I}α=−mgℓsin⁡θ+mbv⁡ℓ\alpha=-m g \ell \sin \theta+\operatorname{mbv} \ell where, I=mℓ2\mathrm{I}=\mathrm{m} \ell^2 ∴d2θdt2=α=−gℓsin⁡θ+bvℓ\therefore \quad \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}=\alpha=-\frac{\mathrm{g}}{\ell} \sin \theta+\frac{\mathrm{bv}}{\ell} for small damping, the solution of the above differential equation will be ∴θ=θ0e−bt2sin⁡(wt+ϕ)\therefore \quad \theta=\theta_0 \mathrm{e}^{-\frac{\mathrm{bt}}{2}} \sin (w t+\phi) ∴\therefore \quad angular amplitude will be =θ.e−bt2=\theta . e^{\frac{-b t}{2}} According to question, in τ\tau time (average life-time), angular amplitude drops to 1e\frac{1}{e} value of its original value (θ)(\theta) ∴θ0e=θ0e−6τ2\therefore \quad \frac{\theta_0}{\mathrm{e}}=\theta_0 \mathrm{e}^{-\frac{6 \tau}{2}} 6τ2=1\frac{6 \tau}{2}=1 ∴τ=2b\therefore \quad \tau=\frac{2}{b}
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