JEE Main 2017 · 9 Apr · Shift 1 · Physics Q12

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A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm $^{-1}$ and oscillates in a damping medium of damping constant $10^{-2}$ kg s $^{-1}$ . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :

2 s

3.5 s

5 s

7 s

Answer and worked solution

Official answer: D

The energy decays as $E=E_0e^{-bt/m}$ , so it halves at
$t=\frac{m\ln2}b=\frac{0.1\times0.693}{10^{-2}}\approx6.9\text{ s},$
closest to 7 s.

Topics:Damped and forced oscillations

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 12