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JEE Main 2021 (17 Mar, Shift 2) · Damped, Forced Oscillations & Superposition

Damped, Forced Oscillations & Superposition question from JEE Main 2021 (17 Mar, Shift 2)

Q6JEE Main 2021 · 17 MAR · SHIFT 2Medium

Damped, Forced Oscillations & Superposition

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion. (take ln 2 = 0.693)

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Answer: (c)

$A = A_0 e^{-bt/2m}$ .
$\frac{6}{12} = e^{-b(120)/(2 \times 1)} \Rightarrow 60b = \ln 2 = 0.693$ .
$b = \frac{0.693}{60} = 1.16 \times 10^{-2}\ \text{kg s}^{-1}$ .
So the answer is (c).

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