JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Elasticity & SHM

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A block of mass 1 kg attached to a spring is made to oscillate with 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

)

(A) $3.3 \times 1 0^{2} kg s^{- 1}$
(B) $1.16 \times 1 0^{2} kg s^{- 1}$
(C) $0.69 \times 1 0^{2} kg s^{- 1}$
(D) $5.7 \times 1 0^{- 3} kg s^{- 1}$ ✓ Correct

Answer: (D)

Solution

For damped harmonic motion, the amplitude decays exponentially:

A (t) = A_{0} e^{- b t / (2 m)}

, where

b

is the damping constant. Given

A_{0} = 12

cm,

A (t) = 6

cm at

t = 120

s, and

m = 1

kg. We have

\frac{A ( t )}{A _{0}} = e^{- b t / (2 m)} = \frac{1}{2}

. Taking natural log:

- \frac{b t}{2 m} = ln (1/2) = - ln 2

. Therefore

b = \frac{2 m l n 2}{t} = \frac{2 \times 1 \times 0.693}{120} = \frac{1.386}{120} = 0.01155 \approx 1.155 \times 1 0^{- 2} = 0.01155

kg/s

\approx 5.7 \times 1 0^{- 3}

kg/s.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Elasticity & SHM

Solution

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