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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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$0 .69 \times 10^{2} {kgs}^{- 1}$

$5 .7 \times 10^{- 3} {kgs}^{- 1}$

$1 .16 \times 10^{- 2} {kgs}^{- 1}$

$3 .3 \times 10^{2} {kgs}^{- 1}$

answer is D.

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Detailed Solution

$A = A_{0} e^{- \frac{b}{2 m} t} Given : A_{0} = 12, A = 6, t = 120 s, m = 1 kg$

$On solving 6 = 12 e^{- \frac{b}{2} 120} \Rightarrow 2 = e^{60 b} \Rightarrow \ln 2 = 60 b \Rightarrow b = \frac{\ln 2}{60} \Rightarrow b = 1 .16 \times 10^{- 2} kg / \sec$

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