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Q.

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 ln2=0.693)

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a

0.69×102 kgs−1

b

3.3×102 kgs−1

c

5.7×10−3 kgs−1

d

1.16×10-2 kgs−1

answer is D.

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Detailed Solution

A=A0e−b2mt  Given :    A0=12, A=6, t=120s , m=1kg On solving       6=12e-b2120 ⇒2=e60b ⇒ln2=60b ⇒b=ln260 ⇒b=1.16×10−2 kg/sec
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