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