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Question

A block of mass 2 kg connected to a spring of spring constant 8 Nm1 is allowed to oscillate on a rough horizontal surface. If the system experiences a damping force =0.230× velocity, then the time required for the amplitude of resulting oscillation to fall to half of its initial value is

A
0.693 s
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B
12 s
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C
0.8 s
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D
14.3 s
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Solution

The correct option is B 12 s
When damping force F=bv where v is velocity
then the amplitude at time t is given by
A=A0ebt2m
Here b=0.23
Thus, t=2mbln(A0A)
=2×20.230ln(2)=12.05 s

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