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Question

A block of mass 200 g executing SHM under the influence of a spring of spring constant k=90Nm−1 and a damping constant b=40gs−1. Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In (1/2)=−0.693)

A
7s
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B
9s
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C
4s
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D
11s
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Solution

The correct option is A 7s
Given data,
mass m=200g
Spring constant k=90Nm1
Damping constant b=40gs1
To calculate: Time taken for the amplitude of vibration to drop to half of the initial value
We know that amplitude at any time t can be given as:
A(t)=A0ebt2m

or T1/2=0.693l×2×0.240×103=6.93s
Time taken for its amplitude of vibrations to drop to half of its initial values is 7s

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