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Question

# A mass of 5 kg is suspended on a spring of stiffness 4000 N/m. The system is fitted with a damper with a damping ratio of 0.2. The mass is pulled down 50 mm and released. Calculate the displacement after 0.3 sec:

A
4.07 mm
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B
4.7 mm
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C
7.4 mm
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D
7.04 mm
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Solution

## The correct option is A 4.07 mm The natural frequency is given by fn=12π√kM⇒fn=12π√40005=4.5Hzωn=2πfn=9π=28.28rad/sThe damped frequency is given by f=fn√1−δ2=4.5×√1−0.04=4.41Hzω=ωn√1−δ2=28.28×√1−0.04=27.71rad/sThe initial displacement is 50mm downwards so C=−50mmand its displacement after 0.3sec is given byx=Ce−δωntcosωt=−50×e−0.2×28.28×0.3cos(27.71×0.3)⇒x=−50×0.183×(−0.443)=4.07mm

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