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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=12πkMfn=12π40005=4.5Hz
ωn=2πfn=9π=28.28rad/s
The damped frequency is given by
f=fn1δ2=4.5×10.04=4.41Hz
ω=ωn1δ2=28.28×10.04=27.71rad/s
The initial displacement is 50mm downwards so C=50mm
and its displacement after 0.3sec is given by
x=Ceδωntcosωt=50×e0.2×28.28×0.3cos(27.71×0.3)x=50×0.183×(0.443)=4.07mm


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