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:
The natural
frequency is given by
fn=12π√kM⇒fn=12π√40005=4.5Hz
ωn=2πfn=9π=28.28rad/s
The damped frequency is given by
f=fn√1−δ2=4.5×√1−0.04=4.41Hz
ω=ωn√1−δ2=28.28×√1−0.04=27.71rad/s
The initial
displacement is 50mm downwards so C=−50mm
and its
displacement after 0.3sec is given by
x=Ce−δωntcosωt=−50×e−0.2×28.28×0.3cos(27.71×0.3)⇒x=−50×0.183×(−0.443)=4.07mm