System Modeling Coursework
Class 19-20: Free (undamped and damped) vibrations
P.R. VENKATESWARAN Faculty, Instrumentation and Control Engineering, Manipal Institute of Technology, Manipal Karnataka 576 104 INDIA Ph: 0820 2925154, 2925152 Fax: 0820 2571071 Email:
[email protected],
[email protected] Web address: http://www.esnips.com/web/SystemModelingClassNotes
WARNING! • I claim no originality in all these notes. These are the compilation from various sources for the purpose of delivering lectures. I humbly acknowledge the wonderful help provided by the original sources in this compilation. • For best results, it is always suggested you read the source material. July – December 2008
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Contents • Modeling of free vibrations – Free vibrations – Forced vibrations – Characteristics of response
• Numerical on the derivations
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Introduction • In theory, the single degree-of-freedom spring-mass system, once set into motion, would continue to move up and down for ever. • In practice all systems are damped, which means that energy is dissipated, and the amplitude of the motion gradually gets smaller and smaller until it stops altogether. • Damping can be introduced from various sources, and is hard to model accurately. July – December 2008
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System with damping • One model, known as viscous damping, is that of a force that is proportional to the velocity of the mass, and which opposes its motion. • This is represented by a dashpot, which is given the symbol shown in Fig.
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Equations of motion • Free body diagram
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Solutions to this equation • The solution to this equation is given by
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Output graphs
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Most important case
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Output graph
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Summary • If ζ > 1, the system is heavily damped, and no vibration occurs. • If ζ = 1, the system is critically damped, and the damping is only just sufficient to prevent vibration. • If ζ < 1, as it is in the majority of cases, there is insufficient damping in the system to prevent vibration and the motion is oscillatory.
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Numerical • Let us return to the system in Figure, and, as before, assume that the weight is given an initial displacement, X0, and then released from rest. Let X0 = 100 mm. Therefore, the initial conditions are that at t = 0, x = 100 mm, and x = 0 . Find the solution for various values of ζ. July – December 2008
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Case (1): Over damped system with ζ = 2 • The displacement is given by • Differentiating with respect to time
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Output for over damped system with ζ = 2
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Case (2): Critically damped system, ζ = 1
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Case (3): Under damped system with ζ = 0.05
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Case (3): Under damped system with ζ = 0.05
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Summary
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Summary
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Numerical - 1
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Solution to Numerical 1
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Numerical - 2
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Numerical - 2
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Numerical 3
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Solution to Numerical 3
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And, before we break… • Reality is often a matter of personal perception as much as objective fact.
Thanks for listening…
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