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  1. An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible.

  2. Let’s analyze this physically. When b = 0 the response is a sinusoid. Damping is a frictional force, so it generates heat and dissipates energy. When the damping constant b is small we would expect the system to still oscillate, but with decreasing amplitude as its energy is converted to heat. Over time it should come to rest at equilibrium.

  3. This document discusses the response of a second-order system, such as the mass-spring-dashpot shown in Fig. 1, to a step function. The modeling of a step response in MATLAB and SIMULINK will also be discussed. Fig. 1. Single-degree-of-freedom mass-spring-dashpot system.

  4. Learning Objectives. Learn to analyze a general second order system and to obtain the general solution. Identify the over-damped, under-damped, and critically damped solutions. Convert complex solution to real solution. Suspended “mass-spring-damper” equivalent system.

  5. ocw.mit.edu › courses › res-8-009-introduction-to-oscillations-and-waves-summerLecture 04: Damped - MIT OpenCourseWare

    How do we model oscillatory phenomena in which air drag causes a decrease in oscillation amplitude? 1.1 Drag and general Damping Forces. To achieve our objective of finding a more accurate model for oscillatory phenomena, we need to first find the correct Newton’s second law equation for such systems.

  6. Example 2. Find the unit impulse response to a critically damped spring-mass-dashpot system having e−pt in its complementary function. Solution. Since it is critically damped, it has a repeated characteristic root −p, and the complementary function is yc = e−pt(c1 + c2t). The function in this family satisfying

  7. Free Response of Critically Damped 2nd Order System For a critically damped system, ζ = 1, the roots are real negative and identical, i.e. ss12==−ζωn (15) The solution form X(t) = A e st is no longer valid. For repeated roots, the theory of ODE’s dictates that the family of solutions satisfying the differential equation is () n (12)

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