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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.
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system, ζ = 1, the roots are real negative and identical, i.e. ss12==−ζωn (15) The solution form X(t) = A e st is no longer valid.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)
Figure 2 shows the response of the series RLC circuit with L=47mH, C=47nF and for three different values of R corresponding to the under damped, critically damped and over damped case.
In this example, we will explore using the free vibration response of a baseball bat suspended from support point O to determine the location of the bat’s center of percussion. 1. Draw an FBD of the bat. 2. Develop the equation of motion (EOM) of the bat in terms of the angle . Linearize this
INTRODUCTION. This tutorial discusses the response of a second-order system to initial conditions, including initial displacement and initial velocity. The mass-spring-dashpot system shown in Fig. 1 is an example of a second-order system.
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.
The response of a system to an impulse looks identical to its response to an initial velocity. The impulse acts over such a short period of time that it essentially serves to give the system an initial velocity. Fig. 3 shows the impulse response of three systems: under-damped, critically damped, and over-damped.
