EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 25, Problem 16P

The motion of a damped spring-mass system (Fig. P25.16) is described by the following ordinary differential equation:

m d 2 x d t 2 + c d x d t + k x = 0 x =

where x = displacement from equilibrium position (m), t = time (s), m = 20-kg mass, and c = the damping coefficient ( N s/m ) . The damping coefficient c takes on three values of 5 (under- damped), 40 (critically damped), and 200 (overdamped). The spring constant k = 20  N/m . The initial velocity is zero, and the initial displacement x = 1 m. Solve this equation using a numerical method over the time period 0 t 15  s . Plot the displacement versus time for each of the three values of the damping coefficient on the same curve.

Chapter 25, Problem 16P, 25.16 	The motion of a damped spring-mass system (Fig. P25.16) is described by the following

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1. Verify Eqs. 1 through 5. Figure 1: mass spring damper In class, we have studied mechanical systems of this type. Here, the main results of our in-class analysis are reviewed. The dynamic behavior of this system is deter- mined from the linear second-order ordinary differential equation: where (1) where r(t) is the displacement of the mass, m is the mass, b is the damping coefficient, and k is the spring stiffness. Equations like Eq. 1 are often written in the "standard form" ď²x dt2 r(t) = = tan-1 d²r dt2 m. M +25wn +wn²x = 0 (2) The variable wn is the natural frequency of the system and is the damping ratio. If the system is underdamped, i.e. < < 1, and it has initial conditions (0) = zot-o = 0, then the solution to Eq. 2 is given by: IO √1 x(1) T₁ = +b+kr = 0 dt 2π dr. dt ل لها -(wat sin (wat +) and is the damped natural frequency. In Figure 2, the normalized plot of the response of this system reveals some useful information. Note that the amount of time Ta between peaks is…

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