SolSec 2.8

# SolSec 2.8 - 2 60 Problems and Solutions Section 2.8(2.72...

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2- 60 Problems and Solutions Section 2.8 (2.72 through 2.77) 2.72*. Numerically integrate and plot the response of an underdamped system determined by m = 100 kg, k = 20,000 N/m, and c = 200 kg/s, subject to the initial conditions of x 0 = 0.01 m and v 0 = 0.1 m/s, and the applied force F ( t ) = 150cos5 t. Then plot the exact response as computed by equation (2.30). Compare the plot of the exact solution to the numerical simulation. Solution: The solution is presented in Matlab: First the m file containing the state equation to integrate is set up and saved as ftp2_72.m function xdot=f(t, x) xdot=[-(200/100)*x(1)-(20000/100)*x(2)+(150/100)*cos(5*t); x(1)]; % xdot=[x(1)'; x(2)']=[-2*zeta*wn*x(1)-wn^2*x(2)+fo*cos(w*t) ; x(1)] % which is a state space form of % x" + 2*zeta*wn*x' + (wn^2)*x = fo*cos(w*t) (fo=Fo/m) clear all; Then the following m file is created and run: %---- numerical simulation --- x0=[0.1; 0.01]; %[xdot(0); x(0)] tspan=[0 10]; [t,x]=ode45('fp2_72',tspan,x0); plot(t, x(:,2), '.'); hold on; %--- exact solution ---- t=0: .002: 10; m=100; k=20000; c=200; Fo=150 ; w=5 wn=sqrt(k/m); zeta=c/(2*wn*m); fo=Fo/m; wd=wn*sqrt(1-zeta^2) x0=0.01; v0= 0.1; xe= exp(-zeta*wn*t) .* ( (x0-fo*(wn^2-w^2)/((wn^2-w^2)^2 . ..

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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 2.8 - 2 60 Problems and Solutions Section 2.8(2.72...

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