Chapter 5For the massm2there are two forces acting on it: the force due to the spring with constantk2is−k2y; and the force due to the spring with constantk1isk1(y−x). So we getm2y0=k1(x−y)−k2y.So the system ism1x0=k1(y−x),m2y0=−k1(y−x)−k2y,or, in operator form,(m1D2+k1)[x]−k1y=0,−k1x+±m2D2+(k1+k2)²[y]=0.Withm1=1,m2=2,k1=4,andk2=10/3, we get(D2+4)[x]−4y,−4x+(2D2+22/3) [y,(5.28)with initial conditions:x(0) =−1,x0(0) = 0,y(0) = 00(0) = 0.Multiplying the second equation of the system given in (5.28) by 4, applying (2
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.