Chapter 4The maximum displacement of the mass is found by determining the Frst time the velocityof the mass becomes zero. Therefore, we havey0(t)=0=±2+0.2√5²e−2√5t−2√5h0.1+±.2√5²tie−2√5t,which givest=22√5(2 + 0.2√5)=1√5(2 + 0.2√5).Thus the maximum displacement isy³1√5(2 + 0.2√5)´=³0.±.2√5²µ1√5(2 + 0.2√5)¶´e−2√5/[√5(2+0.2√5)]≈0.242 (m).11.The equation of the motion of this mass-spring system isy0+0.2y0+ 100y=0,y(0) = 00(0) = 1.Clearly, this is an underdamped motion becauseb2−4mk=(0.2)2−4(1)(100) =−399.96<0.So, we use use equation (16) on page 213 in the text for a general solution. Withα=−b2m=−0.22=−0.1andβ=12m√4mk−
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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 University of California, Berkeley.