ep 2. Now, because the function gx) is a quadratic polynomial, let us assume a a
ID: 3216412 • Letter: E
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ep 2. Now, because the function gx) is a quadratic polynomial, let us assume a articular solution that is also in the form of a quadratic y-Ax +Bx + C. Differential Equations. Exam #2, Summer 2018 (1) Choose exactly 5 questions to answer (not 6) and, in the box below, write the mumber of the question that you choose to omit (If you leave the box empty, only questions 1 through 5 will be gradod.) (2) You may use your textbook and a TI-84 or scientific calculator. Do not bring a cell-phone or paper of any (3) PUT YOUR FINAL ANSWERS ON THIS PAGE [INSIDETHEDOXES Name: I. A 32-pound object stretches a spring 8 feet to reach equilibrium. The object is then pulled to a point that is I foot below the equilibrium position and released from with an upward velocity of 6 feet per second. Assume a damping force of ? 5 . what is the position of the object when it attains its maximum displacement above the equilibrium position? Express your answer to 4 decimal places.Explanation / Answer
Let y(t) be the displacement of mass at any time t.By given information, m=32/32 =1 ,s=8ft,F=32,b=5 .Then by Hooke's law F=ks=>k=F/s = 32/8 =4.
System satisfies the ode my"+by'+ky=0 i.e y"+5y'+4y=0
Auxiliary equation: r^2+5r+4=0 implies r=-1,-4
Thus y(t)=c1e^(-t)+c2e^(-4t)
y(0)=1 implies c1+c2=1
y'(0)=6 implies -c1-4c2=6. So both equation imply c1=10/3 and c2=-7/3.
Thus y(t)=(10e^(-t)-7e^(-4t))/3 .For maximum displacement y'(t)=0 implies -10e^(-t)+28e^(-4t)=0 implies e^(3t)=2.8 implies t=0.1491.So y(0.1491)=1.7847 is the position of mass at time of maximum displacement.
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