Suppose that the mass-spring-dashpot system with m =0, c=9, and k=2 is set in mo

Question

Suppose that the mass-spring-dashpot system with m =0, c=9, and k=2 is set in motion with x(0) = 0 and x'(0)= 5. find its position function x(t).

Explanation / Answer

a) The related differential equation for this problem is x''+(c/m)*x'+(k/m)*x=0 So we have x''+9/10x'+1/5x=0 r^2+9/10*r+1/5=0 10r^2+9r+2=0 r1=(-9+sqrt(81-80))/20=(-9+1)/20=-0.4 r2=(-9-sqrt(81-80))/20=-0.5 Since r1?r2>0 Our damped harmonic oscillator is over-damped. Thus, our general solution is x(t)=Ae^(r1*t)+Be^(r2*t) So x(t)=Ae^(-0.5t)+Be^(-0.4t) x'(t)=-0.5*Ae^(-0.5t)-0.4*Be^(-0.4t) Now we'll plug in our starting conditions: x(0)=A+B=0 ->A=-B x'(0)=-A/2-0.4B=5 5B/10-4B/10=5 0.1B=5 B=50 -> A=-50 x(t)=50e^(-0.4t)-50e^(-0.5t) b) To find out how far it moves to the right, we need to see first when its velocity is 0: i.e. x'(t)=0 -20e^(-0.4t)+25e^(-0.5t)=0 -0.4t=ln(5/4)-0.5t t=10*ln(5/4) At this point in time, the velocity is 0, and so this is as far as it goes. x(10*ln(5/4))= 50*(e^(-4ln(5/4))-e^(-5*ln(5/4)))= 50*((1.25)^-4-(1.25)^-5)= 4.096 m

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