Question

(1 point) A mass m = 4 kg is attached to both a spring with spring constant k = 325 N/m and a dash-pot with damping constant c=4N s/m. The mass is started in motion with initial position Xo = 1 m and initial velocity vo = 9 m/s. Determine the position function z(t) in meters. x(t) = Note that, in this problem, the motion of the spring is underdamped, therefore the solution can be written in the form x(t) = Cie pt cos(wit-ai). Determine C1,w1 ,Qjand p. (assume 0 <«i <27) Graph the function z(t) together with the "amplitude envelope" curves x = -Ciept and x = Cie pt. Now assume the mass is set in motion with the same initial position and velocity, but with the dashpot disconnected ( so c= 0). Solve the resulting differential equation to find the position function u(t). In this case the position function u(t) can be written as u(t) = Cocos(wot - Qo). Determine Co, wo and ao. Co = (assume 0 < «o < 26 ) Finally, graph both function z(t) and u(t) in the same window to illustrate the effect of damping.

Answer 1

m4 Given m = 4kg, k = 325N/m,c=4N.s/m, x, =1. Vo = 9 k 325 . we first find thato, * =* = 325 = 81.25, p = = = 0.5 = p = 0.25 which is smaller than 0,,so the motion is under damped the roots of DE r=-pp- =-0.5+ 70.25-81.25 =-0.51 V-81 = -0.5+9i. r=-0.5+91,-0.5-9i. hence x(t)= cze-0.5t cos(9t) +c2e-054 sin (91) let v(t)=x' (t)=-e-054 (0.54, cos (91)+9c, sin(9t)+0.5c, sin(9t) -9c, cos (9t)) setting t=0 x(0) = ce cos(0)+cze sin(O)=C1 =1=C and v(0) =-e°(0.5c, cos(0)+9c, sin (0)+0.5c, sin (0) - 9c, cos(O)) =9=0.50 -9c2 by solving the equations we get (7 = 1,62 = -0.9444. Thereforex(t)= +0.5 cos(9t)-0.9444e-054 sin (9) removing the damping we get critically damped harmonic motion with the frequency 0, = 181.25 = 9.0138 So the general solution isx(t)= Acos(@t)+B sin(at) =x'(t)=-Q2A sin (@t)+0,B cos(6.1) x(0) = Acos(O) +B sin(0)=1= 4 x'(0) = -0,4 sin (0)+ ®, BcOS(O)=9= 0,B=B= 20 138 = 0.9984. then x(t) = cos(@t-a)= VA+B= 1+0.9969 = 1.4131. as = tan") = tan-1 (0.984) = 0.7845.