Question

2. (24 pts) A 4-foot spring measures 6 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to 1.5 times the instantaneous velocity. If the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s2 for the acceleration due to gravity.) (a) (8 pts) Find the equation of motion. (b) (6 pts) Find the time at which the mass passes the equilibrim position with an upward velocity for the first time. (c) (6 pts) Find the time at which the mass attains its extreme displacement from the equilib- rium position. (d) (4 pts) What is the position of the mass at this instant?

Answer 1

Ky = mg kx2 8x32 = 8x32 2 k 128 lbfft face tumine K = 128 Hift eft=y Damping for ce 1.5 V = Gv Initial condition y(0)=0 jloj (34H/ Equation of motion mdzy dt2 =-ky obv 8 d y = -1284 - 1.5 dl day + 169 + 25 dy + 1.5 dy +164 =0 dtz 8 at d²y 0.1875 dyt 16y at dt 2 (0² + + 0.1875D +16) Auxellery Gym m2 +0.1875 m +16=0 -0.1875 & 10-1875) ² _ 4x16 m = - 0.1875 - Jo.035-64 m = 2 m - -0.1875 + 7.9971 2 Catis) im =-0.09375 +3.9985 i Roots are Imaginary General son med CA cslpt48sing A cos (3.9985 t) + B sin (3.9985+) -0.093754 y(t) e a Pat y (0)=0 y(o) - [A

-0.09375+ [B sin (3-1985+)] ylt) ý (t] -0.093757 -0.09375 B e sin (3-19857) + 3.9985 Bé -0.09375t (as (3.9985t) 7 fils 7 - 3.9985 B 7 3.9985 B 1.75 so Auswey Equation of Motion [1.75 sin (3.9985+) -0.09375t 4 Equation of Motion a © y sin (3.9985+) 0.09375 Be-0-093757 = 3.9985 Beo 6-0.093757 Les $3.9985 tan (3.9985+) 3.9985 0.09375 see Answer s magumum tan (3-9985+) 42.651 3-9985t = tant (42= 651) 3-9985t = 1.547 t = 0.38698 At extreme displacement velocity changes its time t = 0.38698 Sec Auswer yt y (0.38698) displacement -6.09375 x 0-38698) X 1.75 X sin (13.9985 * 0-38698 -0.03628 4 x 1.75 X sin (1.547] y 1.6872 ft Answes @ position at this instant. 598)] 4 maximum desplacement. 1)
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