Question

(7 points) 13. A mass weighing 10 pounds stretches a spring 3 inches. The mass is removed and replaced with a mass weighing 51.2 pounds, which is initially released from a point 4 inches above the equilibrium position with an downward velocity of ft/s. Find the equation of motion, ä(t). (g = 32 ft/s2) (7 points) 14. A mass weighing 4 pounds stretches a spring 2 feet. The system is submerged in a medium which offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Find the equation of motion, ä(t). What type of damped motion is this system?

Answer 1

Solution=13 Using Newton law ma²x = - KX dt spring force al te madhe = 01-0 spering constant m = mass Find K As 10 pound Weight stretches W = kx Ik=W spring 3 inches x should in feet I feet = 12 inches linches = I feet. DC = 3.4t=&ft 12 K = 10 ft = 4 Oft Spring (114) constant As mass is removed and replaced with mass Weighing 51-2 pound W = ng g= 32 ft/82 = 51.2 m=W m m= 1.6 slug 32

+ x=0 dt2 (40 (1.60 X"+ 25x = 0 18 Auxilarly equation 7+25-0 8=+5i Geneual solution X = A cos dwt)+ B sin(at) (w=5) X(t)= A cos 5t + B sin 57 A and Baue constant Initial condition 810)=-4 inches = .4ft- teft 12. X'(0) = 5 ft/sec 4 Put Tt=0 in Eq ② X(0)= A 008 (5X6) + B Sin Coxs) - I = A A=-1 3. differentate ② Werottime Lt). X'(t) = -A5 Sin6t) + 5BCos6t) Put it o X'60)25 = 0 + 5881 5-5B B=1

@ba Equation of motion X (+) = - Lcos 5t) + + sin(st) AR 3 ft Please Note that as beus guidelines 9 have solved Ist q question completely.