Question

2. The system is very underdamped, with only a small amount of friction. A 600 N man leaps up and grabs onto a spring that is hanging vertically. After a long long time, the man comes to rest and the spring has been stretched by 67.3 cm compared to its length before the man grabbed it. WWII a. What is the spring constant? b. Sketch y(t) for the man's hands, at the end of the spring Let y=0 be at the equilibrium position for the spring. Someone pulls downward on the man's feet until they are 20.0 cm closer to the ground and holds him at rest in that position for a moment. Then they release the man at time to. For the rest of the problem, assume that the friction is small enough that we can ignore damping. c. What is the period of the man's motion? d. What is the displacement as a function of time? e. What is the velocity as a function of time? f. What is the acceleration as a function of time? 8. Give the next three times when the man will pass through the equilibrium position.

Answer 1

2a) Spring constant (K) here x = 67.3cm = 0.673m. =) too equilibrium mg = ka =) k ong - 600 - 891.53 N/m m Now amplitude Xo = 20.0an = 0.20m We know that k=mw² where w is angular frequency w=5km =) time period I = 22 > T = 27,1% on = mass = 690 =) T = 2x5 60 =borg 2) T=aa80.0673 IT=1635) 891-53 (d) displacement as a function of time x = Xo sinut Zo 0-20m, L= m = 1891-532.95 3) x = 0.20 sin (3.855t) m 60 rool's

1 a (e) Velocity as a function of time Ved (0.20 sin (3.855)) V= 0.20x3.855 cos (3.855t) V = 0.771 COS(3.855t) m/s (f) acceleration as a function of time. 2 0771 603(3-88st)) a--0771X 3.855 sin (3.8557) a = -2.972 sin (3.855t) m/s2 man pass through equilibrium positon. (8) =) 0.20 sin (3.855€) =0 = 3.855€ = 0,7,27,3 2) +20 SS) 37355 so next three time is 0.8153, 1.6305, Q.4458