Question

2) A 2-pound weight attached to the end of a spring stretches it 6 inches. At t= 0 and from a position 8 inches below the equilibrium position the weight is given a upward velocity of 3/4 ft/s. a) Find the equation of motion. b) What are the amplitude and period of motion? c) How many complete vibrations will the weight have completed at the end of 6t? d) What is the position of the weight at t=5s with respect to the equilibrium position? e) What is the instantaneous velocity at t=5s and in which direction? (25 points)

Answer 1

To =0.5ft 12 @ m= 2 pound, elongation m spring ox: 6 inch .. By Hooke's Tawon OF-kox > mg -.OX = kemg - 2x32:2 rii Ax 0.5 - k=128:8 lb/ft Augulas Speed of S.H.M. - B 128.8 38.02 leads. vm V 2 Amblitude of motion, A= 8inch 8 = 0.67 Jt so' equation of motion : - y=-Acoret {-wesign is used, because : spring is fulled below |_ => 1 Y = -0.67 Co8/8-02)] 30 Amplitude of motion, A = 8 inch = 8 =0.67fcd Period of motion, T= 20 = 2x3-4 = 0.788. - 8.02 © Total number of oscillation in time. Gli second - M = GT = 64,314 = 24:15_ so number of complete. Oscillation = 24

OO @ Put t=s8 in eq. (0 - y=-0.67 Cos (8:02x+5) = 144= -0.49741 e. below the equilibrium @ V=d4 = 0 {-0:67608(8:02+2 %. + 1 = 0.67%8:02 Sin (8:02+) I Put t=58 ; L . 19 = 0.67x8.02 Sin (8.02x5) => 119 = 3.63.fts - It att