Question

A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then stretched an additional 1 inch and released. Find the equation of motion,

Answer 1

Answer:

Step-by-step explanation:

Simple Harmonic Motion:

k = 100 lb/ft

m = 1 lb/32.2 fps²

y = A sin ωt = A sin √(k/m) t

ω = √(k/m) = √(100/(1/32.2))= 56.745 rad/s

f = ω/(2π) = 56.745/(2π) = 9.031 Hz

T = period = 1/f = 1/9.031 = 0.1107 s

y(t) = A sin ωt = sin (9.031)t