Question

A spring with a spring constant kk 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, the amplitude, and the period. neglect friction. then

Answer 1

Given:

k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs 

Solution:

F = -kx 
mx" = -kx 
x" + (k/m)x = 0 

characteristic equation: 
r^2 + k/m = 0 
r = i*sqrt(k/m) 

x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t) 

ω = sqrt(k/m) 
2π/T = sqrt(k/m) 
T = 2π*sqrt(m/k) 
T = 2π*sqrt(0.03106 slugs / 100 lb/ft) 
T = 0.1107 s (period) 

x(0) = 1/12 ft = 0.08333 ft 
x'(0) = 0 

1/12 = Asin(0) + Bcos(0) 
B = 1/12 = 0.08333 ft 

x' = Aω*cos(ωt) - Bω*sin(ωt) 
0 = Aω*cos(0) - (1/12)ω*sin(0) 
0 = Aω 
A = 0 

So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[(2π/0.1107)t]