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

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]