Question

A spring with a spring constant 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.

Answer 1

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

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 is the amplitude.

Equation of motion:
x = 0.08333*cos[(2π/0.1107)t]

Answer 2

I've tried entering that in my online homework...and it says its wrong

Answer 3

Simple Harmonic Motion:
k = 100 lb/ft
m = 1 lb/32.2 fps²

y = A sin ωt = A sin √(k/m) t

A = 1 in

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