Aspring 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]

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

close lines strong magnet less close weaker