Use the approximation that for each time step A spring with

Use the approximation that for each time step. A spring with a relaxed length of 25 cm and a stiffness of 12 N/m stands vertically on a table. A block of mass 79 g is attached to the top of the spring.

Solution

equilibrium position for spring -mass system,

kx = mg

12x = 0.079*9.81

x = 0.0646 m = 6.46 cm

so amplitude A = 30.5 - 25 + 6.46 = 11.96 cm

using equation for SHM: y = Acos(wt)

and mw^2 = k

w = sqrt(12/0.079) = 12.32 rad/s

y = 11.96 cos(12.32t) cm

for t = 0.1 s

r = 11.96 cos(12.32*0.1) = 3.97 cm = 0.0397m down from its initial position.

p = m * dy/dt = - 0.079 * 11.96 * 12.32 * sin(12.32t)

p = - 11.64 sin(12.32t) kg.cm/s

fot t = 0.1 s

p = - 0.11 kg.m/s