Question

a spring with a spring constant of 49 n/m has a 7.2 kg mass hanging from the bottom of it how fast is the mass moving when it passes the equilibrium point

Answer 1

Answer:

ω = (K / M)^1/2   fundamental frequency for Simple Harmonic Motion

ω = (49 kg-m / (sec^2 - m) * 7.2 kg)^1/2 =  2.61 /sec

F = -K A    Hooke's Law and spring constant

A = 7.2 kg * 9.80 m/s^2 / 49 n/m = 1.44 m       amplitude

V =  ω A = 2.61 / sec * 1.44 m = 3.6 m/s        maximum speed