100g weight, spring elongation 8cm.
g = kx find k = 0. 1 * 10/0.08 = 25/2n/m。
The periodic formula of spring oscillator T=2π√m/k
ω=2π/T=√k/m=√25/2*0.25=√50
The elongation of the spring hanging at the equilibrium position of a 250g object is l = g/k = 2.5 * 2/25 = 0.2m.
Find amplitude a through conservation of mechanical energy. (The origin of the X axis is at the equilibrium position)
The conservation equation is formulated by the process that 0.2 1m/s reaches the maximum displacement along the positive direction of the X axis. )
mv？ /2-mg*4cm+k(0.2+0.04)？ /2=k(0.2+A)？ /2-mgA
Be patient. About 0.05 meters
The vibration equation x=0.05cos(√50t+φ).
When t=0s, bring in x=0.04m to find φ.
Cosφ= 0.8φ= 35, and the radian should be 0.6 1.
Finally, the vibration equation x=0.05cos(√50t+0.6 1).