A .6-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of .13 m. Determine a.) the velocity when it passes the equilibrium point, b.) the velocity when it is .1 m from equilibrium. c.) the total energy of the system, and d.) the equation describing the motion of the mass, assuming that x was a maximum at t=0?

Question

George · Accepted Answer

This is WRONG. Did not multiply the 2pi by 3. You just did 2pi

drwls · Answer

a) Maximum velocity, which happens at the equilibrium point, is
Vmax = A*w, where w is the angular frequency, in radians per second.
In this case,
w = 3.0 * 2 pi = 6.28 radians/s
Vmax = 0.13m*6.28 rad/s = 0.816 m/s

b) 0.1 m from equilibrium is 0.1/0.13 = 76.92% of maximum deflection, and will have 0.7692^2 = 59.17% of maximum spring potential energy, which is 59.17% of total energy.
1-0.5917 = 40.83% of the energy will be kinetic energy at 0.1 m stretch.
Velocity wil be sqrt(0.4083)= 63.90% of maximum, or 0.521 m/s

c) (1/2) M*Vmax^2 = maximum KE
Do the numbers
d) X = (amplitude)*cos (wt)
= 0.13 cos(6.28 t)
(It could also be negatve)