A spring of unknown spring constant is attached to a 2.30 kg mass. The mass is pulled horizontally outward by a distance "A" from equilibrium, then released at t = 0. At t = 0.200 seconds, the mass first reaches the equilibrium position, where it is seen to be moving at a velocity of v = −0.950 m/s. (a) What is the period of the motion? (b) What is the frequency of the motion? (c) Calculate the spring constant of the spring. (d) Calculate the amplitude of motion, A.
Period T = 4(0.2000) = 0.8000 seconds.
f = 1/T.
In particular, substitute v = -0.950 m/s, t = 0.2000 seconds and f = 1/T into to get A. Or you could of course note that v = -0.950 m/s when t = T/4. Then and so when v = -0.950. Alternatively, you could note that when x = 0 (equilibrium position), Kinetic Energy .
From release point to equilibrium point: t = T/4.
From equilibrium to opposite position (v = 0): t = T/2.
From opposite point back to equilibrium point: t = 3T/4.
Return to release point: t = T.
Do you see it?