6.5kg mass attached to a spring

Question

 A 6.5 kg mass is attached to a spring (k=64 N/m) and stretched to a distance that is 10 cm away from it’s equilibrium point. The mass is released from this position at time, t=0 sec. Design a data table for 2 whole cycles on the position vs. time graph for the oscillating mass

Answer 1

The graph will be a cosine curve oscillating between 10cm and -10cm. The force will change sign at the midpoint of the movement(when the spring starts pushing the mass backwards) which is in this case 0 so you have to solve the kinematic equation to find the total time(period) of the oscillation, pretty much a plug and chug however for those interested, here's why:

 

From physics:

F = -kx

F = ma

-kx = ma

a = x''

mx'' = -kx

mx'' + kx = 0

Solving using a characteristic equation:

mr^2 + k = 0

r = sqrt(-k/m) = +-i*sqrt(k/m)

sqrt(k/m) is the natural frequency so we'll call it w

 

r = w*i

Therefore the solution is:

c1sin(wt) + c2cos(wt)

using trig identities:

x(t) = Rcos(wt-delta) where delta is the phase shift and R is the forced displacement amplitude.

Therefore the graph should be a cosine graph and the table can be found by plugging