Vertically suspended spring with mass attached. F = ma gives the diff eq: m(d^2y/dt^2) + b(dy/dt) + ky = 0.
y(t) measures the vertical position (upward is positive) of the mass relative to the equilibrium position (how far the mass hangs down if it is not moving).
m = mass of the attached mass
b = damping coefficient of spring (units N*S/m)
k = spring constant (units N/m)
Consider the mass to be released from rest at height A, initial conditions:
y(0) = A and y'(0) = 0
Nondimensionalize the system using t=tcs, y(t)=ycz(s). Choose tc and yc in a way that permits the investigation of the effect of b. Write the nondimensional system using [tex]B[/tex]=b/sqrt(mk)
The Attempt at a SolutionA
Here's what i got so far:
d/dt = (ds/dt)(d/ds) = (1/tc *d/ds)
d^2/dt^2 = (1/tc^2)(d^2/ds^2)
substitute into the system equation on the top gives:
(m/tc2)(d2/ds2(ycz(s))) + (b/tc)(d/ds(ycz(s))) + kycz(s) = 0
(myc/tc2)(d2z/ds2) + (byc/tc)(dz/ds) + kycz(s) = 0
dividing by (myc/tc2) gives:
d2z/ds2 + (btc/m)dz/ds + (tc2k/m)z = 0
Gives 3 dimensionless Pis:
Pi1 = (btc/m)
Pi2 = (tc2k/m)
Unsure about this (am i right?): the corresponding initial conditions are z(0) = A/yc and dz/ds (0) = 0.
If so, this gives Pi3 = A/yc.
Ideas for going further in this problem is that to pick certain Pis and work around to make the system into dimensionless based on dz/ds.