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## Homework Statement

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=t

_{c}s, y(t)=y

_{c}z(s). Choose t

_{c}and y

_{c}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 Solution

AHere's what i got so far:

d/dt = (ds/dt)(d/ds) = (1/t

_{c}*d/ds)

d^2/dt^2 = (1/t

_{c}^2)(d^2/ds^2)

substitute into the system equation on the top gives:

(m/t

_{c}

^{2})(d

^{2}/ds

^{2}(y

_{c}z(s))) + (b/t

_{c})(d/ds(y

_{c}z(s))) + ky

_{c}z(s) = 0

(my

_{c}/t

_{c}

^{2})(d

^{2}z/ds

^{2}) + (by

_{c}/t

_{c})(dz/ds) + ky

_{c}z(s) = 0

dividing by (my

_{c}/t

_{c}

^{2}) gives:

d

^{2}z/ds

^{2}+ (bt

_{c}/m)dz/ds + (t

_{c}

^{2}k/m)z = 0

Gives 3 dimensionless Pis:

Pi

_{1}= (bt

_{c}/m)

Pi

_{2}= (t

_{c}

^{2}k/m)

Unsure about this (am i right?): the corresponding initial conditions are z(0) = A/y

_{c}and dz/ds (0) = 0.

If so, this gives Pi

_{3}= A/y

_{c}.

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.