Simple harmonic motion of two springs

[Eric_meyers]

Homework Statement

"Two light springs have spring constants k1 and k2, respectively, and are used
in a vertical orientation to support an object of mass m. Show that the angular
frequency of small amplitude oscillations about the equilibrium state is
[(k1 + k2)/m]^1/2 if the springs are in parallel, and [k1 k2/(k1 + k2)m]^1/2 if
the springs are in series. "

Homework Equations

f = -kx
w = (k/m)^1/2
frequency = w/(2 * pi)

The Attempt at a Solution

So I set up my fnet for the parallel one

m * x'' = - (k1x + k2x) since in the parallel feature there are two distinct springs exerting two distinct forces

however dividing through I'm left with

x'' = - (k1x + k2x)/m

I'm not quite sure how to use x'' to get to angular frequency or how to remove the negative sign.

 

[Fightfish]

For a simple harmonic oscillation, x"=-(w^2)x

 

[Feldoh]

Solutions to x'' = -[(k1+k2)/m]x can be written in the form x(t) = A*sin(wt+phi) so from what you know about the sine function if you can find the period, you can find the frequency, which will enable you to find the angular frequency...

 

[Eric_meyers]

Oh I think I got it!

x'' + [(k1 + k2)/m]x = 0

And using the characteristic polynomial to solve this second order DE, gives me the solution

x(t) = A cos(wt + phi) if and only if w = [(k1 + k2)/m]^1/2

now for setting up the DE in the series case I'm having some difficulty would it be:

m * x'' = -k1 * k2 x ? Since you could treat both k1 and k2 acting as one k? errr..

 

[Fightfish]

now for setting up the DE in the series case I'm having some difficulty would it be:

m * x'' = -k1 * k2 x ? Since you could treat both k1 and k2 acting as one k? errr..

Sure you got the effective k for springs in series correct?

 

[Eric_meyers]

oh wait, if I want to combine the k in both springs into another constant I'm going to have to take the "center of mass" sort of speak for the spring stiffness - I forget the correct terminology... center of stiffness?

m * x'' = -[(k1 * k2)/(k1 + k2)] x

x '' + (k1 * k2)/[(k1 + k2) * m]x = 0

Using the characteristic equation I again get

x(t) = A cos (wt - phi) if and only if w = {(k1 * k2)/[(k1 + k2) * m]}^1/2

Which of course is the answer.
Thanks