# Homework Help: Simple harmonic motion of two springs

1. Aug 23, 2009

### Eric_meyers

1. The problem statement, all variables and given/known data

"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. "

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

3. 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.

2. Aug 23, 2009

### Fightfish

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

3. Aug 23, 2009

### 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...

4. Aug 23, 2009

### 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..

5. Aug 23, 2009

### Fightfish

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

6. Aug 23, 2009

### 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