Coupled Oscillator

  • Thread starter alex3
  • Start date
  • #1
44
0

Homework Statement


Two masses attached via springs (see picture attachment). [tex]k_n[/tex] represents the spring constant of the [tex]n^{th}[/tex] spring, [tex]x_n[/tex] represents the displacement from the natural length of the spring.

There are two masses, [tex]m_1[/tex] and [tex]m_2[/tex].


2. The attempt at a solution

My problem is formulating the initial forces on these. Here's what I've tried (with reference to the attached picture):

Mass [tex]m_1[/tex] has a force

[tex]F_1 = -k_1 x_1 + k_2 x_2[/tex]

acting on it. We take right as the positive [tex]x[/tex] direction, so mass 1 has the tension in spring 1 acting on it to the left, as well as the tension of spring 2 acting to the right.

Mass [tex]m_2[/tex] has a force

[tex]F_2 = -k_2 x_2 - k_3 x_3 = -x_2 (k_2 + k_3) - x_1 k_3[/tex]

acting on it; spring 2 acts to the left (it's trying to contract), and spring 3, the longest spring, also acts to the left. Here is the assumption I'm unsure about, that the displacement of spring 3, [tex]x_3[/tex], is equal to the sum of the other two springs. I know I'll need to express [tex]x_3[/tex] in terms of [tex]x_1, x_2[/tex] as these correspond to the displacements of the masses, but this solution doesn't work.

3. Solving the equations

I'm OK with this part, I'm using matrix algebra to find the normal modes of the system (the eigenvalues). However, using the above logic I would end up with imaginary angular frequencies:

[tex]
\[ \left(
\begin{array}{cc}
m_1 & 0\\
0 & m_2
\end{array}
\right)
\left(
\begin{array}{c}
\ddot{x_1}\\
\ddot{x_2}
\end{array}
\right)
=
\left(
\begin{array}{cc}
k_1 & -k_2\\
k_3 & k_2 + k_3
\end{array}
\right)
\left(
\begin{array}{c}
x_1\\
x_2
\end{array}
\right)\]
[/tex]

Eventually giving:

[tex]
\[ \left|
\begin{array}{cc}
\frac{k_{1}}{m_{1}}-\omega^{2} & -\frac{k_{2}}{m_{1}}\\
\frac{k_{3}}{m_{2}} & \frac{k_{2}+k_{3}}{m_{2}} - \omega^{2}
\end{array}
\right|\]
[/tex]

And the solutions to the quadratic in [tex]\omega^{2}[/tex] that this produces has imaginary roots, which is not ideal!

So; which initial formulation will help me?
 

Attachments

  • coupled-oscillators-diagram.jpg
    coupled-oscillators-diagram.jpg
    9.9 KB · Views: 375

Answers and Replies

  • #2
44
0
Solution:

[tex]F_{1} = -k_{1}x_{1} - k_{2}x_{1} + k_{2}x_{2}[/tex]
[tex]F_{2} = -k_{3}x_{2} + k_{2}x_{1} - k_{2}x_{2}[/tex]

Thanks all the same!
 

Related Threads on Coupled Oscillator

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
974
  • Last Post
Replies
6
Views
183
Replies
3
Views
698
Replies
11
Views
290
Replies
12
Views
2K
Top