Coupled Oscillator: Solving Initial Forces & Finding Eigenvalues

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
alex3
Messages
43
Reaction score
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] \[ \left(<br /> \begin{array}{cc}<br /> m_1 & 0\\<br /> 0 & m_2<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c}<br /> \ddot{x_1}\\<br /> \ddot{x_2}<br /> \end{array}<br /> \right)<br /> =<br /> \left(<br /> \begin{array}{cc}<br /> k_1 & -k_2\\<br /> k_3 & k_2 + k_3<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{c}<br /> x_1\\<br /> x_2<br /> \end{array}<br /> \right)\][/tex]

Eventually giving:

[tex] \[ \left|<br /> \begin{array}{cc}<br /> \frac{k_{1}}{m_{1}}-\omega^{2} & -\frac{k_{2}}{m_{1}}\\<br /> \frac{k_{3}}{m_{2}} & \frac{k_{2}+k_{3}}{m_{2}} - \omega^{2}<br /> \end{array}<br /> \right|\][/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: 536
Physics news on Phys.org
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!