Coupled Oscillator: Solving Initial Forces & Finding Eigenvalues

AI Thread Summary
The discussion centers on solving a coupled oscillator problem involving two masses connected by springs, focusing on formulating the initial forces and finding eigenvalues. The forces acting on mass m_1 and mass m_2 are expressed in terms of spring constants and displacements, with specific equations provided. The user encounters issues with imaginary angular frequencies when attempting to solve the system using matrix algebra, indicating a potential error in their initial force formulation. A revised solution for the forces is suggested, aiming to correct the approach and achieve real eigenvalues. The thread concludes with a request for further assistance in refining the initial formulation.
alex3
Messages
43
Reaction score
0

Homework Statement


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

There are two masses, m_1 and m_2.2. 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 m_1 has a force

F_1 = -k_1 x_1 + k_2 x_2

acting on it. We take right as the positive x 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 m_2 has a force

F_2 = -k_2 x_2 - k_3 x_3 = -x_2 (k_2 + k_3) - x_1 k_3

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, x_3, is equal to the sum of the other two springs. I know I'll need to express x_3 in terms of x_1, x_2 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:

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

Eventually giving:

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

And the solutions to the quadratic in \omega^{2} 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: 502
Physics news on Phys.org
Solution:

F_{1} = -k_{1}x_{1} - k_{2}x_{1} + k_{2}x_{2}
F_{2} = -k_{3}x_{2} + k_{2}x_{1} - k_{2}x_{2}

Thanks all the same!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top