# Normal Modes and Normal Frequencies

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1. Apr 29, 2016

### RicardoMP

1. The problem statement, all variables and given/known data
I have to determine the frequencies of the normal modes of oscillation for the system I've uploaded.

2. Relevant equations

I determined the following differential equations for the coupled system:
$$\ddot{x_A}+2(\omega_0^2+\tilde{\omega_0}^2)x_A-\omega_0^2x_B = 0$$
$$\ddot{x_B}+2\omega_0^2x_B-\omega_0^2x_A = 0$$
which I confirmed are correct.

3. The attempt at a solution
I assumed that the usual normal modes solutions are:
$$x_A=Ccos(\omega t)$$
$$x_B=C'cos(\omega t)$$
where C and C' are the amplitudes for each of the oscillating masses and $$\omega$$ is the associated normal mode frequency. Therefore, I proceeded by determining the ratio $$\frac{C}{C'}$$ for each equation, after substituting the solutions in the differential equations.
$$\frac{C}{C'}=\frac{\omega_0^2}{-\omega^2 +2\omega_0^2+2\tilde{\omega_0}^2}$$ and $$\frac{C}{C'}=\frac{-\omega^2+2\omega_0^2}{\omega_0^2}$$
My problem now is that, when I try to determine the solutions for $$\omega$$, I arrive to 4 solutions of the form:
$$\omega=\pm \sqrt{\frac{-(-4\omega_0^2 -2\tilde{\omega_0})\pm \sqrt{(-4\omega_0^2 -2\tilde{\omega_0})^2 -4(-4\omega_0^2 -2\tilde{\omega_0})(3\omega_0^3 +4\tilde{\omega_0}\omega_0)}}{2}}$$
which I think it's wrong, since with these solutions I don't even know how to schematically draw the normal modes with these kind of solutions. So is my solving method wrong or is this method only applicable to symmetric systems? Did I get the calculations wrong? Is there another way to solve this problem?

#### Attached Files:

• ###### modosnormais.PNG
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2. Apr 30, 2016

### ehild

What are the data? The masses of A and B, the spring constants. What do the ω 0-s mean?
Assuming your derivation is correct up to *, the last equation perhaps is not. First, ω should not be negative. Solve for ω2. You get two ω2-s, that is, two normal-mode frequencies, which correspond two kinds of motion of A and B, (in the same direction and in opposite directions). Check the powers and simplify the expression for ω2. Show your work. One can not find your mistakes without seeing it.

3. Apr 30, 2016

### RicardoMP

The relevant information for this problem is that the masses are the same $$m_A=m_B$$. The $$\omega_0$$ and $$\tilde{\omega_0}$$ are the natural frequencies associated with each spring. The long springs have constant $$\tilde{k}$$, so $$\tilde{\omega_0}=\sqrt{\frac{\tilde{k}}{m}}$$. The short springs have constant k, so $$\omega_0=\sqrt{\frac{k}{m}}$$. Sorry for the lack of information.
So, assuming that up to * I'm correct (which I verified again), my calculations are the following:
$$(-\omega^2 + 2\omega_0^2+2\tilde{\omega_0^2})(-\omega^2+2\omega_0^2)=(\omega_0^2)^2$$
I call my \omega^2 = a, so $$a^2-2\omega_0^2a-2\omega_0^2a+4\omega_0^4-2\tilde{\omega_0}a+4\tilde{\omega_0}\omega_0=(\omega_0^2)^2$$.
And that's it. I solve the quadratic equation for $$\omega^2$$ and get the last equation in my original post without the large square root.

4. Apr 30, 2016

### RicardoMP

I found at least one mistake, when solving the quadratic equation, dumb me!
The final equation I get, solving the quadratic equation for $$\omega^2$$ is :
$$\omega^2=\frac{(4\omega_0^2+2\tilde{\omega_0}^2)\pm \sqrt{(4\omega_0^2+2\tilde{\omega_0}^2)^2-4(3\omega_0^4+4\omega_0^2 \tilde{\omega_0^2}})}{2}$$
and therefore:
$$\omega^2=\frac{(4\omega_0^2+2\tilde{\omega_0}^2)\pm 2\sqrt{\omega_0^4+\tilde{\omega_0^4}}}{2}$$

5. Apr 30, 2016

### ehild

You have mistakes again with powers and tilde.

6. Apr 30, 2016

### RicardoMP

Yes, just corrected it in the reply above. Is my solving method correct at least? Won't I have 4 solutions for $$\omega$$?

7. Apr 30, 2016

### ehild

It looks correct. Simplify by 2.
You get two values for ω2, and ω can not be negative, so it is 2 solutions for ω.
Remember, the solutions are of the form C*cos(ωt). Do you get different solution with -ω?

Last edited: Apr 30, 2016