Normal Modes and Normal Frequencies

[RicardoMP]

Homework Statement

I have to determine the frequencies of the normal modes of oscillation for the system I've uploaded.

Homework Equations

[/B]
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.

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?
Thank you in advance!

 

[ehild]

Homework Statement

I have to determine the frequencies of the normal modes of oscillation for the system I've uploaded.

Homework Equations

[/B]
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.

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?
Thank you in advance!

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.

 

[RicardoMP]

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.

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.

 

[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:
$$\omega^2=\frac{(4\omega_0^2+2\tilde{\omega_0}^2)\pm 2\sqrt{\omega_0^4+\tilde{\omega_0^4}}}{2}$$

 

[ehild]

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 $$a^2-2\omega_0^2a-2\omega_0^2a+4\omega_0^4-2\tilde{\omega_0}a+4\tilde{\omega_0}\omega_0=(\omega_0^2)^2$$.
.

You have mistakes again with powers and tilde.

 

[RicardoMP]

You have mistakes again with powers and tilde.

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

 

[ehild]

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

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:
$$\omega^2=\frac{(4\omega_0^2+2\tilde{\omega_0}^2)\pm 2\sqrt{\omega_0^4+\tilde{\omega_0^4}}}{2}$$

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?