Solving for Normal Modes in a Non Similar Coupled Pendulum

[aseylys]

Homework Statement

(The picture is having trouble being directed so I attached it)

Hey guys, I'm trying to figure out the normal modes of a pendulum with two different masses. The lengths of the strings are the same and the weight of the strings are negligible.

I can easily find everything if the masses are equal but when the masses are different it kind of screws me up.

Homework Equations

F=ma
F_s=k(x₁-x₂)=kΔx
m₁*a₁=-m₁*g*(x₁/l)-kΔx
m₂*a₂=-m₂*g*(x₂/l)-kΔx

The Attempt at a Solution

x(double dot)_n=a_n

Working it all out I get:

a₁+a₂+(g/l)*(x₁+x₂)+Δx((k/m₁)-(k/m₂))

This doesn't help me get a simple solution like x(t)=A*cos(ωt+$\phi$) because of the addition of the Δx((k/m₁)-(k/m₂))

 

[TSny]

Hello.

OK, so you have two coupled differential equations:

m1*a1=-m1*g*(x1/l)-kΔx
m2*a2=-m2*g*(x2/l)-kΔx

I think you have one sign error in these equations if Δx stands for the same quantity in both equations. (You didn't state what Δx means. Correction: You did state that Δx is x₁-x₂. Can you see which of the equations has a sign error?) If the spring exerts a negative force on one of the masses, will it also exert a negative force on the other mass?

You're going to try to find normal mode solutions in the form

x₁ = A₁cos(ωt + $\phi$)
x₂ = A₂cos(ωt + $\phi$)

Rather than add the two differential equations, try solving them simultaneously for A₁ and A₂ and see what happens.

 

[aseylys]

Oops, I did miss a negative. So it'd be
m1*a1=-m1*g*(x1/l)-kΔx
m2*a2=-m2*g*(x2/l)+kΔx

Ploughing through it and setting some constant (I'm going to pick $\eta$) $\eta$=m_n*x_n

And using $\omega$_g,n²=(g*m_n)/l
So then:
$\eta$₁+$\omega$_g,1²*x₁
$\eta$₂+$\omega$_g,2²*x₂

Then solving them I get (I think):
x₁=A₁cos($\omega$_g,1+$\phi$₁)
x₂=A₂cos($\omega$_g,2+$\phi$₂)

 

[TSny]

Oops, I did miss a negative. So it'd be
m1*a1=-m1*g*(x1/l)-kΔx
m2*a2=-m2*g*(x2/l)+kΔx

That looks good.

Ploughing through it and setting some constant (I'm going to pick $\eta$) $\eta$=m_n*x_n

And using $\omega$_g,n²=(g*m_n)/l
So then:
$\eta$₁+$\omega$_g,1²*x₁
$\eta$₂+$\omega$_g,2²*x₂

I don't follow what you did there.

Then solving them I get (I think):
x₁=A₁cos($\omega$_g,1+$\phi$₁)
x₂=A₂cos($\omega$_g,2+$\phi$₂)

In a normal mode, both particles will have the same frequency. That is, you should have

x₁=A₁cos($\omega$t+$\phi$₁)
x₂=A₂cos($\omega$t+$\phi$₂)

with the same value of $\omega$ for both particles. You need to substitute these expressions into your differential equations.

 

[aseylys]

Maybe I'm missing the concept then. I don't really understand why they'd have the same frequency. If one mass is heavier than the other wouldn't it have a lower frequency?

 

[TSny]

It's essentially part of the definition of "normal mode" that each particle has the same frequency. See:
http://en.wikipedia.org/wiki/Normal_mode.

Also, in a normal mode, the particles move either "in phase" with each other or "out of phase" with each other. This mean that you can assume that the phase constant $\phi$ is the same for each particle. "Out of phase" then means that A₁ and A₂ will have opposite signs. In finding the normal modes, you can actually just let $\phi = 0$.

So, you can make the assumption that the motion of each particle is described by
x₁=A₁cos($\omega$t)
x₂=A₂cos($\omega$t)

If you sub these into your two equations of motion, you will be able to determine the possible normal mode frequencies and whether the particles move in phase or out of phase in each mode.

[EDIT: You can explore the normal modes of a similar system here: http://www.falstad.com/coupled/
You can change to 2 masses and you can vary the individual masses (click on "mouse:pull string"). You can select either normal mode or a combination of the modes (slide the two dots near the middle left side of the screen).]