Coupled mass problem with orthogonal oscillations

[CAF123]

For one eigenvector I got $<-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}>$ How would I get the amplitude of the curve? (If i take a negative and the common factor outside the eigenvector, then I get a negative sin curve oscillating about some point with <1,0,-1> premultiplying the sinusoidal term))

 

[voko]

The vector you obtained means that the center-mass is stationary, and the other two masses are oscillating completely out of phase, with equal amplitude. The amplitude depends on the initial conditions.

 

[CAF123]

The vector you obtained means that the center-mass is stationary, and the other two masses are oscillating completely out of phase, with equal amplitude. The amplitude depends on the initial conditions.

I don't think I was given any IC. For $<1/\sqrt{5}, -\sqrt{2/5}, 1/\sqrt{5}>,$ is this oscillatory motion with the m1 and m3 in phase with same amplitude but the centre mass out of phase with a bigger amplitude (I can't really picture how this would be possible)

For $<-1/\sqrt{5}, \sqrt{2/5}, 1/\sqrt{5}>,$ it would be m1 and m2 out of phase with same amplitude but m2 with a bigger amplitude (Could it be the two masses m1 and m3 are anti-phase so act to compress m3)

 

[voko]

You don't really care about the initial conditions for normal modes, because all they do is just scale the amplitude up or down. For a sketch, this is completely unimportant. What is important is, like ehild said, the ratios of the amplitudes, and these are given by the eigenvectors.

 

[CAF123]

You don't really care about the initial conditions for normal modes, because all they do is just scale the amplitude up or down. For a sketch, this is completely unimportant. What is important is, like ehild said, the ratios of the amplitudes, and these are given by the eigenvectors.

Is my analysis of the other two eigenvectors along the right lines in my previous post?

 

[voko]

Yes, it is correct - assuming the eigenvectors are correct, which I did not check.

 

[CAF123]

Yes, it is correct - assuming the eigenvectors are correct, which I did not check.

Ok, so how does this come across in a graph? Take the first eigenvector. (both masses out of phase and the middle stationary). Do I just draw a normal sine curve at the left of the graph and then somewhere in the middle of the graph, bring the graph to equal 0 and then afterwards to the right of this middle point, draw a sin curve out of phase with the sin curve that I drew to begin with?

So I would be effectively plotting amplitude versus distance for this particular mode.

 

[voko]

You have three masses oscillating with the same frequency. So you just need to provide three graphs of the - one per mass - over their mutual period. Using three different colors, you could arrange them in a single graph.

Sketching the entire system - the masses and the string - is more difficult. You can't do that in a single graph. You could provide a few "snapshots", though. In the 21st century, an animation would be in order, though :)

 

[CAF123]

You have three masses oscillating with the same frequency. So you just need to provide three graphs of the - one per mass - over their mutual period. Using three different colors, you could arrange them in a single graph.

Sketching the entire system - the masses and the string - is more difficult. You can't do that in a single graph. You could provide a few "snapshots", though. In the 21st century, an animation would be in order, though :)

Would that not mean 9 graphs in total? (I.e 3 masses, so one for each mass for each eigenvalue. I have three eigenvalues so 9 graphs?)

 

[voko]

You need just two graphs per mass for the first and second masses, the syn-phase and anti-phase varieties, and just one for the third one. Then you can combine them. Note that you could make the amplitude of the first and the third mass equal in all the graphs, but make sure the amplitude of the second one is in correct ratio with the other amplitudes.

 

[CAF123]

You need just two graphs per mass for the first and second masses, the syn-phase and anti-phase varieties, and just one for the third one. Then you can combine them. Note that you could make the amplitude of the first and the third mass equal in all the graphs, but make sure the amplitude of the second one is in correct ratio with the other amplitudes.

I am not sure I understand the need for 5 graphs. I think I understand why I don't draw as many graphs for m2 (since in one mode it's amplitude is zero) but in the other modes, the ratio of the amplitudes is different (e.g in mode 1 the amplitude ratios are the same for m1 and m2 but for mode 2, the amplitude for m2 is sqrt 2 that of m1, m3.)

 

[voko]

The amplitudes of different modes cannot and should not be compared directly. This is because the amplitude of the mode has an arbitrary multiplicative constant (as does its eignevector, too). It is only amplitudes within one and the same mode that matter. Now observe that in all the modes the amps of m1 and m3 are equal.

 

[ehild]

I was speaking to some people elsewhere and they said you can get the form of the normal mode by just looking at the eigenvectors in front of each of the terms. How so? Could you explain the physical interpretation of these eigenvectors? (i have my general solution in the earlier posts of this thread)

A normal mode is a special motion of the masses in the system: All of them vibrate with the same frequency and in/=out phase. The relative amplitudes constitute the eigen vectors. In this case, you have three 3 dimensional eigenvectors, a, b, c. It is usual to choose them of unit length.

So the eigen vector belonging to the ω₁=√(2T/m ) angular frequency is a=(1/√2, 0, -1/√2). You can see this eigenvector in the attachment. Check the others and make them normalised. I do not see where the √5 came from, and I think you have a sign error in the third eigenvector.

The general motion of the three masses is a linear combination of the normal modes. This motion is also represented by a three-dimensional vector, its components belong to the masses 1,2,3.

(x1,x2,x3)=A(a1,a2,a3)sin(ω₁t+θ₁)+B(b1,b2,b3)sin(ω₂t+θ₂)+C(c1,c2,c3)sin(ω₃t+θ₃).

ehild

 

[CAF123]

So the eigen vector belonging to the ω₁=√(2T/m ) angular frequency is a=(1/√2, 0, -1/√2). You can see this eigenvector in the attachment. Check the others and make them normalised. I do not see where the √5 came from, and I think you have a sign error in the third eigenvector.

Was there an attachment did you say? The minus was a typo and the sqrt{5} was an error in computing the scaling factor to make the vector normalised. It should be 1/2<1, -$\sqrt{2}$,1> for the second eigenvector and 1/2<1,$\sqrt{2}$,1> for the third eigenvector.

[/QUOTE]

 

[ehild]

I attached the picture.

ehild