- #36
##y_p = \sqrt{3}##haruspex said:I don't see how you get that. Please post the detailed steps.
I applied @TSny's method and got the answers quoted in post #28.
You could do that. But it won't lead straightaway to a value for one of the unknown ##C_n##'s.Redwaves said:I have
##y_p =
\sum_{n = 1}^{5} C_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] =
\frac{1}{\sqrt{3}}[C_1 sin (\frac{\pi p}{6}) + C_2 sin (\frac{2 \pi p}{6}) + C_3 sin (\frac{3 \pi p}{6}) + C_4 sin (\frac{4 \pi p}{6}) + C_5 sin (\frac{ 5 \pi p}{6})]##
From there, why I can't choose ##y_p = 0## for ##p = 2## and plug those number in the formula above?
To elaborate on @TSny's response, that will produce a system of five simultaneous equations, as you had much earlier. Fine if you have a matrix inverter app, but messy by hand.Redwaves said:I have
##y_p =
\sum_{n = 1}^{5} C_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] =
\frac{1}{\sqrt{3}}[C_1 sin (\frac{\pi p}{6}) + C_2 sin (\frac{2 \pi p}{6}) + C_3 sin (\frac{3 \pi p}{6}) + C_4 sin (\frac{4 \pi p}{6}) + C_5 sin (\frac{ 5 \pi p}{6})]##
From there, why I can't choose ##y_p = 0## for ##p = 2## and plug those number in the formula above?
The m^{th} normal mode is ##\frac {1} {\sqrt 3} \sin \left( \frac{m \pi p}{6} \right)##, not ##C_m \frac 1 {\sqrt 3} \sin \left( \frac{m \pi p}{6} \right)## So, ##C_m## should not appear. Please correct this.Redwaves said:I got
##y_p (C_m[\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =
\sum_{n = 1}^{5} C_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] \cdot C_m \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})
##
Well, I've been a little sloppy in my wording. The expression ##\frac {1}{\sqrt{3}} \sin \left(\frac{m\pi p}{6} \right)## is not the m^{th} normal mode. Rather, it's the amplitude of the vibration of the p^{th} bead in the m^{th} mode. The m^{th} mode is the collection of all of the ##\frac {1}{\sqrt{3}} \sin \left(\frac{m\pi p}{6} \right)## for all of the values of ##p##; that is, for all of the beads in this mode.Redwaves said:Is the ##m^{th}## normal mode the eigenvector?
It's all good. that wasn't your fault, your explanation is great. It's just me who have a hard time to understand.TSny said:When I said to multiply both sides of the ##y_p## equation by the m^{th} normal mode, I meant to multiply both sides by ##\frac{1}{\sqrt{3}} \sin \left(\frac{m \pi p}{6} \right)##. I should have been more explicit.
Merely being linearly independent would ensure that. Orthogonality means they form an inner product space in which their pairwise dot products are zero, and orthonormality that they are also unit vectors.Redwaves said:all the eigenvectors are orthogonal, which mean that the eigenvectors forms a space of N dimensions
You introduced ##C_n## in post #39, apparently to mean the same as ##B_n##, the amplitudes you are trying to find. Now you are redefining it?Redwaves said:If ##C_n = sin(\frac{n \pi p}{6})##
In post #34, TSny used ##B_n## for the amplitudes of the normalised modes. These are the constants you are trying to find.Redwaves said:I did the same thing as Tsny in post #34, I guess. Otherwise, I don't see why we have to multiply by ##\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})## both side
That is done so that when you sum over p and apply the orthonormality property all the Bs disappear except for ##B_m##. It effectively inverts the matrix equation.Redwaves said:I was using ##C_n##, because I'm reading my book at the same time and it uses ##C_n## instead of ##B_n##. I try to understand what I'm doing. For instance, I still don't know why I have to multiply both side by ##\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})##
I see...haruspex said:That is done so that when you sum over p and apply the orthonormality property all the Bs disappear except for ##B_m##. It effectively inverts the matrix equation.
No, you can't do that. There is a ##C_n## inside the sum over n.Redwaves said:##\sum_{n = 1}^{5} \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right]] \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6}) = 1 ##
You mean ##haruspex said:No, you can't do that. There is a ##C_n## inside the sum over n.
(And the equation above isn't true anyway.)
Swap the order of summation and sum over p.
Why would you think it is, and what has that got to do with my response in post #57?Redwaves said:You mean ##
\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) = \sum_{n = 1}^{5} C_n
## is not true?
That wasn't what I did ?haruspex said:Now you can take the ##C_n## outside the sum over p to produce
##\sum_{n = 1}^{5} [C_n\sum_{p = 1}^{5} \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})\right]]##
Ah, yes, sorry, I see now you nearly had it right. The error is that ##\sum_{p = 1}^{5} \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})\right] = \delta_{m,n}##. It only equals 1 when m=n.Redwaves said:That wasn't what I did ?
I thought ##\sum_{p = 1}^{5} \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})\right] = 1##
Which terms in the sum on the right are not necessarily zero?Redwaves said:This is the kronecker delta? If m ##\neq## n The right hand side is 0. If I can't replace by 1. I don't see what to do with.
##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =\sum_{n = 1}^{5} C_n \delta_{m,n}##
Suppose you let ##m = 2## as an example. What does the right hand side reduce to?Redwaves said:This is the kronecker delta? If m ##\neq## n The right hand side is 0. If I can't replace by 1. I don't see what to do with.
##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =\sum_{n = 1}^{5} C_n \delta_{m,n}##
I just did it. I got ##-\frac{1}{2}## which is correct. I'll try with ##n=3##TSny said:Suppose you let ##m = 2## as an example. What does the right hand side reduce to?
n = 3? or do you mean m = 3?Redwaves said:I just did it. I got ##-\frac{1}{2}## which is correct. I'll try with ##n=3##
isn't the same since ##m=n##?TSny said:n = 3? or do you mean m = 3?
On the right hand side, you have a sum over n. So, you must let n run through all the integers 1, 2, 3, 4, 5. Earlier when you multiplied through by ##\frac {1}{\sqrt {3}} \sin \left( \frac {m \pi p}{6}\right)## you can imagine choosing ##m ## to be anyone of the integers 1...5. So, m is the integer that you can choose freely.Redwaves said:isn't the same since ##m=n##?