# How to find the amplitude of oscillations of a string with 5 beads?

• Redwaves
In summary: I am still puzzled about p. Looks to me like one of your equations in post 1 may have the p in the wrong place.I was thinking maybe I have to find all the eigenvector then get the transpose of the matrix to get the eigenvector for the modes.However, I think I have to find a way with the vector ##(3,0,\sqrt{3},2,2)##That link certainly clarifies matters.
Redwaves said:
for ##n=3##
##\sqrt{3}=B_n[\frac{1}{\sqrt{3}}(1+0-1+0+1)] = B_n \frac{1}{\sqrt{3}}##
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.

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.
##y_p = \sqrt{3}##
##n = 3##
##m = n##

##\sqrt{3} [\frac{1}{\sqrt{3}}(sin(\frac{3 \pi 1}{6}) + sin(\frac{3 \pi 2}{6}) + sin(\frac{3 \pi 3}{6}) + sin(\frac{3 \pi 4}{6}) + sin(\frac{3 \pi 5}{6}))] =

B_3[\frac{1}{\sqrt{3}}(sin(\frac{3 \pi 1}{6}) , sin(\frac{3 \pi 2}{6}) , sin(\frac{3 \pi 3}{6}) , sin(\frac{3 \pi 4}{6}) , sin(\frac{3 \pi 5}{6}))] [\frac{1}{\sqrt{3}}(sin(\frac{3 \pi 1}{6}) , sin(\frac{3 \pi 2}{6}) , sin(\frac{3 \pi 3}{6}) , sin(\frac{3 \pi 4}{6}) , sin(\frac{3 \pi 5}{6}))]##

##\sqrt{3}(\frac{1}{\sqrt{3}}+0-\frac{1}{\sqrt{3}}+0+\frac{1}{\sqrt{3}}) =B_3[\frac{1}{3} + 0 + \frac{1}{3}+0 + \frac{1}{3}] ##

##\sqrt{3}(\frac{1}{\sqrt{3}}) = B_3(1)##

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No, you have misunderstood.
$$y_p = \sum_{n = 1}^{5} B_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right]$$
You were instructed to multiply both sides by a certain expression and sum both sides over p.
Let us see you do that.

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?

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?
You could do that. But it won't lead straightaway to a value for one of the unknown ##C_n##'s.

Let's go slowly. If you haven't done this type of thing before and if you are not yet really comfortable with summation notation, then it can be a bit intimidating.

Start with $$y_p = \sum_{n = 1}^{5} C_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right]$$ The first step is to multiply both sides of this equation by one of the normal modes, say the mth normal mode. What does it look like after this first step? Do not write out all of the individual terms for the summation over ##n## on the right hand side. Keep using the compact summation notation symbol ##\sum_{n = 1}^5##.

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.
The procedure in post #34 uses the orthonormal property of the sin() coefficients to execute the inversion by a much simpler process. It is strongly analogous to Fourier transforms.

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})
##
If it is right so far, I have no idea why. I'm not sure to understand the orthogonality relation. For sure it is probably the main reason why I'm struggling to understand. The fact I didn't see this type of thing in class hurt me a lot.

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})
##
The mth 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.

Is the ##m^{th}## normal mode the eigenvector?

##y_p ([\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 \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})##

Redwaves said:
Is the ##m^{th}## normal mode the eigenvector?
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 mth normal mode. Rather, it's the amplitude of the vibration of the pth bead in the mth mode. The mth 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.

You can collect these into a vector that represents the mth mode:

##\left[\frac {1}{\sqrt{3}} \sin \left(\frac{m\pi 1}{6} \right), \frac {1}{\sqrt{3}} \sin \left(\frac{m\pi 2}{6} \right), \frac {1}{\sqrt{3}} \sin \left(\frac{m\pi 3}{6} \right), \frac {1}{\sqrt{3}} \sin \left(\frac{m\pi 4}{6} \right), \frac {1}{\sqrt{3}} \sin \left(\frac{m\pi 5}{6} \right) \right]##

Yes, this vector is an eigenvector of a matrix that arises in solving the equations of motion for the beads.

When I said to multiply both sides of the ##y_p## equation by the mth 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.

TSny said:
When I said to multiply both sides of the ##y_p## equation by the mth 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.
It's all good. that wasn't your fault, your explanation is great. It's just me who have a hard time to understand.

So far, I understand that all the eigenvectors are orthogonal, which mean that the eigenvectors forms a space of N dimensions. I guess from that I can find any position or velocity by using N eigenvectors.

Last edited:
##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =

\sum_{p = 1}^{5}[\sum_{n = 1}^{5} C_n \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})##

Basically, the right hand side is the sum of all the eigenvector for each mode multiply by a eigenvector of a specific mode (Amplitudes of each bead in a mode), right?

Last edited:
Redwaves said:
all the eigenvectors are orthogonal, which mean that the eigenvectors forms a space of N dimensions
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.

If ##C_n = sin(\frac{n \pi p}{6})##
then,
##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =
\sum_{p = 1}^{5}[\sum_{n = 1}^{5} sin(\frac{n \pi p}{6}) \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})##

##\sum_{p = 1}^{5} \left[ \frac{1}{\sqrt 3} \sin\left(\frac{m\pi p}{6}\right) \right] \left[ \frac{1}{\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] = 1##

I have
##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =
\sum_{p = 1}^{5}\sum_{n = 1}^{5} sin(\frac{n \pi p}{6}) ##

Redwaves said:
If ##C_n = sin(\frac{n \pi p}{6})##
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?

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

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
In post #34, TSny used ##B_n## for the amplitudes of the normalised modes. These are the constants you are trying to find.
In post #39, for no apparent reason, you switched from ##B_n## to ##C_n##.
Now you seem to think that these represent the multipliers ##\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})##. They do not.
Please take it one step at a time and post the whole working:
1. The equation in post #38 (also in post #40). Use ##B_n## to avoid further confusion.
2. Multiply both sides by ##\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})##.
3. Sum both sides over p.
4. Simplify the RHS using the orthonormality property.
This should give an equation in which ##B_m##, and no other Bs, appears on the right. n should not appear at all.

Redwaves
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})##

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})##
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.

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.
I see...

##\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =

\sum_{p = 1}^{5}[\sum_{n = 1}^{5} C_n \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})##

##\sum_{p = 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 ####\sum_{p = 1}^{5} y_p ([\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]) =
\sum_{n = 1}^{5} C_n ##

I don't think this is fine. I'm not sure how to manipulate the sums.

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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 ##
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.

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.
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?

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?
Why would you think it is, and what has that got to do with my response in post #57?
##\sum_{p = 1}^{5}[\sum_{n = 1}^{5} C_n \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})]##
The subscript n tells you you cannot shift the ##C_n## outside the sum over n to produce
##\sum_{p = 1}^{5}C_n[\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})]##
Instead, swap the order of summation:
##\sum_{n = 1}^{5}[ \sum_{p = 1}^{5}C_n \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]]##
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]]##

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]]##
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##

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##
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.

And looking back at the posts, I see you corrected the summation in post #56 after my response in post #57. Apology withdrawn.

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}##

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}##
Which terms in the sum on the right are not necessarily zero?

##n = m##
##n = 1 =m##
##n = 2 =m##
...

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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?

TSny said:
Suppose you let ##m = 2## as an example. What does the right hand side reduce to?
I just did it. I got ##-\frac{1}{2}## which is correct. I'll try with ##n=3##

TSny
Redwaves said:
I just did it. I got ##-\frac{1}{2}## which is correct. I'll try with ##n=3##
n = 3? or do you mean m = 3?

TSny said:
n = 3? or do you mean m = 3?
isn't the same since ##m=n##?

It works! Thanks for your patience. Both of you.
I have a lot to learn. I don't think I can solve a similar problem by myself yet.

Redwaves said:
isn't the same since ##m=n##?
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.

But, yes, when you sum over n on the right, the only term that survives is the term for which n = m.

Redwaves

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