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

[TSny]

Summing the geometric series for the first of those four terms:
$\Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m+n}N)}=\frac{1-e^{i\pi (N\frac {m+n}N)}}{1-e^{i\pi (\frac {m+n}N)}}=\frac{1-e^{i\pi ( m+n)}}{1-e^{i\pi (\frac {m+n}N)}}$
If m+n is even then the numerator vanishes, but it's not going to if m+n is odd.

For m+n odd, the numerator in your final expression is 2.

So, $S1 \equiv \Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m+n}N)} = \frac{2}{1-e^{i\pi (\frac {m+n}N)}}$

Likewise, $S2 \equiv \Sigma_{p=0}^{N-1}e^{-i\pi (p\frac {m+n}N)} = \frac{2}{1-e^{-i\pi (\frac {m+n}N)}}$.

Then, $S1+S2 = 2$.

If m+n is odd, then m-n is also odd.

So, $S3 + S4 = 2$, where $S3 \equiv \Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m-n}N)}$ and $S4 \equiv \Sigma_{p=0}^{N-1}e^{-i\pi (p\frac {m-n}N)}$

Thus, overall, we get $S1 + S2 - S3 - S4 = 0$.

 

[TSny]

So I should have made an error somewhere.
Since, the right answer are $A_2 = -\frac{1}{2}, A_3 = \frac{5-\sqrt{3}}{\sqrt{3}}$ for the amplitudes order by frequency.

What did you get for $A_2$?

 

[Redwaves]

What did you get for $A_2$?

I tried your method, but I don't understand. we didn't use complex numbers.
I finally got an answer from my professor by email. However, I think I'm more confuse now.
He told me I should use the formula to find the eigenvectors of normal modes which is $<n| = \frac{1}{\sqrt{\sum_N}}(sin(\frac{n\pi p}{N+1}, sin(\frac{n\pi p}{N+1},...)$ Where N is the number of beads, p the index of the bead and n the mode.

However, I don't see how to know which mode is the second and third mode order by frequency.

I'm not sure, but I think the eigenvectors is the ratio of the different amplitude. thus, maybe the lowest amplitude is the higher frequency, but I'm just guessing.
As I type that I think what I said makes not sense since I only get the eigenvector for a single mode with are the ratio of different amplitude only in a specific mode. I'm rock solid stuck.

Edit:
Maybe it will sound stupid, but is the modulus of the eigenvector the amplitude?

Edit #2:
I think I have to find a relation between the eigenvector and the amplitude.

 

[TSny]

The n^th normal mode is described by the expression $$\sin\left(\frac{n\pi p}{N+1}\right)$$ This expression gives the relative amplitude of the p^th bead in the n^th normal mode. The initial displacements of the beads in your problem are (y₁, y₂, y₃, y₄, y₅) = (3, 0, √3, 2, 2). These can be expressed as a superposition of the normal mode expressions $$y_p = \sum_{n = 1}^{N} A_n \sin\left(\frac{n\pi p}{N+1}\right) = \sum_{n = 1}^{5} A_n \sin\left(\frac{n\pi p}{6}\right)$$ The $A_n$ are the amplitudes of the normal modes. The normal modes obey an orthogonality relation $$\sum_{p = 1}^{N} \sin\left(\frac{m\pi p}{N+1}\right) \sin\left(\frac{n\pi p}{N+1}\right) = \frac{N+1}{2} \delta_{n,m}$$ or$$ \sum_{p = 1}^{5} \sin\left(\frac{m\pi p}{6}\right) \sin\left(\frac{n\pi p}{6}\right) = 3 \delta_{n,m}$$ The occurrence of the $3$ on the right hand side of the last equation means that the normal mode expressions are not "normalized". You can see that if we choose the normal modes as $$\frac{1}{\sqrt{3}}\sin\left(\frac{n\pi p}{6}\right)$$ where we have included a "normalization factor" of $\frac{1}{\sqrt{3}}$, then we have $$ \sum_{p = 1}^{5} \left[ \frac{1}{\sqrt 3} \sin\left(\frac{m\pi p}{6}\right) \right] \left[ \frac{1}{\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] = \delta_{n,m}$$ With these normalized normal modes, the right hand side is just $\delta_{n,m}$ with a coefficient of 1. For the case where $m = n$, $$ \sum_{p = 1}^{5} \left[ \frac{1}{\sqrt 3} \sin\left(\frac{n \pi p}{6}\right) \right]^2 = 1 \,\,\,\,\,\,\,$$ for each normalized normal mode ($n = 1, 2, ..., 5)$.

Using these normalized modes, the initial displacements $y_p$ can be expressed as $$y_p = \sum_{n = 1}^{5} B_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right]$$ where I'm using $B_n$ to denote the amplitude of the n^th normalized normal mode.

You are asked to find the values of $B_n$ for $n = 2$ and $n=3$. To do this, multiply both sides of this equation by the m^th normal mode $\frac 1 {\sqrt 3} \sin\left(\frac{m\pi p}{6}\right)$ and sum over the index $p$. Using the orthogonality condition, you should end up with an expression for calculating $B_m$ in terms of the given values of $y_p$.

 

[Redwaves]

That's what I tried, but I didn't get the right answer. For example, I have $B_n = 0$ for $n = 2$ and $B_n = 3$ for $n = 3$

for $n=3$
$\sqrt{3}=B_n[\frac{1}{\sqrt{3}}(1+0-1+0+1)] = B_n \frac{1}{\sqrt{3}}$
thus, if I multiply both side by $\frac{1}{\sqrt{3}}$
I have $1 = B_n(\frac{1}{3})$

 

[haruspex]

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.

 

[Redwaves]

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.

$y_p = \sqrt{3}$
$n = 3$
$m = n$

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

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

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

 

[haruspex]

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

 

[Redwaves]

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?

 

[TSny]

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 m^th 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 $\sum_{n = 1}^5$.

 

[haruspex]

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.

 

[Redwaves]

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.

 

[TSny]

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 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]

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

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

 

[TSny]

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 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.

You can collect these into a vector that represents the m^th 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.

 

[TSny]

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.

 

[Redwaves]

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.

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.

 

[Redwaves]

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

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?

 

[haruspex]

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.

 

[Redwaves]

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

 

[haruspex]

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?

 

[Redwaves]

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

 

[haruspex]

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 $\frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})$

 

[haruspex]

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.

 

[Redwaves]

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.

 

[haruspex]

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

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.

 

[Redwaves]

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?

 

[haruspex]

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?
You had on the right:
$\sum_{p = 1}^{5}[\sum_{n = 1}^{5} C_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})]$
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
$\sum_{n = 1}^{5} [C_n\sum_{p = 1}^{5} \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \cdot \frac{1}{\sqrt{3}} sin (\frac{m \pi p}{6})\right]]$

 

[Redwaves]

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