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

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

 

[haruspex]

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.

 

[Redwaves]

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.

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

 

[haruspex]

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?

 

[Redwaves]

$n = m$
$n = 1 =m$
$n = 2 =m$
...

 

[TSny]

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]

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]

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?

 

[Redwaves]

n = 3? or do you mean m = 3?

isn't the same since $m=n$?

 

[Redwaves]

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.

 

[TSny]

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.