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

[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 Apology withdrawn.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.