The discussion revolves around finding the amplitude of oscillations for a string with 5 beads, specifically for modes 2 and 3. Participants are exploring the terminology and equations related to coupled oscillators and the behavior of the system described in their textbooks.
The discussion is ongoing, with participants providing insights and questioning the definitions and assumptions related to the problem. Some have suggested that the original poster clarify their understanding of the variables and the physical setup, while others are exploring the implications of the equations presented.
There is uncertainty regarding the definitions of $$\kappa$$ and $$p$$, as well as the initial conditions necessary to determine the amplitudes for the modes. Participants are also considering the implications of fixed endpoints on the string and how this affects the equations they are using.
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##?