The discussion focuses on calculating the amplitudes of oscillations for a string with 5 beads, specifically for modes 2 and 3. The relevant formulas include $$A_n = \sin(\kappa p)$$ and $$\kappa = \frac{n\pi p}{N+1}$$, where $$N$$ represents the number of beads. Participants emphasize the importance of understanding the variables involved, particularly $$p$$ and $$\kappa$$, and the necessity of using initial conditions to determine the amplitudes accurately. The conversation highlights the complexity of the problem, particularly in formulating the equations correctly and understanding the relationship between modes and amplitudes.
PREREQUISITESStudents of physics, particularly those studying wave mechanics, as well as educators and researchers interested in the dynamics of coupled oscillators and their applications in real-world systems.
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##?