Intuition for normal modes of a beaded string These questions are inspired by the following the paper http://www.soton.ac.uk/~stefano/courses/PHYS2006/chapter7.pdf on 'Normal Modes of a Beaded String'.
Problem Statement
Given a recurrence relation of $-\omega^2A_p=\omega_0^2(A_{p+1}-2A_p+A_{p-1})$ for the displacement amplitude of vibration of the $i^{th}$ bead at frequency $\omega$, we aim to find an expression for the $n^{th}$ mode frequency.
Solution (Taken from the paper)
Suppose that we have already found a mode for the string. If we shift the string, by translational invariance it will be the same. Therefore, if $A_n$ gives a frequency mode $\omega$, the shifted $A_n'$ also gives a frequency mode of $\omega$. We then have that
$$A_n'=A_{n+1}$$
Now let’s look for a translation invariant mode, which reproduces itself when we
do the shift. Since a mode is arbitrary up to an overall scale, this means
$$A_n'=A_{n+1}=hA_n$$
for some constant h, so that the new amplitudes are proportional to the old ones. Applying this relation repeatedly gives
$$A_n=h^nA_0$$
After substituting into the initial recurrence relation we have
$$-\omega^2h^nA_0=\omega_0^2(h^{n-1}A_{0}-2h^nA_0+h^{n+1}A_{0})$$
$$\omega^2=\omega_0^2\left(2-h-\frac{1}{h}\right) \ \ \ \ (\star)$$
This shows that $h$ and $\frac{1}{h}$ give the same normal mode frequency. Conversely, if the frequency $\omega$ is fixed, the amplitudes $A_n$ must be an arbitrary linear combination of the amplitudes for $h$ and $\frac{1}{h}$
$$A_n=\alpha h^n+\beta h^{-n}$$
Letting $h=e^{i\theta}$ we have
$$\omega^2=4\omega_0^2\sin^2(\frac{\theta}{2})$$
My Issues
(1) Since a mode is arbitrary up to an overall scale, this means $A_n'=A_{n+1}=hA_n$. I am not sure what the intuition behind this step actually is? If you shift the beads to left by one why must we scale the amplitude accordingly?
(2)How does $(\star)$ show that $h$ and $\frac{1}{h}$ give the same normal frequency and what does it even mean to say this?
(3)Why does fixing the frequency $\omega$ mean that the amplitudes must be a linear combination of $h$?
 A: *

*The definition of two equivalent normal modes is given in the caption of Figure 7.3: A normal mode remains "the same" if all amplitudes are multiplied by a (complex) constant. You can get an intuition as follows. If one bead configuration exactly coincides with another one after some time evolution (scaling by $h=e^{i \theta}$) or after zooming out vertically (scaling by $h$ real), then they are considered equivalent, or "the same"

*The equation is invariant if you exchange $h\leftrightarrow1/h$. It means that if $A_n = h^n A_0$ satisfies the recurrence equation, then $A_n = h^{-n} A_0$ will too. An intuitive way to see this property is by noting that if a translation (to the left) results in an equivalent mode $A_{n+1} = h A_n$, then a translation to the right $A'_n = A_{n-1}$ would also give "the same" normal mode but it holds $A'_n = A_{n-1} = A_{n}/h$ . Moreover, to obtain $\omega>0$ it is required that $h=e^{i \theta}$. This is not a convenience as said in your link, it is a must, because any other real or complex number $h$ will lead to a complex $\omega$, which is not correct

*The amplitudes are a linear combination of $h^n$ and $h^{-n}$, not $h$. This is because the wave equation is linear. So, if you found two solutions for the same mode frequency $A_p$ and $B_p$, then their linear combination $\alpha A_p + \beta B_p$ will also be a solution. It is confusing for the author to call $A_n$ both the translation invariant modes and the general superposition, because $\alpha h^n + \beta h^{-n}$ is not translation invariant $h A_n = \alpha h^{n+1} + \beta h^{-n+1} \neq \alpha h^{n+1} + \beta h^{-n-1} = A_{n+1}$. For $h=e^{i \theta}$ this means that the two normal modes with translation invariance $e^{i n \theta}$ and $e^{-i n \theta}$ are two traveling waves in opposite directions. For a finite number of beads this means the solutions are standing waves ($\alpha = -\beta$) with a node at the origin and another node in the last bead

Hope it helps
