This post is about a problem on wave propagation in a solid rod. I took it from Savchenko et al 1981. Before I continue the discussion, I have to present you the problem statement, which goes as follows.
a. Given a harmonic force acting on a free end of a rod of length $L$. A standing wave is formed due to multiple reflections. Determine locations of nodes (points with 0 stress) in the standing wave of deformations. Determine the amplitude of the external force if the amplitude of the stress in the standing wave in the rod is $\sigma_0$ and the rod cross-section is $S$.
I decided to write up the solution here due to the amount of time I spent figuring it out. Another reason is that it is easy to get an answer close to the correct one by following a wrong solution method. Which might lead someone to think that there is a typo in the book. I know I was considering this possibility myself, even went to check the newer edition, but no luck, the problem answers and statements are identical in both editions. Therefore I continued to break my head over this problem. My mistake was that I somehow considered only two first reflections assuming that would be enough. But when I realized that there are many more relevant reflections I managed to get the correct answer.
I start by considering the stress at the end of the rod where the harmonic force is applied:
\begin{equation} \sigma=\frac{F}{S}=\frac{F_0 \sin \omega t}{S} \end{equation}Let's consider a point in the rod at a distance $x$ from the end of the rod opposite to the one where the external force $F$ is applied (see the figure above). The stress at the point $x$ can be computed as a superposition of deformation stresses after 0, 1, 2, 3, ..., $n$, ... reflections from the rod edges. Let's denote these stress values as $P_0, P_1, ...$ and the speed of the propagation of deformations in the rod as $c$. Then $P_0$ (the stress due to the wave after 0 reflections) at $x$ can be expressed using the time delay it takes for the disturbances to propagate from the end where the force is applied to the point $x$ as follows:
\begin{equation} P_0=\frac{F_0}{S}\sin \omega \left(t - \frac{L-x}{c} \right) = \Im\left( \frac{F_0}{S} e^{i\omega\left(t - \frac{L-x}{c} \right)} \right) = \Im(\hat{P}_0) \end{equation}where $\Im(a + i b) = b$ is the imaginary part function of a complex number. Then for $P_1$ the deformation would travel the distance $L+x$, so the expression for $P_1$ is as follows:
\begin{equation} P_1=-\frac{F_0}{S}\sin \omega \left(t - \frac{L+x}{c} \right) = \Im\left( -\frac{F_0}{S} e^{i\omega\left(t - \frac{L+x}{c} \right)} \right) = \Im(\hat{P}_1) \end{equation}The negative sign appearing in the above equation for $P_1$ is due to the reflection from the rode edges where we consider that compression deformations are transformed into stretching deformations and inversely. Therefore, following the above considerations we can express stresses at point $x$ after 2,3, ... , $n$ reflections from the rod edges.
\begin{equation} \hat{P}_2=\frac{F_0}{S} e^{i\omega\left(t - \frac{3L-x}{c} \right)} \end{equation}\begin{equation} \hat{P}_3=-\frac{F_0}{S} e^{i\omega\left(t - \frac{3L+x}{c} \right)} \end{equation}...
From the above we can group $\hat{P}_n$ terms for even and odd values of $n$.
The sum of $\hat{P}_n$ for even $n$ is computed as:
\begin{split} \hat{P}_{\rm even} & = \frac{F_0}{S} \left( e^{i\omega\left(t - \frac{L-x}{c} \right)} + e^{i\omega\left(t - \frac{3L-x}{c} \right)} + e^{i\omega\left(t - \frac{5L-x}{c} \right)} + ...\right) \\ & = \frac{F_0}{S}e^{i\omega \left( t + \frac{x}{c} \right) }\sum_{k=0}^{+\infty}e^{-\frac{i\omega(2k+1)L}{c}}\\ & = \frac{F_0}{S}e^{i\omega \left( t + \frac{x}{c} \right) }\frac{e^{-\frac{i \omega L}{c}}}{1 - e^{-\frac{2 i \omega L}{c}} } \end{split}Similarly the sum of $\hat{P}_n$ for odd $n$ is computed as:
\begin{equation} \hat{P}_{\rm odd} = -\frac{F_0}{S}e^{i\omega \left( t - \frac{x}{c} \right) }\frac{e^{-\frac{i \omega L}{c}}}{1 - e^{-\frac{2 i \omega L}{c}} } \end{equation}Finally by adding the odd and even terms we get
\begin{split} \hat{P} &= \hat{P}_{\rm odd} + \hat{P}_{\rm even}\\ &=\frac{F_0}{S} \left( e^{i\omega \left( t + \frac{x}{c} \right) } - e^{i\omega \left( t - \frac{x}{c} \right) } \right) \frac{e^{-\frac{i \omega L}{c}}}{1 - e^{-\frac{2 i \omega L}{c}} } \\ &=\frac{F_0}{S}\frac{\sin\frac{\omega x}{c}}{\sin\frac{\omega L}{c}}e^{i \omega t} \end{split}$\hat{P}$ is the deformation stress in the standing wave, and the stress amplitude can be computed as follows:
\begin{equation} A(x)=\left| \hat{P}\; \right|=\frac{F_0}{S} \left| \frac{\sin\frac{\omega x}{c}}{\sin\frac{\omega L}{c}} \right| \end{equation}So the condition for nodes $x_j$ of the standing wave, i.e. $A(x_j) = 0$, is as follows
\begin{equation} \sin\frac{\omega x_j}{c} = 0 \Rightarrow \frac{\omega x_j}{c} = j \pi \Rightarrow x_j = \frac{\pi c}{\omega}j\,,\; j=0,1,2,3,... \end{equation}Using the expression for the wavelength $\lambda = c\cdot T = \frac{2 \pi c}{\omega}$ the expression for $x_j$ becomes:
\begin{equation} \boxed{ x_j = \frac{\lambda}{2}j } \end{equation}To determine the amplitude of the external harmonic force we can use the condition
$$ \max_{x} A(x)=\sigma_0 $$Using the expression for $A(x)$ we get:
\begin{equation} \max_{x} A(x)=\max_{x}\left( \frac{F_0}{S} \left| \frac{\sin\frac{\omega x}{c}}{\sin\frac{\omega L}{c}} \right| \right)=\frac{F_0}{S} \frac{1}{\left|\sin\frac{\omega L}{c}\right|} \end{equation}By comparing the two equations above we get the expression for $F_0$:
\begin{equation} \boxed{ F_0 = \sigma_0 S \left|\,\sin\frac{\omega L}{c}\,\right| } \end{equation}Using the equation above we can plot a resonance curve (see the figure below), i.e. $\frac{\sigma_0 S}{F_0}$ as a function of frequency $\omega$:
\begin{equation} \frac{\sigma_0 S}{F_0} = \frac{1}{\left|\,\sin\frac{\omega L}{c}\,\right|} \end{equation}
b. The last question of the problem is: for which values of the frequency $\omega$ the harmonic oscillations would be sustained in the rod without external forcing?
Let us reformulate the question. Determine frequencies of sine standing waves (also known as modes) that could arise in the rod assuming that the stress at the edges ($x=0$ and $x=L$) is 0. Standing waves with nodes at $x=0$ and $x=L$ would have the whole number of half wavelengths fit in the rod. This condition can be expressed as follows:
\begin{equation} L = n \frac{\lambda}{2} = n\frac{c \pi}{\omega_{0,n}} \Rightarrow\boxed{\omega_{0,n}=\frac{\pi c}{L}n\,,\: n = 1, 2, 3,...} \end{equation}These frequencies $\omega_{0,n}$, also known as natural frequencies or eigenfrequencies of an oscillator $\omega_0$, coincide with the maxima of the resonance curve. The modes are explained very well by R. Feynman in his lecture on modes (https://www.feynmanlectures.caltech.edu/I_49.html#Ch49-F3).
