Physics problem: calculating the tension of ropes tying together timbers of a raft

This problem comes from my usual source of inspiration, the book by Savchenko et al., from the section on floating bodies and Archimedes' force. I spent a considerable amount of time figuring this one out, mostly because the answers in the edition I have and in the newer edition do not match. Another reason I could not get the correct answer is that I did not think that we should account that there are actually two ropes one at the front and the second is at the back of the raft.

Here is the problem statement: Consider two rows of wooden timbers tied together into a raft, as shown in the figure. Half of the top row of the timbers is above the water surface. Determine the minimum tension of the two ropes holding the timbers together, if the mass of each timber is $m$. The raft is very wide.

Solution

Let's denote water density as $\rho$ and the volume of each timber as $V$. Let's refer to forces, acting between the timbers and between a tibmer and a rope as $N$ and $F$ respectively. See the figure below (the Latex code for the fiigure can be found here).

We can express the equillibrium of a bottom and of a top timbers separately using vertical components of forces acting on them:

$$\tag{1} \begin{cases} 2F - N\sqrt{3} - mg + \rho V g = 0 \\ -2F + N\sqrt{3} - mg + \frac{1}{2} \rho V g = 0 \end{cases}\\ \Rightarrow -6F + 3N\sqrt{3} - mg = 0$$

Note that we have $2F$ in each of the equations above for the force the ropes push timber downwards or upwards because we have two ropes one at the front and another at the back of the raft.

Let's consider vertical components of the forces acting on the upper part of one of the ropes. Assuming that we have around $n$ timbers in each row and that the mass of the rope is negligible, we get:

$$\begin{equation}\tag{2} n \cdot F = 2 T\cos\alpha \end{equation}$$

Where $\alpha$ - is the angle of the tension force with the vertical at the left and right edges of the raft. The tension arises from the bottom part of the rope pulling the upper part down. Since the $n$ is large, according to the problem statement, we can deduce:

$$\begin{equation}\tag{3} F = 2 \frac{T\cos\alpha}{n} \rightarrow 0 \end{equation}$$

So the equation (1) simplifies to the following: $$\begin{equation}\tag{4} 3 N\sqrt{3} - mg = 0 \end{equation}$$

Now let's consider the right part of the timbers as a system and write down the equillibrium condition for horizontal components of the forces acting on it. I highlighted in red the timbers in the selected system as well as relevant forces:

$$\begin{equation}\tag{5} \frac{N}{2} + P_1 + P_2 - 4T = 0\\ \Rightarrow 4T = \frac{N}{2} + P_1 + P_2 \end{equation}$$

where $P_1 \geq 0$, and $P_2 \geq 0$ - horizontal forces exerted by the neighbour timbers from the top and the bottom rows respectively onto the above selected system of the timbers.

Therefore,

$$\begin{equation}\tag{6} 4 T \geq \frac{N}{2} \Rightarrow T_{\min} = \frac{N}{8} \Rightarrow N = 8 T_{\min} \end{equation}$$

Then substituting the above expression for $N$ into equation (4) we get:

$$\begin{equation}\tag{7} 24 T_{\min}\sqrt{3} - mg = 0 \\ \Rightarrow \boxed{T_{\min} = \frac{mg}{24\sqrt{3}} = \frac{mg\sqrt{3}}{72}} \end{equation}$$

The above result is the same as in the newer edition of the book. At this point, I am not sure how the answer in the old edition of the book was obtained, because if we forget about one rope, the $T_{\min}'$ would be two times greater then the correct one, but in the older edition the tension is $4 T_{\min}$ for some reason....

Stability of a rectangular box floating on a surface of a liquid

This problem, as many of the previous discussed here, is coming from the Savchenko et al book. It took me around one month to wrap my head around this one. I have to admit that I improved (or refreshed) my knowledge of Archimedes' force thanks to this one. Specifically, this problem highlights the fact that Archimedes' force is applied to the center of mass of the water replaced by a floating object. Therefore, even if the object's density is uniform, it can happen that the torques of the gravity force and the Archimedes' force are not compensated even though the modules of the forces are equal. This fact makes more interesting the question about stability of a floating object in a liquid.

Problem statement

Given a rectangular body of length $a$, width $a$, and height $b$. Its density is $\rho$ and it floats in a liquid of density $\rho_0$. Determine the ratio $\frac{a}{b}$ for which the object's equillibrium state is stable. It is assumed that the side $b$ of the box is vertical in the equillibrium position.

Discussion

It is obvious that the object's equllibrium is stable with respect to vertical displacements, because if we push it down, the Archimedes' force increases and pushes it back. Similarly, if we pull the object up, the Archimedes' force becomes smaller and the gravity pushes it back to its initial equillibrium position where the gravity force is compensated by the Archimedes' force. Therefore, further down we consider only stability of the object with respect to rotations.

We won't consider the case with the box completely submerged into the liquid as in this case the center of mass of the box coincides with the center of mass of the liquid pushed out by the box, due to the uniform density of the box, and therefore the net torque (exerted by the gravity and Archimedes' forces) will always be zero. I guess this case is not stable with respect to rotations, as the box won't be restored to its original vertical position. Zero gravity, like in the outer space, all orientations are equivalent.

Here we can speculate a bit. Intuitively we would expect that the larger base (i.e. greater $a$) with respect to the height $b$ would give us a more stable equillibrium. Hence we can expect the result to be in a form $\frac{a}{b} > r$ or $\frac{a}{b} \geq r$.

Let's denote the height of submerged part of the box as $h$ and the pivot point as $O$. The pivot point is the intersection of the vertical axis of symmetry of the box and the horizontal plane at the level of the surface of the liquid (the updated level after the box is submerged), see the figure below. We consider rotations at a small angle $\varphi$ around this pivot point $O$.

The latex source for the plot can be found here.

There are two ways to find out if the vertical upright orientation of the box is stable: using the change in the height of the center of mass of the system liquid and box, i.e., if the above rotations lower the center of mass of the system then its vertical orientation is not stable, otherwise it is stbale, and using the net torque exerted by the gravity and Archimedes' forces, i.e., in the stable case the torque would act in the direction opposite to the direction of rotation,otherwise it is not stable. We'll use both methods and compare the results.

But first let's prepare a few useful expressions. If we denote as $A$ the center of mass of the liquid pushed out by the box (this is also the point where the Archimedes' force is applied to the box). Let's determine its coordinates $\left(x_{A\varphi}, y_{A\varphi}\right)$ when the box is rotated, in the system of coordinates with the center at the point $O$ and the $y$-axis pointing downward and the $x$-axis pointing to the right. It is easy to see that the mass ($m_A=\rho_0 a^2 h$) of the liquid pushed out by the box does not change during the rotation as well as the module of the Archimede's force $F_A=m_A g$. By definition, expressions for the coordinates of the center of mass of the liquid pushed out by the box can be written as:

$$\begin{cases} m_A x_{A\varphi} = m_A x_{A'\varphi} + m_L x_L + m_R x_R \\ m_A y_{A\varphi} = m_A y_{A'\varphi} + m_L y_L + m_R y_R \end{cases}$$

where

$$\begin{align} m_L = -m_R\\ x_L = -x_R\\ y_L = -y_R\\ \end{align}$$

and the $A'\varphi$ is the center of mass of the fictious liquid volume when it is rotated from the upright position as a solid body, $m_L$ and $m_R$ are the masses of the prisms (triangles in 2D, see the above figure) to be removed from and added to the fictious liquid body respectively to get the real shape of the liquid pushed out by the box during rotation. $\left(x_R, y_R\right)$ and $\left(x_L, y_L\right)$ are the centers of mass of the prisms of masses $m_R$ and $m_L$ respectively (red and blue triangles in the figure, the blue one has negative mass and the red one has positive mass). Since $m_L$ is the mass to be removed, it is negative in the equations to avoid confusion with the signs. We use these triangles (prismes in 3D) to facilitate the calculation of the position of the center of mass of the liquid pushed out by the box $A\varphi$. It is easy to compute coordinates of the center of mass of the fictious rectangular volume $B_1B_2RL$ and then to find coordinates of $A\varphi$ (of the complex shape of the liquid pushed out by the rotated box) we compute the center of mass of a system consisting of $B_1B_2RL$, $m_R$ and the negative mass $m_L$. The center of mass of such a system is $A\varphi$. If we substitude the relations between $m_L$ and $m_R$ into the equations for $x_{A\varphi}$, $y_{A\varphi}$ we get the following expressions:

$$\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \end{cases}$$

It is easy to see that

$$\begin{align} \frac{m_R}{m_A} & = \frac{\rho_0 \frac{1}{2}\cdot \frac{1}{2} a \cdot \frac{1}{2} a \tan\varphi\cdot a}{\rho_0 a^2 h} = \ \frac{a}{8h}\tan\varphi \approx \frac{a}{8h}\varphi \\ x_{A'\varphi} & = -\frac{h}{2}\sin\phi \approx -\frac{h}{2}\phi\\ y_{A'\varphi} & = \frac{h}{2}\cos\phi \end{align}$$

We keep exact values for the $y$ coordinates here as the changes in $y$ due to the rotation are on the order of $\sim\varphi^2$ and the variations of $x$ coordinates are on the order of $\sim\varphi$.

From the geometry and the properties of the centre of mass of a triangle we can determine $x_R$ and $y_R$: $$\begin{cases} x_R & = \frac{a}{2\cos\varphi}\cdot\frac{1}{2} + \frac{1}{3} \frac{a}{2\cos\varphi} \cos2\varphi = \frac{a}{4\cos\varphi}\left(1 + \frac{\cos2\varphi}{3}\right)\\ y_R & = \frac{1}{3}\left(\frac{a}{2\cdot2\cos\varphi}\right)\sin2\varphi = \frac{1}{6}a\sin\varphi \end{cases}$$

For small angles $\varphi$ (which is the case in the problem at hand), we get the following expressions for $x_R$ and $y_R$:

$$\begin{cases} x_R & \approx \frac{a}{3} \\ y_R & \approx \frac{a}{6}\varphi \end{cases}$$

Using the above relation we can express $\left(x_{A\varphi}, y_{A\varphi}\right)$ as functions of $h$, $a$, and $\varphi$:

$$\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \approx -\frac{h}{2}\phi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{3} = \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right)\\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \approx \frac{h}{2}\cos\varphi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{6}\varphi = \ \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 \end{cases}$$

We did not replace $\cos\varphi$ with one for the $y$ coordinate since the accuracy of the terms is $\varphi ^ 2$.

Let's note the coordinates of the center of mass of the pushed out liquid for $\varphi=0$:

$$\begin{cases} x_{A0} = 0\\ y_{A0} = \frac{h}{2} \end{cases}$$

It is easy to derive how the coordinates of the center of mass of the box ($C$) change during rotation, as it is always in the middle of the box:

$$\begin{cases} x_{C\varphi} = \left( \frac{b}{2} - h \right)\sin\varphi\\ y_{C\varphi} = \left( h - \frac{b}{2} \right)\cos\varphi \end{cases}$$

Finally, let's express $h$ using equillibrium between the Archimedes' ($F_A$) and the gravity ($F_C = m_C g$) forces:

$$\begin{align} F_C-F_A=0 \Rightarrow \rho a^2 b - \rho_0 a^2 h = 0 \Rightarrow h=\frac{\rho}{\rho_0}b \end{align}$$

Method 1: analysis of torques

Without the loss of generality we can consider clockwise rotation as shown in the plot. Torques acting in the counter-clockwise direction are positive and torques acting in the clockwise direction are negative. With this convention, for stability it is required that the net torque ($\tau$) should act in the direction opposite to the rotation (i.e. anti-clockwise). Therefore the condition for stability is $\tau > 0$.

Let's consider two cases and impose the stability condition:

a) the center of mass is below the liquid surface $h\geq b/2$

If $x_{A\varphi} \geq 0$, then $\tau = F_C |x_{C\varphi}| + F_A |x_{A\varphi}|$, which is always positive. Both $\tau_C$ and $\tau_A$ act to stabilize the box against rotation.

If $x_{A\varphi} < 0$, then stability condition is:

$$\begin{align} \tau = F_C |x_{C\varphi}| - F_A |x_{A\varphi}| > 0 \end{align}$$

Since $F_C = F_A$ (as well as $m_A = m_C$), for stability we need the following:

$$\begin{align} |x_{C\varphi}| - |x_{A\varphi}| > 0 &\Rightarrow - \left( \frac{b}{2} - h \right)\varphi - (-1)\cdot\frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -\frac{b}{2} + h + \frac{1}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -2bh + 4h^2 + \frac{a^2}{3} - 2 h^2 > 0 \\ & \Rightarrow -6bh + a^2 + 6 h^2 > 0 \\ & \Rightarrow -6b^2\frac{\rho}{\rho_0} + a^2 + 6 \left(\frac{\rho}{\rho_0} b\right)^2 > 0 \\ & \Rightarrow \left(\frac{a}{b}\right)^2 > 6\frac{\rho}{\rho_0} - 6 \left(\frac{\rho}{\rho_0} \right)^2 \\ & \Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}$$

b) the center of mass is above the liquid surface $h < b/2$

In this case the gravity force acts to destabilize the box with its torque acting in the direction of rotation. So the stability in this case is possible when $|x_{A\varphi}| > |x_{C\varphi}|$ and $x_{A\varphi} \cdot x_{C\varphi} > 0$ (i.e. A and C are from the same side from the pivot point $O$) which is equivalent to:

$$\begin{align} \tau = -F_C |x_{C\varphi}| + F_A |x_{A\varphi}| > 0 &\Rightarrow \ - (-1) \cdot \left(h - \frac{b}{2}\right)\varphi + \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ &\Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}$$

Let's check if the conditions we imposed do actually make sense.

$$\begin{align} h < b/2 \Rightarrow \rho/\rho_0 < 1/2 \Rightarrow a/b > \sqrt{3/2} \Rightarrow \frac{a^2}{3} - \frac{b^2}{2} > 0 \Rightarrow x_{A\varphi} > 0 \end{align}$$

remembering that we consider only $\varphi > 0$.

Method 2: minimum of the potential energy of the system

I like this method more as it requires less logic branching. Let's denote as $y$ the $y$-coordinate of the center of mass of the system liquid and box. Since the positive direction of the axis is downwards, the change $\Delta y < 0$ during the rotation of the box would mean that the center of mass of the system is lifted and therefore the system's potential energy would be increasing and the system itself will tend to minimize its potential energy, therefore $\Delta y < 0$ is our stability condition. By the definition of the center of mass we have:

$$\begin{align} (M - m_A + m_C) y = m_C y_{C\varphi} - m_A y_{A\varphi} + M y_0 \end{align}$$

where $M$ is the mass the liquid would have when the immersed part of the box is replaced by the liquid. $y_0$ is the $y$-coordinate of the center of mass of $M$.

Let's examine the change in the $y$-coordinate of the center of mass of the system during rotation:

$$\begin{align} \Delta y = \frac{1}{M - m_A + m_C}\left ( m_C \Delta y_{C} - m_A \Delta y_{A} \right) \Rightarrow \Delta y \propto m_C \Delta y_{C} - m_A \Delta y_{A} \end{align}$$

Noting that $m_C = m_A$, we get:

$$\begin{align} \Delta y \propto \Delta y_{C} - \Delta y_{A} \end{align}$$

Therefore the condition for the system stability is equivalent to:

$$\begin{align*} \Delta y < 0 \Rightarrow \Delta y_{C} - \Delta y_{A} < 0 \end{align*}$$

From the relations prepared above we can express $\Delta y_{C}$ and $\Delta y_{A}$ as follows:

$$\begin{align*} \Delta y_{A} & = y_{A\varphi} - y_{A0} = \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 - \frac{h}{2} = \frac{a^2}{24h}\varphi^2 - \frac{h}{4} \varphi^2 \\ \Delta y_{C} & = y_{C\varphi} - y_{C0} = \left(h - \frac{b}{2}\right)\left(\cos\varphi - 1 \right) \approx -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} \end{align*}$$

Now plugging the above expressions for $\Delta y_{C}$ and $\Delta y_{A}$ into the stability inequality, we get:

$$\begin{align*} & -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} - \frac{a^2}{24h}\varphi^2 + \frac{h}{4} \varphi^2 < 0 \Rightarrow \\ & \Rightarrow \frac{b}{4}\varphi^2 - \frac{h}{4} \varphi^2 - \frac{a^2}{24h}\varphi^2 < 0 \Rightarrow \\ & \Rightarrow 6 (b - h) h - a^2 < 0 \Rightarrow 6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right) b^2 - a^2 < 0 \Rightarrow \\ & \Rightarrow \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{align*}$$

In conclusion, we are happy that the minimum potential energy and the torque methods give the same result.

$$\begin{equation*} \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{equation*}$$

How to set a correct keyboard layout in i3 window manager

Recently I rediscovered for myself the i3 desktop environment, got tired of windows spread here and there, but I am using an Apple aluminum keyboard with french letters and numeric keypad (just used to it), and I just hated that the layout did not correspond to what is marked on the buttons out of the box. To try to find a solution I googled and tried many things. I eventually ended up using

localectl

to set the correct keyboard layout. So below are the steps which might help you out:

	# check the current layout settings
	localectl status
	# System Locale: LANG=en_CA.UTF-8
	# VC Keymap: ca-multix
	# X11 Layout: ca
	# X11 Model: apple
	# X11 Variant: multix
	# the command to set keyboard layout is
	# localectl set-x11-keymap LAYOUT [MODEL [VARIANT [OPTIONS]]]
	# so we need to figure out what to put
	# for LAYOUT MODEL and VARIANT for my keyboard
	# to list all available layouts
	localectl list-x11-keymap-layouts
	# from the output of the above command
	# I figured that LAYOUT=ca in my case
	# now there is a command to list available models
	# since I was interested in apple, I added a grep
	localectl list-x11-keymap-models | grep -i apple
	# apple
	# apple_laptop
	# applealu_ansi
	# applealu_iso
	# applealu_jis
	# from the above I figured that for me MODEL=apple
	# would be a good first guess.
	# Now we need to figure out the value for the VARIANT
	# for LAYOUT=ca
	localectl list-x11-keymap-variants ca
	# eng
	# fr-dvorak
	# fr-legacy
	# ike
	# multix
	# I tried using fr-dvorak and it was a disaster,
	# it completely messed up my keyboard,
	# using trial and error approach I realized that VARIANT=multix
	# for my Apple keyboard with numeric keypad.
	# So, finally I have all the ingredients
	# to set my layout as follows
	localectl set-x11-keymap ca apple multix
	# and use cmd+shift+e to exit i3 and then
	# login again and enjoy a nicely working keyboard in i3))
	## UPDATE:
	# I needed the left option/alt key to behave as altgr, it is just easier to type the curly and
	# square brackets on the apple keyboard, to achieve this you would need to add the
	# following argument to the localectl command 'lv3:alt_switch' as below:
	localectl set-x11-keymap ca \
	apple \
	multix
	# I used the following to swap left and right alts
	setxkbmap -option lv3:lalt_switch -option lv3:ralt_alt
	# to save the above setxkbmap options for switching alts, modify /etc/default/keyboard as follows:
	cat /etc/default/keyboard
	'''
	# KEYBOARD CONFIGURATION FILE
	# Consult the keyboard(5) manual page.
	XKBMODEL="apple"
	XKBLAYOUT="ca"
	XKBVARIANT="multix"
	XKBOPTIONS="lv3:lalt_switch,lv3:ralt_alt"
	BACKSPACE="guess"
	'''
	# To persist the switch of the left-right alts in i3
	# I had to add the following to the ~/.config/i3/config
	exec --no-startup-id setxkbmap -option lv3:lalt_switch -option lv3:ralt_alt

view raw localectl-examples.sh hosted with ❤ by GitHub

Sound waves in an ocean column

This problem is about infrasound oscillations in an ocean column, it is taken from Savchenko et. al. (1981). And the question is as follows.

What ocean depth would allow development of physiologically dangerous infrasound oscillations at a linear frequency of $\nu=7\;{\rm Hz}$.

My approach is to examine modes allowed by the ocean column assuming that we have a displacement node at the ocean floor and a displacement antinode at the ocean surface. That is when a forced oscillation is excited by some external forcing (e.g wind) only the modes corresponding to the natural (eigen) frequencies would be reinforced by a resonance. I assume that we have an antinode at the surface as the surface is much less constrained than the deeper water layers.

Using a general equation for the modes (also known as standing waves, which could be combined to approximate any solution to the wave equation, provided the standing wave components satisfy imposed boundary conditions). The displacement $\Psi_m(z, t)$ in such a standing wave is expressed as follows, assuming $z=0$ at the bottom of the ocean.

$$\begin{equation} \Psi_m(z, t) = A_m \sin \left( k_m z \right) \sin\left(\omega_m t + \alpha_m \right) \end{equation}$$

Where $k_m=\frac{\omega_m}{c}=\frac{2\pi\nu_m}{c}$ and $c$ - is the speed of sound in the ocean. It is interesting to note here that all the points in the standing wave oscillate at the same frequency $\omega_m$.

These $\Psi_m(z, t)$ automatically satisfy the node condition at the ocean floor. Now we are interested in the frequencies $\nu_m=\nu=7\; {\rm Hz}$ and impose the antinode condition at the ocean surface, i.e. $z=H_n$:

$$\begin{equation} \sin \left( k_m H_n \right) = 1 \Rightarrow k_m H_n = \frac{\pi}{2} + n \pi \end{equation}$$

where $n = 0,1,2,3,...$.

From the above we can determine $H_n$, by setting $\nu_m = \nu$:

$$\begin{split} k_m H_n = \frac{\pi}{2} + n \pi & \Rightarrow \\ \frac{2\pi\nu_m}{c} H_n = \frac{\pi}{2} + n \pi & \Rightarrow \\ H_n=\frac{c}{4 \nu} \left(2 n + 1 \right) \end{split}$$

The above results approximately (assuming $c\approx 1500 \; {\rm m/s}$) in the following values for the ocean depth

$$\begin{split} & H_0 = 50\;{\rm m} \\ & H_1 = 150\;{\rm m} \\ & H_2 = 250\;{\rm m} \\ & H_3 = 350\;{\rm m} \\ & H_4 = 450\;{\rm m} \\ & ... \end{split}$$

Which looks ok to me, but unfortunately this is not the answer given in the book, and it really annoyed me for some time. And I am still not sure why but the authors advance that $H_n = \frac{c}{4 \nu} \left(4 n + 1 \right)$ instead of the $H_n = \frac{c}{4 \nu} \left(2 n + 1 \right)$ obtained above. It seems like the depths $H_n = \frac{c}{4 \nu} \left(4 n - 1 \right)$ are not suitable for some reason. If someone can explain this to me, that would be great ... But for now I am leaving it here hoping that maybe they simply forgot to consider these solutions.

Wave propagation, reflection, and superposition inside a solid rod

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).