Physics/A/String vibration

“Among vibrating bodies there are none that occupy a more prominent position than Stretched Strings. From the earliest times they have been employed for musical purposes ... . To the mathematician they must always possess a peculiar interest as a battle-ﬁeld on which were fought out the controversies of D’Alembert, Euler, Bernoulli, and Lagrange relating to the nature of the solutions of partial diﬀerential equations. To the student of Acoustics they are doubly important." --"Lord Rayleigh:[32, Vol. I, Chap. VI].



https://openstax.org/books/university-physics-volume-1/pages/5-5-newtons-third-law#:~:text=Newton's%20third%20law.-,Newton's%20Third%20Law%20of%20Motion,the%20force%20that%20it%20exerts.&text=F%20%E2%86%92%20AB%20%3D%20%E2%88%92%20F%20%E2%86%92%20BA%20

openstax.org/books/university-physics-volume-1/pages/5-5-newtons-third-law

Intro
This discussion was motivated by the fact that the flow of energy associated with Poynting vector does not seem physical. This is not the only case of a conservation law seems to leave unanswered questions about the details. Here we consider the conservation of energy in a stretched string. In this case, there is a unique answer to the question, "where is the potential energy". What is odd is how difficult that question is to answer. Rowland published what seems to be acceptable but approximate solutions for some special cases.

DEBUG
This discussion focuses on making a subset of those approximate solutions more accessible to novice lovers of physics.

It is worth pointing out that stax exchange seems to be wrong on this topic: https://physics.stackexchange.com/questions/414521/does-a-vibrating-string-produce-changes-in-tension-in-the-tring

Three tricks:
XXX

First derivative revisited
Most readers have probably seen a definition of the derivative that can be found at:
 * Wikipedia:Simple:Special:Permalink/7230410.

$$f^\prime(x+\ell/2)\approx \frac{f(x+\ell) - f(x)}{\ell}$$

$$f^\prime(x)\approx \frac{f(x+\ell/ 2) - f(x- \ell / 2)}{\ell}$$

Second derivative
Finite difference coefficient

$$\begin{align} f^{\prime\prime}(x) &\approx \frac{f^\prime(x+\ell/ 2) - f^\prime(x- \ell / 2)}{\ell}  \\ & \approx\frac{f(x+\ell)-2f(x) + f(x-\ell)}{\ell^2}\end{align}$$

Let "1" be a small parameter

 * Let $$\ell=1$$

introducing kappa as spring constant
$$\vec f= m\ddot\vec r=-\kappa_s (r-a) \hat r$$

$$\vec r=\vec\ell+\xi\hat x + \eta\hat y$$

$$\vec r =x\hat x + y\hat y =(1+\xi)\hat x + \eta\hat y$$

$$r \equiv\left|\vec r\right| =\sqrt{(1+\xi)^2 + \eta^2}$$

$$\hat r = \frac \vec r r =\frac{(1+\xi)\hat x + \eta\hat y} {\sqrt{(1+\xi)^2 + \eta^2}}$$

$$\vec f=-\kappa_s (r-a) \hat r =-\kappa_s \left(\vec r-a\hat r\right)$$

Used https://www.symbolab.com/

$$\hat r \cdot \hat x =\frac{1+\xi} {\sqrt{(1+\xi)^2 + \eta^2}}=1-\tfrac 1 2 \eta^2 + \mathcal O \psi^3$$

$$\hat r \cdot \hat y =\frac{\eta} {\sqrt{(1+\xi)^2 + \eta^2}}=\eta\left(1-\xi-\xi^2-\tfrac 1 2 \eta^2+\mathcal O \psi^3\right)$$

$$\frac{\vec f}{-\kappa_s}=(1+\xi)\hat x + \eta\hat y -a\left(1-\tfrac 1 2 \eta^2 +\mathcal O\psi^3\right)\hat x -a\eta\left(1-\xi-\xi^2-\tfrac 1 2 \eta^2 +\mathcal\psi^3\right)\hat y$$

3


$$\begin{array} {lllll} & &\quad\mathcal O\psi^0 &\quad\mathcal O \psi^1 & \quad\mathcal O \psi^2\\ f_x&=&-\kappa_s(1-a)       &-\kappa_s\xi           &-\tfrac 1 2 \kappa_s a \eta^2  &+\quad\mathcal O \psi^3 &+\ldots \\ f_y&=&                     &-\kappa_s(1-a)\eta     & -\kappa_sa\eta\xi    &-\kappa_sa\left(\eta\xi^2+\xi\eta^2\right) &+\ldots \end{array}$$

algebra for 3
x1: $$\frac{f_x}{-\kappa_s}=$$ $$\left(\vec r-a\hat r\right)\cdot\hat x$$ equals $$(1+\xi) $$ minus $$ a\left(1-\tfrac 1 2 \eta^2 + \mathcal O z^3\right)$$

x2: $$\frac{f_x}{-\kappa_s}= (1+\xi) - a\left(1-\tfrac 1 2 \eta^2 + \mathcal O z^3\right)$$

x3: $$\frac{f_x}{-\kappa_s}= (1+\xi) - a\left(1-\tfrac 1 2 \eta^2 + \mathcal O z^3\right)=(1-a) + \xi + \tfrac 1 2 \eta^2a +\ldots$$

x4: $$f_x =-\kappa_s(1-a) -\kappa_s\xi -\tfrac 1 2 \kappa_s a \eta^2+\ldots$$

y1: $$\frac{f_y}{-\kappa_s}=$$ $$\left(\vec r-a\hat r\right)\cdot\hat y$$ equals $$\eta$$ minus $$ a\eta\left(1-\xi-\xi^2-\tfrac 1 2 \eta^2+\mathcal O z^3\right)$$

y2: $$\frac{f_y}{-\kappa_s}= \eta- a\eta\left(1-\xi-\xi^2-\tfrac 1 2 \eta^2+\mathcal O z^3\right)$$

y3: $$\frac{f_y}{-\kappa_s}= (1-a)\eta +a\xi\eta+a\xi^2\eta+a\tfrac 1 2 \eta^3+\ldots$$

y4 $$f_y= -\kappa_s(1-a)\eta - \kappa_sa\xi\eta+ \kappa_sa\left(\xi^2\eta+\tfrac 1 2 \eta^3\right)\ldots$$

Constructing the wave equation
Here we adopt the convention that $$\ell=1$$ unit of length. We label masses with the variable $$X$$ that represents each mass by an integer $$(X=0,\pm 1, \pm 2,\ldots)$$. To obtain a wave equation we focus on the three consecutive integers, $$(A,B,C)$$. When the string is at equilibrium (i.e., zero wave amplitude), we can also use non-integral values of $$X$$ to form a coordinate system that labels points in space between the masses, as shown in the top of Figure 2.

For non-zero wave amplitude, each mass can move away from its equilibrium point by $$\xi$$ in the x direction and $$\eta$$ in the y-direction (Rowland, et al, use the symbol $$\zeta$$ to describe motion in the other transverse direction.)

Defining $$\psi$$ to be the vector associated with this displacement, we have,

$$\vec\psi_B =\xi_B\hat x + \eta_B\hat y $$

We denote

$$m \ddot\vec\psi \equiv m\ddot\xi\hat x + m\ddot\eta\hat y $$

The convention used by OpenStax Physics is that $$\vec F_{AB} = \vec F_\text{by-A-on-B}$$ refers to the force on object A by object B. To keep the the notation in Figure 2 compact, we define the displacement vector from A to B as:

$$\vec\psi_{AB}=\psi_B-\psi_A\approx \left.\frac{d\vec\psi}{dX}\right|_{X=A+1/2} (\text{if }B=A+1)$$

$$$$ $$$$   $$$$   $$$$

$$\begin{array}{rcl} \vec f_{AB} &=-\kappa_s \left[\vec\psi^{\,\prime}_{X-1/2}-a\hat\psi^\prime_{X-1/2} \right] &=\vec f\bigl(\vec\psi^{\,\prime}_{X-1/2}\bigr)\\

\vec f_{CB} &=+\kappa_s \left[\vec\psi^{\,\prime}_{X+1/2}-a\hat\psi^\prime_{X+1/2} \right] &=\vec f\bigl(\vec\psi^{\,\prime}_{X+1/2}\bigr)\\

\end{array}$$

$$ \Sigma\vec f_B =\vec f_{AB} - \vec f_{BC} = \vec f\bigl(\vec\psi^{\,\prime}_{X-1/2}\bigr)-\vec f\bigl(\vec\psi^{\,\prime}_{X+1/2}\bigr) =-\vec f^{\,\prime}\bigl(\vec\psi^{\,\prime}_{X}\bigr)$$

Wave equation
$$\mathbb F(\xi^\prime,\eta^\prime)=\frac{d}{dX} $$ $$\,\mathbb F$$

$$\begin{array}{rccc} &-f_x(\xi^\prime,\eta^\prime)= &+\kappa_s(1-a) &+\kappa_s\xi^\prime &+\tfrac 1 2 \kappa_s a \left(\eta^\prime\right)^2 +\ldots \\ \Sigma F_x= &-f^\prime_x(\xi^\prime,\eta^\prime)= & 0 &+\kappa_s\xi^{\prime\prime} &+\kappa_s a \eta^\prime\eta^{\prime\prime} +\ldots \\&&&&\\ &-f_y(\xi^\prime,\eta^\prime)= &+(1-a)\kappa_s\eta & -\kappa_sa\eta\xi &+\ldots \\ \Sigma F_y= &-f^\prime_y(\xi^\prime,\eta^\prime)= & (1-a)\kappa_s\eta^{\prime\prime}&+\underbrace{\kappa_sa\left(\eta^\prime\xi+\eta\xi^\prime\right)}_\text{drop}&+\ldots \end{array}$$

Leave as exercise for the readers to verify the Table (with both compact and PDE forms). And also to realte X to x.

Equation for kappas and a
Define $$\kappa_T=(1-a)\kappa_s\le\kappa_s$$ =>  $$\kappa_T=\kappa_s-a\kappa_s$$  => $$a\kappa_s=\kappa_s-\kappa_T$$

$$m\ddot\xi=\kappa_s\xi^{\prime\prime}+a\kappa_s\eta^\prime\eta^{\prime\prime}$$

$$m\ddot\xi=\kappa_s\xi^{\prime\prime}+ (\kappa_s-\kappa_T)\eta^\prime\eta^{\prime\prime}$$

$$m\ddot\eta = \kappa_T\eta^{\prime\prime}$$

UNDER CONSTRUCTION: Allowing ℓ≠1
The dimensional analysis conventions introduced in OpenStax University Physics permit us to show that and  are equivalent to results obtained in reference

DEBUG3
From from reference RowlandEJP:

$$\ddot\xi \approx c_L^2\xi^{\prime\prime} + \left(c_L^2-c_T^2\right)\eta^\prime\eta^{\prime\prime}$$

$$\ddot\eta = c_T^2\eta^{\prime\prime} + \left(c_L^2-c_T^2\right)\left( \underbrace{\tfrac 3 2 {\eta^\prime}^2\eta^{\prime\prime}}_{\text{new}} + \underbrace{\xi^\prime\eta^{\prime\prime} + \eta^\prime\xi^{\prime\prime} }_{\text{dropped}}

\right)$$

Symbolic computation probably renders this exercise unnecessary, but one way to "guess" the wave equation when $$\ell\ne1$$ is to use dimensional analysis, though if you want certainty it might be better to repeat all the steps with the extra term included. Using

Product rule
From Phasor_algebra:

$$\Re e\left(\psi_1\right)\Re e\left(\psi_2\right) =\tfrac 1 2 |\psi_1|\,|\psi_2|\cos(\Phi_1-\Phi_2)\;+ \;\tfrac 1 2 |\psi_1|\,|\psi_2|\cos(\Phi_1+\Phi_2)$$

$$$$ $$$$  $$$$   $$$$  $$$$  $$$$     $$$$  -

Variables
$$\vec r = x \hat x + y\hat y$$ * position with unit vectors

$$X_j$$ * coordinate variable parallel to string

Equilibrium

$$K\;\text{ not }\;k$$ * spring constant

$$a$$ *relaxed spring length

$$\ell$$ * equilibrium spring length (no wave present)

$$\tau$$ * equilibrium tension in string

$$\rho=m/\ell$$ * linear mass density at equilibrium

$$ss$$ *

$$c_T=\omega_T/k_T=\sqrt{\tau/\rho}$$ * transverse wave speed

$$c_L$$ *longitudinal wave speed

Perturbation

$$\xi$$ * x-deviation from equilibrium (longitudinal)

$$\eta$$ * y-deviation from equilibrium (transverse)

$$\ddot\xi=\partial^2\xi/\partial t^2,\quad \eta'=\partial\xi/\partial x,\quad \dot\xi'=\partial^2 \xi /\partial x\partial t, \quad...$$

$$\epsilon = \kappa + u$$ * energy density (kinetic+potential) kappa is mine. I don't like their e or k_e.

Avoid

$$\zeta$$ * zeta is used for third dimension (polarized waves)

$$ss$$ *

$$ss$$ *

Appendix
$$\tfrac 1 2 \int \rho\Phi dV_\text{ol}=\dots$$ elsewhere

Temp: