# 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].

### 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: Figure 1. A mass ${\color {Red}m}$ is attached to a spring of length $r$ . The mass has been displaced from its equilibrium point by $\xi {\hat {x}}+$ direction and by $\eta {\hat {y}}$ direction. Both $\xi$ and $\eta$ are assumed small for a wave in the linear regime. The vector ${\vec {\ell }}$ represents the distance between consecutive lumped masses when the wave amplitude is zero. It is assumed that $\ell >a$ because the string is under tension, and $a$ is the length of the relaxed spring of spring constant $\kappa _{s}$ . In expressing Newton's second law, the acceleration is written as ${\ddot {\vec {r}}}$ using the convention that a dot denotes differentiation with respect to time.

XXX

#### First derivative revisited

Most readers have probably seen a definition of the derivative that can be found at:

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

{\begin{aligned}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{aligned}} #### Let "1" be a small parameter

Let $\ell =1$ ### Calculation of force in Figure 1

#### introducing kappa as spring constant

${\vec {f}}=m{\ddot {\vec {r}}}=-\kappa _{s}(r-a){\hat {r}}$ ${\vec {r}}={\vec {\ell }}+\xi {\hat {x}}+\eta {\hat {y}}$ ${\vec {r}}=x{\hat {x}}+y{\hat {y}}=(1+\xi ){\hat {x}}+\eta {\hat {y}}$ $r\equiv \left|{\vec {r}}\right|={\sqrt {(1+\xi )^{2}+\eta ^{2}}}$ ${\hat {r}}={\frac {\vec {r}}{r}}={\frac {(1+\xi ){\hat {x}}+\eta {\hat {y}}}{\sqrt {(1+\xi )^{2}+\eta ^{2}}}}$ ${\vec {f}}=-\kappa _{s}(r-a){\hat {r}}=-\kappa _{s}\left({\vec {r}}-a{\hat {r}}\right)$ ${\hat {r}}\cdot {\hat {x}}={\frac {1+\xi }{\sqrt {(1+\xi )^{2}+\eta ^{2}}}}=1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}\psi ^{3}$ ${\hat {r}}\cdot {\hat {y}}={\frac {\eta }{\sqrt {(1+\xi )^{2}+\eta ^{2}}}}=\eta \left(1-\xi -\xi ^{2}-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}\psi ^{3}\right)$ ${\frac {\vec {f}}{-\kappa _{s}}}=(1+\xi ){\hat {x}}+\eta {\hat {y}}-a\left(1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}\psi ^{3}\right){\hat {x}}-a\eta \left(1-\xi -\xi ^{2}-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {\psi }}^{3}\right){\hat {y}}$ #### 3

shownbelow

(?)

${\begin{array}{lllll}&&\quad {\mathcal {O}}\psi ^{0}&\quad {\mathcal {O}}\psi ^{1}&\quad {\mathcal {O}}\psi ^{2}\\f_{x}&=&-\kappa _{s}(1-a)&-\kappa _{s}\xi &-{\tfrac {1}{2}}\kappa _{s}a\eta ^{2}&+\quad {\mathcal {O}}\psi ^{3}&+\ldots \\f_{y}&=&&-\kappa _{s}(1-a)\eta &-\kappa _{s}a\eta \xi &-\kappa _{s}a\left(\eta \xi ^{2}+\xi \eta ^{2}\right)&+\ldots \end{array}}$ #### algebra for 3

Extended content

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: ${\frac {f_{x}}{-\kappa _{s}}}=(1+\xi )-a\left(1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}z^{3}\right)=(1-a)+\xi +{\tfrac {1}{2}}\eta ^{2}a+\ldots$ 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 $f_{y}=-\kappa _{s}(1-a)\eta -\kappa _{s}a\xi \eta +\kappa _{s}a\left(\xi ^{2}\eta +{\tfrac {1}{2}}\eta ^{3}\right)\ldots$ ## Constructing the wave equation Figure 2: Free-body force diagram for lumped mass model for vibrating string. We label the three masses shown with the consecutive integers $(A,B,C).$ 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:

${\vec {\psi }}_{AB}=\psi _{B}-\psi _{A}\approx \left.{\frac {d{\vec {\psi }}}{dX}}\right|_{X=A+1/2}({\text{if }}B=A+1)$    ${\begin{array}{rcl}{\vec {f}}_{AB}&=-\kappa _{s}\left[{\vec {\psi }}_{X-1/2}^{\,\prime }-a{\hat {\psi }}_{X-1/2}^{\prime }\right]&={\vec {f}}{\bigl (}{\vec {\psi }}_{X-1/2}^{\,\prime }{\bigr )}\\{\vec {f}}_{CB}&=+\kappa _{s}\left[{\vec {\psi }}_{X+1/2}^{\,\prime }-a{\hat {\psi }}_{X+1/2}^{\prime }\right]&={\vec {f}}{\bigl (}{\vec {\psi }}_{X+1/2}^{\,\prime }{\bigr )}\\\end{array}}$ $\Sigma {\vec {f}}_{B}={\vec {f}}_{AB}-{\vec {f}}_{BC}={\vec {f}}{\bigl (}{\vec {\psi }}_{X-1/2}^{\,\prime }{\bigr )}-{\vec {f}}{\bigl (}{\vec {\psi }}_{X+1/2}^{\,\prime }{\bigr )}=-{\vec {f}}^{\,\prime }{\bigl (}{\vec {\psi }}_{X}^{\,\prime }{\bigr )}$ ## Wave equation

List of subpages

$\mathbb {F} (\xi ^{\prime },\eta ^{\prime })={\frac {d}{dX}}$ $\,\mathbb {F}$ ${\begin{array}{rccc}&-f_{x}(\xi ^{\prime },\eta ^{\prime })=&+\kappa _{s}(1-a)&+\kappa _{s}\xi ^{\prime }&+{\tfrac {1}{2}}\kappa _{s}a\left(\eta ^{\prime }\right)^{2}+\ldots \\\Sigma F_{x}=&-f_{x}^{\prime }(\xi ^{\prime },\eta ^{\prime })=&0&+\kappa _{s}\xi ^{\prime \prime }&+\kappa _{s}a\eta ^{\prime }\eta ^{\prime \prime }+\ldots \\&&&&\\&-f_{y}(\xi ^{\prime },\eta ^{\prime })=&+(1-a)\kappa _{s}\eta &-\kappa _{s}a\eta \xi &+\ldots \\\Sigma F_{y}=&-f_{y}^{\prime }(\xi ^{\prime },\eta ^{\prime })=&(1-a)\kappa _{s}\eta ^{\prime \prime }&+\underbrace {\kappa _{s}a\left(\eta ^{\prime }\xi +\eta \xi ^{\prime }\right)} _{\text{drop}}&+\ldots \end{array}}$ 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 $\kappa _{T}=(1-a)\kappa _{s}\leq \kappa _{s}$ =>  $\kappa _{T}=\kappa _{s}-a\kappa _{s}$ =>  $a\kappa _{s}=\kappa _{s}-\kappa _{T}$ $m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+a\kappa _{s}\eta ^{\prime }\eta ^{\prime \prime }$ $m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+(\kappa _{s}-\kappa _{T})\eta ^{\prime }\eta ^{\prime \prime }$ $m{\ddot {\eta }}=\kappa _{T}\eta ^{\prime \prime }$ ## 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 <sub><big><big>

## 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 $\ell \neq 1$ 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

### Wave-wave interactions

#### 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)

$ss$ *

$ss$ *

## Footnotes

• Need to look at: P. M. Morse and K. U. Ingard, Theoretical Acoustics ~McGraw–Hill, New York, 1968 referenced in Rowland 1999.

## Appendix

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

Temp: {{#lst:String vibration|toc}}