# String vibration

Transverse string wave longitudinal motion

## ARticle

### Intro

Transverse string wave lumped mass without text

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

### Three tricks:

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

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

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

#### Second derivative

{\displaystyle {\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 ${\displaystyle \ell =1}$

### Calculation of force in Figure 1

${\displaystyle {\vec {f}}=m{\ddot {\vec {r}}}=-\kappa _{s}(r-a){\hat {r}}}$

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

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

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

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

${\displaystyle {\vec {f}}=-\kappa _{s}(r-a){\hat {r}}=-\kappa _{s}\left({\vec {r}}-a{\hat {r}}\right)}$

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

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

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

#### 3

shownbelow

(?)

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

#### algebra for 3

Extended content

x1: ${\displaystyle {\frac {f_{x}}{-\kappa _{s}}}=}$ ${\displaystyle \left({\vec {r}}-a{\hat {r}}\right)\cdot {\hat {x}}}$ equals ${\displaystyle (1+\xi )}$ minus ${\displaystyle a\left(1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}z^{3}\right)}$

x2: ${\displaystyle {\frac {f_{x}}{-\kappa _{s}}}=(1+\xi )-a\left(1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}z^{3}\right)}$

x3: ${\displaystyle {\frac {f_{x}}{-\kappa _{s}}}=(1+\xi )-a\left(1-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}z^{3}\right)=(1-a)+\xi +{\tfrac {1}{2}}\eta ^{2}a+\ldots }$

x4: ${\displaystyle f_{x}=-\kappa _{s}(1-a)-\kappa _{s}\xi -{\tfrac {1}{2}}\kappa _{s}a\eta ^{2}+\ldots }$

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

y2: ${\displaystyle {\frac {f_{y}}{-\kappa _{s}}}=\eta -a\eta \left(1-\xi -\xi ^{2}-{\tfrac {1}{2}}\eta ^{2}+{\mathcal {O}}z^{3}\right)}$

y3: ${\displaystyle {\frac {f_{y}}{-\kappa _{s}}}=(1-a)\eta +a\xi \eta +a\xi ^{2}\eta +a{\tfrac {1}{2}}\eta ^{3}+\ldots }$

y4 ${\displaystyle f_{y}=-\kappa _{s}(1-a)\eta -\kappa _{s}a\xi \eta +\kappa _{s}a\left(\xi ^{2}\eta +{\tfrac {1}{2}}\eta ^{3}\right)\ldots }$

## 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 ${\displaystyle (A,B,C).}$

Here we adopt the convention that ${\displaystyle \ell =1}$ unit of length. We label masses with the variable ${\displaystyle X}$ that represents each mass by an integer ${\displaystyle (X=0,\pm 1,\pm 2,\ldots )}$. To obtain a wave equation we focus on the three consecutive integers, ${\displaystyle (A,B,C)}$. When the string is at equilibrium (i.e., zero wave amplitude), we can also use non-integral values of ${\displaystyle X}$ 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 ${\displaystyle \xi }$ in the x direction and ${\displaystyle \eta }$ in the y-direction.[2] Defining ${\displaystyle \psi }$ to be the vector associated with this displacement, we have,

${\displaystyle {\vec {\psi }}_{B}=\xi _{B}{\hat {x}}+\eta _{B}{\hat {y}}}$

We denote

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

The convention used by OpenStax Physics[3] is that ${\displaystyle {\vec {F}}_{AB}={\vec {F}}_{\text{by-A-on-B}}}$ 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:

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

       


${\displaystyle {\begin{array}{rcl}{\vec {f}}_{AB}&=-\kappa _{s}\left[{\vec {\psi }}_{X-1/2}^{\,\prime }-a{\hat {\psi }}_{X-1/2}^{\prime }\right]&={\vec {f}}{\bigl (}{\vec {\psi }}_{X-1/2}^{\,\prime }{\bigr )}\\{\vec {f}}_{CB}&=+\kappa _{s}\left[{\vec {\psi }}_{X+1/2}^{\,\prime }-a{\hat {\psi }}_{X+1/2}^{\prime }\right]&={\vec {f}}{\bigl (}{\vec {\psi }}_{X+1/2}^{\,\prime }{\bigr )}\\\end{array}}}$

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

## Wave equation

List of subpages

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

${\displaystyle {\begin{array}{rccc}&-f_{x}(\xi ^{\prime },\eta ^{\prime })=&+\kappa _{s}(1-a)&+\kappa _{s}\xi ^{\prime }&+{\tfrac {1}{2}}\kappa _{s}a\left(\eta ^{\prime }\right)^{2}+\ldots \\\Sigma F_{x}=&-f_{x}^{\prime }(\xi ^{\prime },\eta ^{\prime })=&0&+\kappa _{s}\xi ^{\prime \prime }&+\kappa _{s}a\eta ^{\prime }\eta ^{\prime \prime }+\ldots \\&&&&\\&-f_{y}(\xi ^{\prime },\eta ^{\prime })=&+(1-a)\kappa _{s}\eta &-\kappa _{s}a\eta \xi &+\ldots \\\Sigma F_{y}=&-f_{y}^{\prime }(\xi ^{\prime },\eta ^{\prime })=&(1-a)\kappa _{s}\eta ^{\prime \prime }&+\underbrace {\kappa _{s}a\left(\eta ^{\prime }\xi +\eta \xi ^{\prime }\right)} _{\text{drop}}&+\ldots \end{array}}}$

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 ${\displaystyle \kappa _{T}=(1-a)\kappa _{s}\leq \kappa _{s}}$  =>  ${\displaystyle \kappa _{T}=\kappa _{s}-a\kappa _{s}}$  =>  ${\displaystyle a\kappa _{s}=\kappa _{s}-\kappa _{T}}$

${\displaystyle m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+a\kappa _{s}\eta ^{\prime }\eta ^{\prime \prime }}$  and  ${\displaystyle m{\ddot {\eta }}=\kappa _{T}\eta ^{\prime \prime }}$

${\displaystyle m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+(\kappa _{s}-\kappa _{T})\eta ^{\prime }\eta ^{\prime \prime }}$

#### Allowing ${\displaystyle \ell \neq 1}$

Finding the longitudinal component of a transverse wave.

Note that ${\displaystyle \kappa _{T}\leq \kappa _{s}}$. Let ${\displaystyle \psi _{0}}$ be real, and define,

${\displaystyle \psi =\psi _{0}e^{i(kX-\omega t)}}$

${\displaystyle \omega ^{2}={\frac {\kappa }{m}}={\frac {\kappa \ell ={\text{tension}}}{m/\ell ={\text{density}}}}={\frac {\kappa \ell ^{2}}{m}}}$

Therefore, ${\displaystyle m{\ddot {\eta }}=\kappa _{T}\eta ^{\prime \prime }}$  and ${\displaystyle m{\ddot {\xi }}=\kappa _{s}\xi ^{\prime \prime }+(\kappa _{s}-\kappa _{T})\eta ^{\prime }\eta ^{\prime \prime }}$, becomes:

${\displaystyle {\ddot {\eta }}=\omega _{T}^{2}\eta ^{\prime \prime }}$  and ${\displaystyle {\ddot {\xi }}=\omega _{s}^{2}\xi ^{\prime \prime }+(\omega _{s}^{2}-\omega _{T}^{2})\eta ^{\prime }\eta ^{\prime \prime }}$

Also: ${\displaystyle \partial /\partial X=\ell ^{-1}\partial /\partial X\rightarrow \eta ^{\prime }=\eta ^{\prime }/\ell \rightarrow k=k\ell }$

CONFUSED!!!!!!!

EQUATION 11 IN THE PAPER SAYS:${\displaystyle {\ddot {\xi }}\approx c_{L}^{2}\xi ^{\prime \prime }+\left(c_{L}^{2}-c_{T}^{2}\right)\eta ^{\prime }\eta ^{\prime \prime }}$

### Wave-wave interactions

#### Product rule

From Phasor_algebra:

${\displaystyle \Re e\left(\psi _{1}\right)\Re e\left(\psi _{2}\right)={\tfrac {1}{2}}|\psi _{1}|\,|\psi _{2}|\cos(\Phi _{1}-\Phi _{2})\;+\;{\tfrac {1}{2}}|\psi _{1}|\,|\psi _{2}|\cos(\Phi _{1}+\Phi _{2})}$

                


### Variables

Variables

${\displaystyle {\vec {r}}=x{\hat {x}}+y{\hat {y}}}$ * position with unit vectors

${\displaystyle X_{j}}$ * coordinate variable parallel to string

Equilibrium

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

${\displaystyle a}$ *relaxed spring length

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

${\displaystyle \tau }$ * equilibrium tension in string

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

${\displaystyle ss}$ *

${\displaystyle c_{T}=\omega _{T}/k_{T}={\sqrt {\tau /\rho }}}$ * transverse wave speed

${\displaystyle c_{L}}$ *longitudinal wave speed

Perturbation

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

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

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

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

Avoid

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

${\displaystyle ss}$ *

${\displaystyle ss}$ *

## Templates

### Aligned wikitable

Caption text
Example Example Example
Example Example Example
Example Example Example

### Equation

We create a four-term expression using (4) and (5). Then we move the terms involving energy density and the Poynting vector to the LHS to obtain:

${\displaystyle f=ma}$

(10)

### Key equation

${\displaystyle e=mc^{2}}$

(11)

### Outline

1. Lumped sum model (linear) for transverse and longitudinal waves.
2. Need image
3. Calculate field energy density and pose simple question between two choices. One is partially wrong and the other is completely wrong.
4. See if I can derive the nonlinear interaction obtained by Rowland

## Footnotes

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

## TRASH

trash

${\displaystyle m{\frac {d^{2}{\vec {r}}_{j}}{dt^{2}}}=-\kappa _{s}\left(\ell _{j,j-1}-a\right)\left[{\frac {{\vec {r}}_{j}-{\vec {r}}_{j-1}}{\ell _{j,j-1}}}\right]+\kappa _{s}\left(\ell _{j,j+1}-a\right)\left[{\frac {{\vec {r}}_{j+1}-{\vec {r}}_{j}}{\ell _{j+1,j}}}\right]}$

${\displaystyle m{\frac {d^{2}{\vec {r}}_{j}}{dt^{2}}}=-\kappa _{s}\left(\ell _{j,j-1}-a\right)\left[{\frac {{\vec {r}}_{j}-{\vec {r}}_{j-1}}{\ell _{j,j-1}}}\right]+\kappa _{s}\left(\ell _{j,j+1}-a\right)\left[{\frac {{\vec {r}}_{j+1}-{\vec {r}}_{j}}{\ell _{j+1,j}}}\right]}$

${\displaystyle \ell _{j,j-1}=\left|{\vec {r}}_{j}-{\vec {r}}_{j-1}\right|={\sqrt {(x_{j}-{\vec {x}}_{j\pm 1})^{2}+(y_{j}-{\vec {y}}_{j-1})^{2}}}}$

${\displaystyle {\vec {\ell }}_{j,j-1}={\vec {r}}_{j}-{\vec {r}}_{j-1}}$

FOOTNOTE${\displaystyle {\vec {\gamma }}_{j}=\xi _{j}{\hat {x}}\,+\,\eta _{j}{\hat {y}}\,+\,\zeta _{j}{\hat {z}}}$

${\displaystyle {\vec {\gamma }}_{j}=\xi _{j}{\hat {x}}\,+\,\eta _{j}{\hat {y}}}$

### 2

${\displaystyle {\vec {F}}_{76}={\vec {F}}_{\text{on 7}}^{\,{\text{ by 6}}}}$

${\displaystyle {\vec {\ell }}_{76}=\ell {\hat {x}}+{\vec {\gamma }}_{7}-{\vec {\gamma }}_{6}}$

${\displaystyle \ell =1}$

${\displaystyle {\vec {\ell }}_{76}={\hat {x}}+{\vec {\gamma }}_{7}-{\vec {\gamma }}_{6}=(1+\Delta \xi _{76}){\hat {x}}+(\Delta \eta _{76}){\hat {y}}}$

Recall that for small ${\displaystyle \epsilon }$,

${\displaystyle {\sqrt {1+\epsilon }}=1+{\tfrac {1}{2}}\epsilon -{\tfrac {1}{8}}\epsilon ^{2}+{\tfrac {1}{16}}\epsilon ^{3}-{\tfrac {5}{128}}\epsilon ^{4}+{\tfrac {7}{256}}\epsilon ^{5}-\ldots ,,}$

${\displaystyle \ell _{76}={\sqrt {1+2\Delta \xi _{76}+(\Delta \xi _{76})^{2}+(\Delta \eta _{76})^{2}}}=1+\Delta \xi _{76}+{\tfrac {1}{2}}(\Delta \eta _{76})^{2}+{\mathcal {O}}(\Delta \xi )^{2}+{\mathcal {O}}(\Delta \eta )^{4}+...}$

1. David R Rowland 2011 Eur. J. Phys. 32 1475 Rowland, David R. "The potential energy density in transverse string waves depends critically on longitudinal motion." European journal of physics 32.6 (2011): 1475.
2. Rowland, et al, use the symbol ${\displaystyle \zeta }$ to describe motion in the other transverse direction.
3. OpenStax University Physics Volume I Section 5.5: Newton s third law of motion 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.