*Notes to a video lecture on http://www.unizor.com*

__Rope Energy__

Assume that we have a rope stretched along the X-axis (horizontally).

Its end (

*) is forcefully oscillated up and down according to a formula*

**x=0***.*

**D(t)=A·cos(ω·t)**This causes a rope's piece positioned at distance

*from the forcefully oscillated end (*

**x***) to perform harmonic up and down oscillations along the Y-axis (vertically) according to a formula*

**x=0**

**y(x,t) = A·cos(ω·t−k·x)**where

*is vertical deviation of the rope's piece at distance*

**y(x,t)***from the source of oscillations as a function of time*

**x***,*

**t***is an amplitude of oscillations along the Y-axis,*

**A***is an angular speed of oscillations,*

**ω***is time,*

**t***is a rope's characteristic.*

**k**The physical meaning of

*will be clear, if the formula for oscillations at distance*

**k***from the rope's end is presented as*

**x***, where*

**y=A·cos(ω·(t−x/v))***is a speed of wave propagation along a rope and, therefore,*

**v***is a time delay between oscillations at their origin*

**x/v***and oscillations at distance*

**x=0***from the origin.*

**x**Then we can construct different relations between

parameter

*,*

**k**speed of wave propagation

*,*

**v**period of oscillation

*,*

**τ**frequency of oscillations

*and*

**f**wavelength

**λ**as follows:

**k = ω/v**

**v = λ/τ**

**τ = 1/f**

**v = λ·f**

**2π·f = ω**

**k = ω/v = ω/(λ·f) = 2π/λ**Previous lecture was about modeling the oscillations of a horizontally stretched rope with oscillations of an infinite number of vertically oriented tiny springs, each attached to an independent infinitesimal piece of a rope.

The characteristics of this approach to model rope's oscillations are as follows:

1. Mass attached to each tiny spring equals to

*d*.

**m=μ·**d**x**2. Location of each tiny spring is

*- its distance from the rope's end.*

**x**3. All springs have the same elasticity

**k**d_{e}=ω²·μ·**x**4. All springs are initially stretched by

*, so their amplitude will be the same as that of the rope.*

**A**5. All springs will have the same angular speed

*, the same as waves on a rope.*

**ω**6. After initial stretch the springs will be let go, but not simultaneously; the time delay of a spring at distance

*from the beginning will correspond to time needed for a wave on a rope to reach that distance, that is the time delay will be equal to*

**x***.*

**x/v=x/(λ·f)=2π·x/(λ·ω)**7. As a result, a spring at distance

*from the beginning oscillates according to a formula*

**x***[*

**y(x,t) = A·cos(ω·t−k·x) =**

= A·cos= A·cos

*]*

**ω·(t−2π·x/(λ·ω))**where

**k = 2π/λ**Modeled as described above, rope oscillations are identical to oscillations of all those tiny springs. The point-masses at the top of these springs will oscillate exactly the same as if they are attached to a rope.

The above characteristics of the springs, that we have used to model the propagating waves on a rope, are sufficient to analyze the distribution of energy, using what we have already analyzed about springs.

Our task is to find what energy is carried by one single wave of a rope. For this we will calculate the energy carried by a single tiny spring that models oscillations of an infinitesimal piece of a rope

*d*and integrate it along the wavelength of oscillations

**x***.*

**λ**For this purpose we consider the tiny springs along the X-axis within any segment of the length

*, for example, from*

**λ***to*

**x=0***.*

**x=λ**Each individual tiny spring at fixed X-coordinate

*carries potential and kinetic energies, and these energies are changing with the time, that is they are functions of time for each fixed spring at distance*

**x***from origin.*

**x**Consider a potential energy of an individual tiny single spring

*for any fixed*

**U(x,t)***as a function of time*

**x***.*

**t**This potential energy depends only on its vertical deviation

*which determines the degree of a stretch.*

**y(x,t)**Earlier we derived a formula for potential energy of a spring with elasticity

*stretched by*

**k**_{e}*from a neutral position as*

**d***.*

**½k**_{e}·d²In our case distance a spring is stretched is

*.*

**d=y(x,t)**The elasticity coefficient was derived above as a function of required angular speed

*of oscillations for a particular mass*

**ω***d*attached to a spring:

**m=μ·**d**x***.*

**k**d_{e}=ω²·μ·**x**So, the potential energy of a spring at distance

*from the origin of oscillations at time*

**x***is*

**t***,*

**U(x,t) = ½k**

= ½ω²·μ·d_{e}·y²(x,t) == ½ω²·μ·

**x·A²·cos²(ω·t−k·x)**where

*.*

**k = 2π/λ**For each moment in time

*(that is,*

**t***is fixed and participates in the integration by*

**t***as a constant) the potential energy of all springs within a segment [*

**x***] of a distance from the origin of oscillation is a definite integral of the above expression by*

**0,λ***from*

**x***to*

**x=0***.*

**x=λ**Let's find integral

*,*

**I = ∫**d_{0}^{λ}cos²(ω·t−k·x)**x**where

*, and multiply the result by a constant*

**k = 2π/λ***.*

**½ω²·μ·A²**To calculate

*, we substitute*

**I**

**z = ω·t−k·x***d*

**z = −k·**d**x***d*

**x = −**d**z/k**We have to express the limits of integration in terms of

*:*

**z**if

*,*

**x=0**

**z=ω·t**if

*,*

**x=λ**

**z=ω·t−k·λ=ω·t−2π**Now our integral looks like

**I = −∫**d_{0}^{λ}cos²(z)**z/k**Calculating the indefinite integral

**∫cos²(z)·**d**z = ½z + ¼sin(2z) + C**Using Newton-Leibnitz formula for definite integral, we have to substitute upper limit for

*, then lower limit and subtract the latter from the former:*

**z**

**½(ω·t−2π)+¼sin(2·(ω·t−2π)) −**

− ½ω·t−¼sin(2ω·t) = −π− ½ω·t−¼sin(2ω·t) = −π

Note, this integral is independent of time

*. It means that combined potential energy of all tiny springs on a segment of the length*

**t***does not change with time, as one spring is stretched or squeezed more, another is less, so the total sum remains constant. For a rope it means that a single rope's wave has potential energy in all its pieces not changing with time.*

**λ**Therefore,

**I = π/k = π·λ/(2π) = ½λ**Consequently, the potential energy of all tiny springs on a segment of the length

*will be*

**λ**

**U**

= ¼ω²·μ·A²·λ_{λ}= ½ω²·μ·A²·I == ¼ω²·μ·A²·λ

Let's consider a kinetic energy of all tiny springs on a segment of the length

*with attached to them infinitesimal masses.*

**λ**The general formula for kinetic energy is

*,*

**K = ½M·V²**where

*is mass and*

**M***is speed of an object.*

**V**For a single tiny spring with an infinitesimal mass attached to it we have

*- product of linear mass density*

**M = μ·**d**x***by the length of a piece of a rope attached to the top of a spring*

**μ***d*

**x***- the first derivative of vertical position of the mass attached to a spring by time*

**V =**∂**y(x,t)/**∂**t**Since we know the expression for vertical movements of the top of a spring

*,*

**y(x,t) = A·cos(ω·t−k·x)**where

**k = 2π/λ**we can find the speed of vertical movement at any moment of time:

*∂*

**y(x,t)/**∂**t = −ω·A·sin(ω·t−k·x)**The kinetic energy of a point-mass attached to this tiny spring is

**K(x,t) = ½M·V² =**

= ½μ·d= ½μ·

**x·ω²·A²·sin²(ω·t−k·x)**As you see, the expression for

*resembles the expression for*

**K(x,t)***with the only difference that this time we have*

**U(x,t)***sin²()*instead of

*cos²()*.

When we integrate this expression by

*from*

**x***to*

**0***(that is, across all tiny springs that cover a distance equaled to a wavelength*

**λ***), we will have a similar result - the total kinetic energy of all point-masses attached to springs on a segment of the length*

**λ***does not depend on time*

**λ***and equals to the same value*

**t**

**K**_{λ}= ¼ω²·μ·A²·λIt's quite remarkable that potential and kinetic energies of one wavelength of a rope are the same and independent of time.

Now we can calculate the combined energy carried by a single piece of a rope of the length equaled to a wavelength of oscillations:

**E**_{λ}= U_{λ}+ K_{λ}= ½ω²·μ·A²·λThe above amount of energy is carried by one wavelength of a oscillations. That takes an amount of time equaled to a period of oscillations

*. Therefore, the oscillations produce an average power*

**τ**

**P**_{ave}= E_{λ}/τ = ½ω²·μ·A²·λ/τSince

*(speed of wave propagation), we conclude*

**λ/τ = v**

**P**_{ave}= E_{λ}/τ = ½ω²·μ·A²·vThe following picture schematically represents a wave (red line) and the level of its kinetic energy (blue line), which is based on the shape of function

**sin²(ω·t−k·x)**The potential energy graph is based on

*and its graph would be similar, just shifted to be in anti-phase with kinetic energy.*

**cos²(ω·t−k·x)**Their sum is constant because

*sin²()+cos²()=1*.

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