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

__Rope & Springs__

A wave on a rope, initiated by some forceful oscillations of its end, propagates along a rope, thereby transferring the initial boost of energy to distant locations.

Our task is to determine how much energy is carried by a single wave on a rope - a part of a rope from crest to crest or from trough to trough, the length of which is the

**wavelength**

*of rope's oscillations.*

**λ**Let's define the X-axis along the direction the rope is stretched (we will call it "horizontal"). The forced oscillations of the rope's end will be along the Y-axis ("vertical") that is perpendicular to X-axis.

Previous lecture was dedicated to analysis of the vertical displacement of each piece of a rope as a function

*of its distance from the endpoint of a rope*

**y(x,t)***and time*

**x***, if the rope's end (*

**t***) is forced to perform harmonic oscillations of the form*

**x=0***. Here*

**y(0,t)=A·sin(ω·t)***is an*

**A****amplitude**and

*is the*

**ω****angular speed**of oscillations.

The result of this analysis was a

**wave equation**that function

*must satisfy*

**y(x,t)**

**(μ/T)·**∂**²y(x,t)/**∂**²t =**∂**²y(x,t)/**∂**x²**where

*is linear mass density of a rope (mass per unit of length), assumed to be uniform;*

**μ***is rope's tension;*

**T**The solution to this equation with initial condition

*was suggested in the previous lecture as*

**y(0,t)=A·sin(ω·t)**

**y(x,t) = A·sin(ω·t−k·x)**where

*and*

**A***are defined above and constant*

**ω***must satisfy the condition*

**k**

**k²/ω² = μ/T**For the purposes that will be clear later we would like to start time (

*) not when a rope's end is at horizontal level (*

**t=0***), but when it's lifted up to a full extent of an amplitude*

**y=0***, that is our initial condition will be*

**A***.*

**y(0,t)=A·cos(ω·t)**From the energy perspective this modification does not change anything because we will be dealing with an energy of a single wave from crest to crest or from trough to trough, which is the same, whether the wave is based on function

*sin()*or

*cos()*, as long as amplitude

*and angular speed*

**A***are the same.*

**ω**Then the solution to a wave equation that satisfies our initial condition can be

**y(x,t) = A·cos(ω·t−k·x)**The approach we will take is based on the analogy between sinusoidal (harmonic) movement up and down of each infinitesimal piece of a horizontally stretched rope and harmonic oscillation of the point-mass object on a vertically oriented spring.

Assume that each infinitesimal piece of a rope of the length

*d*and mass

**x***can be considered as an independent point-mass on some tiny spring, and all these pieces are oscillating in a manner resembling and totally identical by all parameters to movements of these pieces as when they are parts of a rope - the same oscillations in terms of the amplitude and angular speed as well as timing of these oscillations.*

**μ·**d**x**If all the motion parameters of each infinitesimal piece of a rope are the same as those of the corresponding point-mass on a spring, the forces must be the same and, consequently, the kinetic and potential energy are the same.

Since we have addressed the energy of a point-mass on a spring in earlier lectures (see "Energy of Waves"), we will use these results to calculate the energy of each infinitesimal piece of a rope, which, in turn, will allow to determine the energy of a single wave by integrating all the individual infinitesimal quantities of energy.

Let's start with the equation describing the vertical movement of a rope at each its point that satisfies the wave equation and the initial condition we have set

**y(x,t) = A·cos(ω·t−k·x)**where

*is an amplitude (known, defined by external force that causes oscillations);*

**A***is an angular speed (known, defined by external force that causes oscillations),*

**ω***is time;*

**t***is a known distance of the point of a rope that we analyze from its end that is forced to harmonically oscillate to make waves along the rope;*

**x***is a coefficient that depends on the properties of a rope (linear mass density and tension) as well as the angular speed of oscillations*

**k***and defined by an equation*

**ω***,*

**k²/ω²=μ/T**that is a known quantity equaled to

*.*

**ω·√μ/T**The physical meaning of the coefficient

*was discussed in the previous lecture, that derived the formula for a speed of propagation of a wave*

**k***:*

**v***.*

**v = ω/k**At the same time speed of wave propagation

*is the wavelength*

**v***divided by period or, since a period is an inverse of frequency*

**λ***,*

**f***,*

**v=λ·f=λ·ω/(2π)**from which we derive

**k = ω/v = 2π/λ**

**v = ω·λ/(2π)**From the formula for a speed of waves propagation we can derived another form of the solution to a wave equation

*[*

**y(x,t) = A·cos(ω·t−k·x) =**

= A·cos= A·cos

*]*

**ω·(t−k·x/ω)***[*

**=**

= A·cos= A·cos

*]*

**ω·(t−x/v)**The expression

*is, obviously, the time it takes for the wave to go from the place of its origination (*

**x/v***) to any fixed distance*

**x=0***, that is a time delay of oscillations at point*

**x***from oscillations at point*

**x***.*

**x=0**So, oscillations at any point

*of a rope are the same in terms of amplitude*

**x***and angular speed*

**A***, as at point*

**ω***, just with some delay*

**x=0***, which depends on a known distance from the wave origination*

**x/v***and a known speed of wave propagation*

**x***.*

**v**Obviously, we can make a tiny spring with infinitesimal piece of a rope attached to it that has the same oscillations as that same piece being a part of a rope.

We do know the mass attached to a spring, that is the mass of an infinitesimal piece of a rope

*d*, it's

**x***d*.

**m=μ·**d**x**We can arrange the oscillations of a spring to have the same amplitude

*as each piece of a rope. All we need to do is to initially stretch a spring by*

**A***.*

**A**We also need to have a spring with proper elasticity

*to make sure the angular speed of oscillations of a spring is equal to angular speed of oscillations of each piece of a rope*

**k**_{e}*.*

**ω**NOTE: we use here symbol

*for elasticity to differentiate it from*

**k**_{e}*used in the solution for a wave equation*

**k***.*

**y(x,t)=A·cos(ω·t−k·x)**We know from previous lectures that for a spring the angular speed of oscillations relates to elasticity

*and mass of an object attached to it*

**k**_{e}*as*

**m***.*

**ω² = k**_{e}/mIn our case of a tiny spring with mass attached to it equaled to

*d*we can determine the elasticity from this mass and required angular speed

**m=μ·**d**x***:*

**ω**

**k**d_{e}= ω²·μ·**x**and choose a spring with this exact elasticity.

So, taking a tiny spring with the above elasticity and stretching it by initial length

*from its neutral state will produce the oscillations with an amplitude*

**A***and angular frequency*

**A***.*

**ω**What remains to be done is to arrange a proper timing, so the oscillations of the tops of springs will go in waves similarly to waves on a rope.

To accomplish that, after initial stretching of all springs by an amplitude

*, we have to release these springs with a time delay required for a wave to reach any particular point on a rope. For a distance*

**A***of a particular spring from the beginning, knowing the speed of wave propagation on a rope*

**x***, we will just release the spring after*

**v***time interval from the moment of releasing the first spring.*

**x/v**The movement of the point-mass attached to a spring with described above parameters and delayed start is described by an equation

*[*

**y(x,t) = A·cos***]*

**ω·(t−x/v)**In the above equation for a spring at distance

*amplitude*

**x***and angular speed of oscillations*

**A***are defined by initial condition of oscillating the rope*

**ω**

**y(0,t) = A·cos(ω·t)**The speed of propagation

*is a derived parameter and, therefore, is known as well.*

**v**As specified above,

**v = √T/μ ,**where

*is a linear mass density of a rope and*

**μ***is a rope's tension - all known parameters.*

**T**Since our task is to determine an energy of a wavelength

*of a rope, we have to integrate the energy of all springs by*

**λ***on a segment of the length*

**x***. Periodicity of oscillations assures that all such segments will have the same energy.*

**λ**Let's summarize the characteristics of tiny springs that should oscillate exactly like a rope.

1. Mass attached to each tiny spring equals to

*d*.

**m=μ·**d**x**2. Each tiny spring is located at a distance

*from the beginning.*

**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=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−2π·x/(λ·ω))**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.

Next lecture will use all this to find potential and kinetic energy of a single wave on an oscillating rope.