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4 The mechanical wave equation
4.1 Properties of mechanical waves _{}^{}
For mechanical waves, the "waving" is done by bits of mass  atoms. molecules, or connected sets of particles, which meeans tha mechanical waves can exist only within a physical material (called the "medium" in which the wave propagates). In case of a wave on a string the disturbance of the wave is the physical displacement of matter from the undisturbed (equilbrium) position and the disturbance has dimensions of distance (with SI units of meter). In pressure waves such as sound waves the disturbance may be masured as the change in the density of medium or the change in the pressure within the medium, so they are expessed a density fluctuation (with SI units of kg/m^{3}) or a pressure change (with N/m^{2}).
So one common aspect of all mechanical waves is that they require a physical medium. For any medium, the material's "inertial" property and its "elastic" property have a critical impact on the propagation of mechanical wave.
The inertial property of mateial is related mass (for discrete media) or mass density (for continuous) of the material. Because "inertia" describes the tendency of all mass to resist acceleration, material of high mass is more difficult to get moving than material with low mass. The mass density affects the velocity of propagation of mateial waves as well as the ability of a source to couple energy into a wave within the medium (called the "impedence" of the medium).
The elastic property of material is related to the restoring force that tends to return displaced particles to their equilibrium positions. So a medium can be elastic in the same way a rubber band or a spring is elastic  when we stretch it, dispacing particles from their eqilibrium positions, the medium produces restoring forces. The strength of those restoring forces joins the mass density in determining the speed of propagation, the dispersion, and the impedence of a medium.
We should understand the difference between the motion of individual particles and the motion of wave itself. For mechanical waves, the net displacement of material produced by the wave over one cycle or over millions cycles, is zero. So what actually moves at the speed of the wave, as we'll see in Section 4.4, is energy.
Although the displacement of individual particles within is small, the direction of the particles' movement is important, because the direction is often used to categorize the wave as either transverse or longitudinal. In a transverse wave, the particles move in a direction that is perpendicular to the direction of wave motion. In a longitudinal wave, the particles of medium move in a direction parallel and antipararell to the direction of wave motion. What type of wave is set up in a given medium depends on the source and on the direction of the restoring force.
4.2 Waves on a string _{}^{}
We can use Newton's second law to relate tension forces on a segment of a string to the acceleration of that segment. Consider a segment of a string with uniform linear density which has ben displaced from the equilibrium (horizontal) poition, as shown in Fig. 4.1. The string has elasticity, so as the segment is displaced, there are tension forces acting on each end, whch tend to return to its equilibrium position.
As the string stretches, the mass per unit length must decrease, but we're going to consider that the change in liner density is negligible. We'll also consider vertical displacements that are small relative to the horizontal extent of the disturbance, so the angle 𝜃 that any portion of the horizontal direction is small.^{(1)} And we'll assume the effects of all other forces (such as gravity) are negligible compared with those of the tension forces. As the segment of string is displacement from the its equilibrium position, the tension force 𝑻_{1} acts on the left end the tension force 𝑻_{2} acts on the right end. To see some interesting physics consider the the 𝑥 and 𝑦components of the tension forces 𝑻_{1} and 𝑻_{2} shown in Fig. 4.2.
𝑇_{1.𝑥} = ∣𝑻_{1}∣ cos 𝜃_{1}, 𝑇_{1.𝑦} = ∣𝑻_{1}∣ sin 𝜃_{1}, 𝑇_{2.𝑥} = ∣𝑻_{2}∣ cos 𝜃_{1}, 𝑇_{2.𝑦} = ∣𝑻_{2}∣ sin 𝜃_{1}.
The next step is to use Newton's second law,
(4.12) ¢²𝐹_{𝑥} = ∣𝑻_{1}∣ cos 𝜃_{1} + ∣𝑻_{2}∣ cos 𝜃_{2} = 𝑚𝑎_{𝑥}, ¢²𝐹_{𝑦} = ∣𝑻_{1}∣ sin 𝜃_{1} + ∣𝑻_{2}∣ sin 𝜃_{2} = 𝑚𝑎_{𝑦}.
As long as the string segment oscillates up and down along the 𝑦direction, we can take 𝑎_{𝑥} = 0, and if the amplitude of the oscillation is small, then 𝜃_{1} and 𝜃_{2} are both small and using approximations and settng 𝑎_{𝑥} to zero gives
cos 𝜃_{1} ≈ cos 𝜃_{2} ≈ 1, ¢²𝐹_{𝑥} = ∣𝑻_{1}∣ cos 𝜃_{1} + ∣𝑻_{2}∣ cos 𝜃_{2} = 0, which means that
(4.3) ∣𝑻_{1}∣ ≈ ∣𝑻_{2}∣,
so the magnitudes of the tension forces are approximately equal, but the directions of those forces are not equal. as shown
Leftend slope = tan 𝜃_{1} = 𝛥𝑦_{1}/𝛥𝑥_{1}, Rightend slope = tan 𝜃_{2} = 𝛥𝑦_{2}/𝛥𝑥_{2}.
Consider only an infinitesimal piece of the string by letting 𝛥𝑥_{1} or 𝛥𝑥_{2} approach zero:
𝛥𝑦_{1}/𝛥𝑥_{1} ¡æ [¡Ó𝑦/¡Ó𝑥]_{left}, 𝛥𝑦_{2}/𝛥𝑥_{2} ¡æ [¡Ó𝑦/¡Ó𝑥]_{right}
Eqs. (4.4) says that the slopes of the ends of the segment are equal to the tangent of 𝜃_{1} and 𝜃_{2}, for small angles those tangents may be approximated by sines:
tan 𝜃_{1} = sin 𝜃_{1}/cos 𝜃_{1} ≈ sin 𝜃_{1}, tan 𝜃_{2} = sin 𝜃_{2}/cos 𝜃_{2} ≈ sin 𝜃_{2}. Thus
¢²𝐹_{𝑦} = ∣𝑻_{1}∣[¡Ó𝑦/¡Ó𝑥]_{left} + ∣𝑻_{2}∣[¡Ó𝑦/¡Ó𝑥]_{right} = 𝑚𝑎_{𝑦} and since 𝑎_{𝑦} = ¡Ó^{2}𝑦/¡Ó𝑡^{2}
𝑚 ¡Ó^{2}𝑦/¡Ó𝑡^{2} = ∣𝑻_{1}∣[¡Ó𝑦/¡Ó𝑥]_{left} + ∣𝑻_{2}∣[¡Ó𝑦/¡Ó𝑥]_{right} since ∣𝑻_{1}∣ = ∣𝑻_{2}∣ = 𝑇:
¡Ó^{2}𝑦/¡Ó𝑡^{2} = 𝑇/𝑚 {[¡Ó𝑦/¡Ó𝑥]_{right}  [¡Ó𝑦/¡Ó𝑥]_{left}} = 𝑇/𝑚 𝛥(¡Ó𝑦/¡Ó𝑥) which means
¡Ó^{2}𝑦/¡Ó𝑡^{2} = 𝑇/𝑚 𝛥(¡Ó𝑦/¡Ó𝑥) = 𝑇/𝜇𝛥𝑥 𝛥(¡Ó𝑦/¡Ó𝑥), But, or small 𝛥𝑥,
𝛥(¡Ó𝑦/¡Ó𝑥)/𝛥𝑥 = ¡Ó^{2}𝑦/¡Ó𝑥^{2}
which means
¡Ó^{2}𝑦/¡Ó𝑡^{2} = 𝑇/𝜇 (¡Ó^{2}𝑦/¡Ó𝑥^{2}) or
(4.5) ¡Ó^{2}𝑦/¡Ó𝑥^{2} = 𝜇/𝑇 (¡Ó^{2}𝑦/¡Ó𝑡^{2}).
One interesting observation about this form of wave equation is that since the displacement (𝑦) in this is actually a physical displacement,^{(2)} We can determine the phase speed of wave comparing the multiplicative tem in Eq. (2.5) with that in Eq. (4.5).
1/𝑣^{2} = 𝜇/𝑇, so
(4.6) 𝑣 = ¡î(𝑇/𝜇).
As expected, the phase speed of the string wave depends both on the elastic (𝑇) and on the inertial (𝜇) properties of the string (which is the medium of propagation in this case). The phase speed of the waves was simply called "𝑣", but now we know that speed to be 𝑣 = ¡î(𝑇/𝜇).
Example 4.1 Compare the displacement velocity, and acceleration for a tranverse harmonic wave on a string.
If the displacement 𝑦(𝑥, 𝑡) = 𝐴 sin(𝑘𝑥  𝜔𝑡), the transverse velocity 𝑣_{1} = ¡Ó𝑦/¡Ó𝑡 = 𝐴𝜔 cos(𝑘𝑥  𝜔𝑡), and the transverse acceleration is 𝑎_{𝑡} = ¡Ó^{2}𝑦/¡Ó𝑡^{2} = 𝐴𝜔^{2} sin(𝑘𝑥  𝜔𝑡) as in Fig. 4.3.
One additional concept is the effect of changes in the linear mass density (𝜇) or tension (𝑇) of the string. We can see the effect of allowing the string characteristics to change with distance by treating both the density and tenion as functions of 𝑥: 𝜇 = 𝜇(𝑥) and 𝑇(𝑥), So this modified version of the wave equation:
¡Ó^{2}𝑦/¡Ó𝑡^{2} = 1/𝜇(𝑥) ¡Ó/¡Ó𝑥 [𝑇(𝑥) ¡Ó𝑦/¡Ó𝑥].
Notice that 𝑇(𝑥) cannot be moved out of spatial derivative, and the ratio of tension to the density may no longer constant. In general, the spatial portions of the solution to this equation are nonsinusoidal. This has an interesting analogy with quantum waves in inhomogenuous medium.
4.3 Pressure waves _{}^{}
By "pressure wave" we mean any wave in a medium a mechanical source causes a physical displacement and compression or refraction of the material in the direction in which the wave is propagating. We can see an illustration of how a pressure wave works in Fig. 4.4. The "distubance" of such waves involves three things: the longitudinal displacement of material, changes in density of the material, and variation of the pressure within material. So pressure waves could also be called "density waves" or even "longitudinal displacement waves".
Though we're considering onedimensional wave motion, pressure waves exist in a threedimensional medium, so instead of considering the linear mass density 𝜇, it's the volumetric mass density 𝜌 that will provide the inertial characteristic of the medium. In this case, we'll assume that the pressure and density variations are small relative to the equilibrium values and consier only longitudinal displacement.
We can start by defining the pressure (𝛲) at any location in terms of the equilibrium pressure (𝛲_{0}) and density (𝜌) at any location can be written in terms of the equilibrium density (𝜌_{0}) and the incremental change in pressure or density produced by the wave (𝑑𝛲 or 𝑑𝜌):
𝛲 = 𝛲_{0} + 𝑑𝛲, and 𝜌 = 𝜌_{0} + 𝑑𝜌.
Before relationg these quantities to he acceleration of material in the medium using Newton's second law, we had better to familiarize ourselves with the terminology and equations of volume compressibility. The compressibility of a substance is the inverse of its "bulk modulus", which relates an incremental change in pressure (𝑑𝛲) to the fractional change in density (𝑑𝜌/𝜌_{0}) of the material:
(4.78) 𝐾 ¡Õ 𝑑𝛲/(𝑑𝜌/𝜌_{0}) or 𝑑𝛲 = 𝐾(𝑑𝜌/𝜌_{0}).
With this relationship in hand, we can consider Newton's second law for the segment of material being displaced and compressed or rarefied by the wave. So consider the pressure from the surrounding material acting on the left and the right side of the segment, as shown in Fig. 4.5. Setting the sum of the 𝑥direction forces equal to the acceleration in the 𝑥direction gives
(4.9) ¢²𝐹_{𝑥} = 𝛲_{1}𝐴  𝛲_{2}𝐴 = 𝑚𝑎_{𝑥},
where 𝑚 is the mass of segment.
𝑚 = 𝜌_{0}𝐴 𝑑𝑥.
Using the symbol 𝜓 to represent the displacement of the material due to the wave, the accelerlation of the 𝑥direction can be written as
𝑎_{𝑥} = ¡Ó^{2}𝜓/¡Ó𝑡^{2}.
Substituting these expressions for 𝑚 and 𝑎_{𝑥} into Newton's second law (Eq. (4.9)) gives
¢²𝐹_{𝑥} = 𝛲_{1}𝐴  𝛲_{2}𝐴 = 𝜌_{0}𝐴 𝑑𝑥 ¡Ó^{2}𝜓/¡Ó𝑡^{2}.
𝛲_{1}𝐴  𝛲_{2}𝐴 = (𝛲_{0} + 𝑑𝛲_{1})𝐴  (𝛲_{0} + 𝑑𝛲_{2})𝐴 = (𝑑𝛲_{1}  𝑑𝛲_{2})𝐴.
Change in overpressure = 𝑑𝛲_{2}  𝑑𝛲_{1} = ¡Ó(𝑑𝛲)/¡Ó𝑥 𝑑𝑥,
which means
¡Ó(𝑑𝛲)/¡Ó𝑥 𝑑𝑥 𝐴 = 𝜌_{0}𝐴 𝑑𝑥 ¡Ó^{2}𝜓/¡Ó𝑡^{2} or 𝜌_{0} ¡Ó^{2}𝜓/¡Ó𝑡^{2} = ¡Ó(𝑑𝛲)/¡Ó𝑥.
But 𝑑𝛲 = 𝑑𝜌𝐾/𝜌_{0} (Eq. (4.8)), so
(4.10) 𝜌_{0} ¡Ó^{2}𝜓/¡Ó𝑡^{2} = ¡Ó[(𝐾/𝜌_{0})𝑑𝜌]/¡Ó𝑥.
The next step is to relate the change in density (𝑑𝜌) to the displacements of the left and right ends of the segment (𝜓^{1} and 𝜓^{2}). To do that, the mass of the segment is the same before and after the segment is compressed. Because 𝑚 = 𝜌𝑉, 𝑉_{1} = 𝐴 𝑑𝑥 before compression and 𝑉_{2} = 𝐴(𝑑𝑥 + 𝑑𝜓) after compression as in Fig. 4.4. Thus,
𝜌_{0}𝑉_{1} = (𝜌_{0} + 𝑑𝜌)𝑉_{2}, 𝜌_{0}(𝐴 𝑑𝑥) = (𝜌_{0} + 𝑑𝜌)𝐴(𝑑𝑥 + 𝑑𝜓).
The change in displacement (𝑑𝜓) over distance dx can be written as
𝑑𝜓 = ¡Ó𝜓/¡Ó𝑥 𝑑𝑥, so
𝜌_{0}(𝐴 𝑑𝑥) = (𝜌_{0} + 𝑑𝜌)𝐴(𝑑𝑥 + ¡Ó𝜓/¡Ó𝑥 𝑑𝑥], 𝜌_{0} = (𝜌_{0} + 𝑑𝜌) (1 + ¡Ó𝜓/¡Ó𝑥) = 𝜌_{0} + 𝑑𝜌 + 𝜌_{0} ¡Ó𝜓/¡Ó𝑥 + 𝑑𝜌 ¡Ó𝜓/¡Ó𝑥.
Since the last term is small relative to 𝜌_{0}, a reasonable approximation we can write
𝑑𝜌 ≃ 𝜌_{0} ¡Ó𝜓/¡Ó𝑥.
which we can insert into Eq. (4.10), giving
𝜌_{0}¡Ó^{2}𝜓/¡Ó𝑡^{2} = ¡Ó[(𝐾/𝜌_{0})(𝜌_{0} ¡Ó𝜓/¡Ó𝑥)]/¡Ó𝑥 = ¡Ó[(𝐾(¡Ó𝜓/¡Ó𝑥)]/¡Ó𝑥.
Rearranging makes this into an equation with a familiar form:
(4.11) 𝜌_{0} ¡Ó^{2}𝜓/¡Ó𝑡^{2} = 𝐾 ¡Ó^{2}𝜓/¡Ó𝑥^{2} or ¡Ó^{2}𝜓/¡Ó𝑥^{2} = 𝜌_{0}/𝐾 ¡Ó^{2}𝜓/¡Ó𝑡^{2}.
By comparing the classical wave equation Eq. (2.5) with that in Eq.(4.11). Setting these factors equal to one another gives
(4.12) 1/𝑣^{2} = 𝜌_{0}/𝐾 so 𝑣 = ¡î(𝐾/𝜌_{0}).
So the phase speed of the pressure wave depends on the elastic, that is, the bulk modulus (𝐾) and the inertial density (𝜌_{0}) of the medium.
Example 4.2 Determine the speed of sound in air.
Sound is a type of pressure wave, so we can use Eq. (4.12) to determine the speed of sound in air.
(4.13) 𝑣 = ¡î(𝐾/𝜌_{0}) = ¡î[𝑑𝛲/(𝑑𝜌/𝜌_{0})/𝜌_{0}] = ¡î(𝑑𝛲/𝑑𝜌).
The quantity 𝑑𝛲/𝑑𝜌 can be related to equilibrium pressure (𝛲_{0}) and density (𝜌_{0}) using the adiabatic gas law. The adiabatic law means that we're assuming that the regions of compression and rarefaction produced by the sound wave will not lose or gain energy by heating as the wave oscillates. That's a good assumption for sound waves in air under typical conditon, because the flow of energy by conduction (molecules colliding and transferring kinetic energy) occurs over distances comparable to the mean free path. (the average distance molecules travel between collisions). That distance is several orders of magnitude smaller than the distance between regions of compression of compression and rarefaction (that is, half a wavelength) in sound waves. Thus the wave action may indeed be considered to be an adiabatic process.
To apply the adiabatic gas law, we can write the relationship bwtween pressure (𝛲) and volumn (𝑉) as
(4.14) 𝛲𝑉^{𝛾} = constant,
in which 𝛾 represents the ratio of specific heats as constant pressure and constant volume and the approximately value is 1.4 for air under typical condition. Since volume is inversely proportiol to density 𝜌, Eq. (4.14) can be written as
𝛲 = (constant)𝜌^{𝛾} so 𝑑𝛲/𝑑𝜌 = (constant)𝛾𝜌^{𝛾1} = 𝛾 (constant)𝜌^{𝛾}/𝜌 = 𝛾 𝛲/𝜌.
Inserting this into Eq. (4.13) gives
𝑣 = ¡î(𝛾 𝛲/𝜌)
For typical values for air of 𝛲 = 1 ⨯ 10^{5} Pa and 𝜌 = 1.2 kg/m^{3}, this yields
𝑣 = ¡î1.4 1 ⨯ 10^{5}/1.2] = 342 m/s.
which is very close to the measured value (according to wikipedia, 343 m/s at 20 ^{∘}C).
4.4 Energy and power of mechanical waves _{}^{}
What is propagating in a mechanical wave is "energy". The mechanical energy of a system consists of kinetic energy (as "energy of moton") and potential energy (as "energy of position"). The kinetic energy of a moving object is proportional to the mass of the object and the square of the object's speed, while the object's potential energy on its position in a region in which external forces act upon the object.
Potential enrgy comes in several types such as gravitational energy in a gravitational field and elastic potential energy by an elastic force. As we may recall, a conservative force is a force for which the work done by the force as the object changes depends only on the change in position and not on the path taken by the object. Gravity and elastic forces are conservative, while forces such a friction and drag are noncoservative, because for such dissipative forces, longer paths convert more mechanical energy into internal energy, and we don't get that energy back by getting over the same path in the opposite direction. It's only the change in the potential energy that has any physical significance.
Applying the concepts of kinetic and potential energy to a mechanical wave such as a transverse wave on a string is straightforwaed. The most common approach is to find expressions for the kinetic and potential energies of one small segment of the string, which give the density (that is, the energy per unit length). For harmonic waves, the kinetic energy (𝐾𝐸) of a small segment of the string depends on the segment's mass (𝑚) and the square of the segment's tranverse velocity (𝑣_{𝑡}):
𝐾𝐸_{segment} = 1/2 𝑚𝑣_{𝑡}^{2}.
With a linear mass density of 𝜇 and a lenghth of 𝑑𝑥 is 𝑚 = 𝜇𝑑𝑥, so
𝐾𝐸_{segment} = 1/2 (𝜇 𝑑𝑥)𝑣_{𝑡}^{2}.
The trnasverse velocity of the segmet is 𝑣_{𝑡} = ¡Ó𝑦/¡Ó𝑡, so
(4.15) 𝐾𝐸_{segment} = 1/2 (𝜇 𝑑𝑥) (¡Ó𝑦/¡Ó𝑡)^{2}.
In Fig. 4.3 we can expect the kinetic energy is greatest for those segment passing through the equiibrium position. And to determine the potential energy of segment, it's essential that we keep track of the streching of the string as the segment, because the potential energy is related to the work done by the tension forces that cause the segment to stretch. A sketch of the situation is shown in Fig. 4.6 and we can approximate the segment of length (𝑑𝑠) as 𝑑𝑠 = ¡î(𝑑𝑥^{2} + 𝑑𝑦^{2}).
If we let 𝑑𝑥 approach zero, 𝑑𝑦 can be written as
𝑑𝑦 = ¡Ó𝑦/¡Ó𝑥 𝑑𝑥, 𝑑𝑠 = ¡î[𝑑𝑥^{2} + (¡Ó𝑦/¡Ó𝑥 𝑑𝑥)^{2}] = 𝑑𝑥¡î[1 + (¡Ó𝑦/¡Ó𝑥)^{2}].
This expression can be simplified using the binomial theorem, which says that (1 + 𝑥)^{𝑛} ≈ 1 + 𝑛𝑥
as long as 𝑥 is small relative to one. So we can write
𝑑𝑠 = 𝑑𝑥[1 + (¡Ó𝑦/¡Ó𝑥)^{2}]^{1/2} ≈ 𝑑𝑥[1 + 1/2 (¡Ó𝑦/¡Ó𝑥)^{2}] ≈ 𝑑𝑥 + 1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥.
This means that the segment is stretched by an amount 𝑑𝑠  𝑑𝑥, which is
Amount of strech = 𝑑𝑠  𝑑𝑥 = 1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥.
To find out the work done by elastic (tension) force in stretching the string, we can utilize the fact that the work equals the component of the force multiplied by the amount of stretch. In this case the elastic force is the tension (𝑇) of the spring, so the work is
Work = 𝑇[1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥].
This work is the change in potential energy (𝑃𝐸) of the segment, so the potential energy of segment:
(4.16) 𝑃𝐸_{segment} = 𝑇[1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥].
The total mechanical energy (𝑀𝐸) of any segment of the string is the sum of the segment's knetic and potential energies, which is
𝑀𝐸_{segment} = 1/2 (𝜇 𝑑𝑥) (¡Ó𝑦/¡Ó𝑡)^{2} + 𝑇[1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥].
So the mechanical energy density (the energy per unit length) can be found by dividing the expression by 𝑑𝑥:
(4.17) 𝑀𝐸_{unit length} = 1/2 𝜇 (¡Ó𝑦/¡Ó𝑡)^{2} + 𝑇[1/2 (¡Ó𝑦/¡Ó𝑥)^{2}].
For wavefunction 𝑦(𝑥, 𝑡) = 𝑓(𝑥  𝑣_{phase}𝑡) travelling in the positive direction, as Eq. (2.7)
(4.18) ¡Ó𝑦/¡Ó𝑥 = 1/𝑣_{phase} ¡Ó𝑦/¡Ó𝑡, so
𝑀𝐸_{unit length} = 1/2 𝜇 (¡Ó𝑦/¡Ó𝑡)^{2} + 𝑇[1/2 (1/𝑣_{phase} ¡Ó𝑦/¡Ó𝑡)^{2}] = 1/2 (𝜇 + 𝑇/𝑣_{phase}^{2}) (¡Ó𝑦/¡Ó𝑡)^{2}.
But by Eq. (4.6) 𝑣_{phase} = ¡î(𝑇/𝜇) which means 𝜇 = 𝑇/𝑣_{phase}^{2}, the above equation gives
𝑀𝐸_{unit length} = 1/2 (𝑇/𝑣_{phase}^{2} + 𝑇/𝑣_{phase}^{2}) (¡Ó𝑦/¡Ó𝑡)^{2} = (𝑇/𝑣_{phase}^{2}) (¡Ó𝑦/¡Ó𝑡)^{2}.
Then ¡Ó𝑦/¡Ó𝑡 is the tranverse speed 𝑣_{1}, and we have
(4.19) 𝑀𝐸_{unit length} = (𝑇/𝑣_{phase}^{2}) 𝑣_{1}^{2}.
Example 4.3 What are the kinetic, potential, and mechanical energy of a segment of string of length 𝑑𝑥 with wavefunction 𝑦(𝑥, 𝑡) = 𝛢 sin(𝑘𝑥  𝜔𝑡)?
Tranverse velosity 𝑣_{1} = ¡Ó𝑦/¡Ó𝑡 = 𝛢𝜔 cos(𝑘𝑥  𝜔𝑡) and the slope of wavefunction ¡Ó𝑦/¡Ó𝑡 = 𝛢𝑘 cos(𝑘𝑥  𝜔𝑡), so by Eqs. (4.15) and (4.16)
(4.20) 𝐾𝐸_{segment} = 1/2 (𝜇 𝑑𝑥) (¡Ó𝑦/¡Ó𝑡)^{2} = 1/2 𝜇𝛢^{2}𝜔^{2} cos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥
𝑃𝐸_{segment} = 𝑇[1/2 (¡Ó𝑦/¡Ó𝑥)^{2} 𝑑𝑥] = 𝑇[1/2 𝛢^{2}𝑘^{2} cos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥].
Because 𝑣_{phase} = ¡î(𝑇/𝜇) and 𝑣_{phase} = 𝜔/𝑘, 𝑇 = 𝜇𝜔^{2}/𝑘^{2}. Thus
𝑃𝐸_{segment} = (𝜇 𝜔^{2}/𝑘^{2}) 1/2 𝛢^{2}𝑘^{2} cos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥 or
(4.21) 𝑃𝐸_{segment} = 1/2 𝜇𝛢^{2}𝜔^{2} cos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥.
If we compare Eqs. (4.21) with (4.20), the segment's kinetic and potential energies are identical. So the total energy density:
(4.22) 𝑀𝐸_{segment} = 𝜇𝛢^{2}𝜔^{2} cos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥.
In case of a transvers wave on a string it's the length of a segment relative to the equilbrium length. The stretched length depend on the slope (¡Ó𝑦/¡Ó𝑥) of he segment as in Fig. 4.6. We can see the total energy of the string segments as a function of 𝑥 in Fig. 4.7. In case of a transverse string wave, the segments's kinetic and potential energy both reach their maximum value at the same time when the segment passes through equilbrium position. To find energy in an entire wavelength of the wave, we can integrate Eq. (4.22) over a distance of one wavelength (𝜆):
𝑀𝐸_{one wavelength} = ¡ò_{0}^{𝜆} 𝜇𝛢^{2}𝜔^{2} pcos^{2}(𝑘𝑥  𝜔𝑡)𝑑𝑥,
which can be done by selecting a fixed time such as 𝑡 = 0 and using the dfinite integral
¡ò_{0}^{𝜆} cos^{2}[2¥ð/𝜆 𝑥]𝑑𝑥 = 𝜆/2.
Thus the mechnical energy in each wavelength of a tranverse string wave is
𝑀𝐸_{one wavelength} = 1/2 𝜇𝛢^{2}𝜔^{2}𝜆.
Notice that the mechanical enrgy is proportional to he square of the maximum displacement (𝛢). In case of a pressure wave the energy is proportional to the square of the maximum overpressure of the wave.
Now we can find the power of the wave. Power is defined as the rate of change of energy and has SI units of joules per second or watts, and power of a propagating wave tells us the amount of energy that passes a given location per unit time. Since the product of 𝑀𝐸_{unit wavelength} is the number of joules in the wave per meter of distance and the phase speed (𝑣_{phase}) is the number of meters the the wave moves per second gives the power of the wave:
𝑃 = (𝑀𝐸_{unit wavelength})𝑣_{phase}.
The mechanical energy density is given in Eq. (4.19), so
𝑃 = [(𝑇/𝑣_{phase}^{2}) 𝑣_{𝑡}^{2}] 𝑣_{phase} = (𝑇/𝑣_{phase}) 𝑣_{1}^{2}.
Since 𝑣_{phase} = ¡î(𝑇/𝜇),
𝑃 = [𝑇/¡î(𝑇/𝜇)] 𝑣_{𝑡}^{2} or
(4.23) 𝑃 = [¡î(𝜇𝑇)]𝑣_{𝑡}^{2}.
The quantity ¡î(𝜇𝑇) represents the "impedence" of the medium in which the wave propagates (usually denoted by 𝛧).
Example 4.4 Find the power in a transverse mechnical wave with wave function 𝑦(𝑥, 𝑡) = 𝛢 sin(𝑘𝑥  𝜔𝑡).
For the harmonic wave the tansverse velocity is 𝑣_{phase} = ¡î(𝑇/𝜇) and 𝑣_{phase} = 𝜔/𝑘, so 𝑇 = 𝜇𝜔^{2}/𝑘^{2}. Thus
𝑃 = ¡î(𝜇𝑇) 𝑣_{𝑡}^{2} = ¡î[𝜇 (𝜇 𝜔^{2}/𝑘^{2})] 𝑣_{𝑡}^{2} = 𝜇 𝜔/𝑘 𝑣_{𝑡}^{2}.
But for this wave 𝑣_{phase} = 𝜔𝛢 cos(𝑘𝑥  𝜔𝑡), so
𝑃 = 𝜇 𝜔/𝑘 [𝜔𝛢 cos(𝑘𝑥  𝜔𝑡)]^{2} = 𝜇 𝜔^{3}/𝑘 [𝛢^{2} cos^{2}(𝑘𝑥  𝜔𝑡)].
To find the average power, recall that the average value of cos^{2} over many cyclesis 1/2, so
𝑃_{avg} = 𝜇 𝜔^{3}/𝑘 [𝛢^{2}(1/2)] = 1/2 𝜇𝛢^{2}𝜔^{2}(𝜔/𝑘) or
(4.24) 𝑃_{avg} = 1/2 𝜇𝛢^{2}𝜔^{2}𝑣_{phase} = 1/2 𝑍𝛢^{2}𝜔^{2},
where impedance 𝑍 = ¡î(𝜇𝑇) = ¡î(𝜇^{2}𝜔^{2}/𝑘^{2}) = 𝜇𝑣_{phase}.
4.5 Wave impedance, reflection, and transmission _{}^{}
To understand the physical significance of impedance, consider the force that must be exerted by the source of a mechnaical wave in order to produce the initial displacement of the material of the medium. In case of a transverce mechanical wave, the source must overcome the vertical component of the tension force. If the displacement of the string from the equilibrium positin is 𝑦. the angle of the string with respect to the horizontal is 𝜃, the vertical componet of the tension force is
𝐹_{𝑦} = 𝑇 sin 𝜃 ≈ 𝑇 ¡Ó𝑦/¡Ó𝑥.
For any single traveling wave, we can use Eq. (4.18)
𝐹_{𝑦} = 𝑇 (1/𝑣_{phase}) ¡Ó𝑦/¡Ó𝑡, and since ¡Ó𝑦/¡Ó𝑡 = 𝑣_{𝑡}
𝐹_{𝑦} = 𝑇 (1/𝑣_{phase}) 𝑣_{𝑡} = (𝑇/𝑣_{phase}) 𝑣_{𝑡} or since 𝑣_{phase} = ¡î(𝑇/𝜇),
(4.25) 𝐹_{𝑦} = [𝑇/¡î(𝑇/𝜇)] 𝑣_{𝑡} = ¡î(𝜇𝑇) 𝑣_{𝑡}.
Thus the force needed to generated the wve is proportional to the transverse velocity of the string. Usng 𝐹_{𝑦,source} = 𝐹_{𝑦} helps make the meaning of impedance clearer:
(4.25) 𝑍 = ¡î(𝜇𝑇) = 𝐹_{𝑦,source}/𝑣_{𝑡}.
So, for a mechanical wave, the impedance tells you the amount of force necessicty to produce a given transverse velocity of the material displaced by the wave. In SI units, 𝑍 gives number of newtons needed to cause the material to move at a speed of m/s. But the higherimpedeance string has more power than a lowerimpedace string with the same transverse velocity as we can see in Eq. (4.21).
The analysis of the effect of changing media on a mechnical wave from Eq. (4.25) is the following:
(1) The medium produces a drag force on the source of the wave.
(2) hat drag force is proportional to the transverse velocity produced in the medium by the wave, and in the opposite direction (so 𝐹_{𝑦} 𝛼 𝑣_{𝑡}).
(3) The constant of proportionality between the force and the tansverse velocity is the impedance (𝑍) of the medium.
If the wave source is at the left end of the string and the right end is some finitedistance away, at the right end , the string might be clamped to a wall, it might be free, or it might be attached to another string with different linear density 𝜇 or elastic force, tension 𝑇. These situation are shown in Fig. 4.8.
And take a look at the situations shown in Fig. 4.9.. The first case shows an infinite string and that string presents a drag force of 𝑍𝑣_{1} to the source of the wave. Here there is no aaditional waves. The below cases are with dashed portions of string which are eliminated. What could we attach to the string at the right end that would produce a drag force equal to that of the missing portion of the string? The second case with a hanging mass would not be proportional to the transverse velocity of the incoming wave but the transverse acceleration according to Newton's second law. The third case with a spring would be proportional to the displacement accordng to Hooke's law. The last case with a "dashpot"  a mechanical device which resists motion using viscous friction would be proportional to the velocity but acts in the opposite direction. On replacing the missing portion of the infinite spring with a dashpot, the drag force has the same dependence on 𝑣_{1}. We can think of that as adjusting impedance of the dashpot to match the impedance of the string. The dashpot is described also as a "purely resistive" device because if we terminate a finite string with a dashpot with the matching impedance with the string, the dashpot act a perfect absorber of the wave energy.
Now imagin hooking the right end of the string to another string, as shown Fig. 4. 10. If the second string has different values of linear mass density 𝜇_{2} or tenstion 𝛵_{2}, and thus different impedance 𝑍_{2}, we can determine how the wave will behave as follows. But it's necessary to enforce two boundary condition:
(1) the string is continuous, so the displacement (𝑦) must be the same on either side of the interface.
(2) the tangential force (𝑇 ¡Ó𝑦/¡Ó𝑥) must be the same on either side of the interface.
Applying these boundary conditions leads to the following equation (we can see the details on thetextbook's website):
𝑦_{reflected} = (𝑍_{1}  𝑍_{2})/(𝑍_{1} + 𝑍_{2}) 𝑦_{incident},
where the factor (𝑍_{1}  𝑍_{2})/(𝑍_{1} + 𝑍_{2}) is called the amplitude reflection coefficient (𝑟):
(4.27) 𝑟 = (𝑍_{1}  𝑍_{2})/(𝑍_{1} + 𝑍_{2}).
So if 𝑟 = 1, the reflected wave has the same amplitude as the incident wave, and if 𝑟 = 0, there is no reflected wave. Alternatively 𝑟 can be negative and if 𝑟 = 1, for example, the reflected wave has the same amplitude as the incident wave.
If we want know the "tranmitted wave" which propagates past the interface, a simliar analysis show that
𝑦_{transmitted} = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2}) 𝑦_{incident},
So calling the amplitude transmission coefficient 𝑡, we can write
(4.28) 𝑡 = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2}).
This tells us how the amplitue of the transmitted wave compares with the amplitude of the incident wave. If 𝑡 = 1, the transmitted wave has the same amplitude as the incident wave. But if 𝑡 = 0, the amplude of the transmitted wave is zero. For any interfce, 𝑡 = 1 + 𝑟, so so the values of 𝑡 can range from 0 to +2.
To apply the equation for 𝑟 and 𝑡 to a string with the right end clampled to a fixed position, consider that this is equivalent to making 𝑍_{2} = ¡Ä. In this case
𝑟 = (𝑍_{1}  𝑍_{2})/(𝑍_{1} + 𝑍_{2}) = (𝑍_{1}  ¡Ä)/(𝑍_{1} + ¡Ä) = 1 and 𝑡 = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2}) = 2𝑍_{1}/(𝑍_{1} + ¡Ä) = 0.
So here none of the incident wave's amplitude is transmitted past the interface, and the relected wave ia an copy of the incident wave.
In the case which the right end of the string is left free by letting 𝑍_{2} = 0, the reflected wave has the same amplitude as the incident wave (𝑟 = 1).
Example 4.5 Consider a transverse pulse with maximum displacement of 2 cm propagating in the positive xdirection on a string with mass density 0.15 g/cm and tension 10 N. What happens if the pulse encounter a short section of string with twice the mass density and the same tension?
A sketch of this situation is shown in Fig. 4.11. 𝑍_{1} = 𝑍_{light} and 𝑍_{2} = 𝑍_{heavy}.
𝑍_{1} = ¡î(𝜇_{light}𝛵_{light}) = ¡î[(0.0015kg/m){10N)] = 0.387 kg/s,
𝑍_{2} = ¡î(𝜇_{heavy}𝛵_{heavy}) = ¡î[2(0.0015kg/m){10N)] = 0.548 kg/s.
𝑡 = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2}) = (2)(0.387)/(0.387 + 0.548) = 0.83.
Thus in propagating from the light string to the heavy string the amplitude of the pulseis reduced to 83% of its original value at first (left) interface. But at the second (right) interface the impedances are 𝑍_{1} = 0.548 kg/s and 𝑍_{2} = 0.387kg/s. This means thetransmission coefficient at the right interface
𝑡 = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2}) = (2)(0.548)/(0.548 + 0.387) = 1.2,
so the amplitudeis is reduced by a factor 0.83 times 1.2 and the final one is about 97% of its original value of 2 cm.
Consider what happens if we substract the amplitude reflection coefficient 𝑟 from the amplitude transmission 𝑡:
𝑡  𝑟 = 2𝑍_{1}/(𝑍_{1} + 𝑍_{2})  (𝑍_{1}  𝑍_{2})/(𝑍_{1} + 𝑍_{2}) = (2𝑍_{1}  𝑍_{1} + 𝑍_{2})/(𝑍_{1} + 𝑍_{2}) = (𝑍_{1} + 𝑍_{2})/(𝑍_{1} + 𝑍_{2}) = 1 or 𝑡 = 1 + 𝑟.
In case of 𝑍_{2} much larger than 𝑍_{1} 𝑟 approaches 1 and 𝑡 = 1 + 𝑟 = 1 + (1) = 0, so the amplitude of transmitted wave should be zero. But in case of 𝑍_{1} much larger than 𝑍_{2} 𝑟 approaches +1 and 𝑡 = 1 + 1 = 2. How can the amplitude of the transmitted wave possibly be 2? To understand the answer we have to consider the energy being carried by the reflected and transmitted waves. Recalling Eq. (4.24) that power of the wave 𝑃 is proportional to 𝑍𝐴^{}: 𝑃 ¡ð 𝑍𝐴^{2}. Thus the ratio of the power in the transmitted wave to the power to the incident wave, called the power transmission coefficient 𝑇 is
(4.29) 𝑇 = 𝑃_{transmitted}/𝑃_{incident} = 𝑍_{2}𝐴^{}_{transmitted}/𝑍_{1}𝐴^{}_{incident} = (𝑍_{2}/𝑍_{1}) 𝑡^{2},
since 𝑡 is the ratio of the transmitted to the incident amplitude. But the reflected wave travels in the same medium as the incident wave, so the power reflection coefficient 𝑅 is
(4.30) 𝑅 = 𝑃_{reflected}/𝑃_{incident} = 𝑍_{1}𝐴^{}_{reflected}/𝑍_{1}𝐴^{}_{incident} = (𝑍_{1}/𝑍_{1}) 𝑟^{2} = 𝑟^{2},
So, for example, in case of 𝑍_{2} = 0, the power is entirely in the reflected wave since 𝑅 = 1 with noneof the power in the transmitted wave since 𝑇 = 0. Since 𝑅 represented the fraction of the power of the incoming wave that is reflected and 𝑇 representedthe fraction that is transmitted, the sum 𝑅 + 𝑇 must equal one.
(1) Small enough to allow us to make approximations such as cos 𝜃 ≈ 1 and sin 𝜃 ≈ tan 𝜃, which are good to within 10% if 𝜃 is less than 25^{∘}.
(2) In Chapter 1 "displacement" may refer to any deviation from equilibrium, but in this case, 𝑦 is the actual distance from equilibrium position.
* Textbook: D. Fleisch & J. Kinnaman A Student's Guide to Waves (Cambridge University Press 2015) 

