Solved Problems III (Wave Phenomenon)

1.      A tension F is working on a rope, the mass per unit length of the rope is ρ_o [kg/m]; the influence of the gravitational force is neglected. A transverse harmonic wave propagates along the rope with frequency $f={}^{2\pi }/{}_{\omega }$ . The wave propagates in the positive x-direction; the disturbance is in the y-direction. The complex representation of harmonic disturbance in general is given by: $Y(x,\omega )=A{{e}^{\pm ikx}}$ with propagation number k is given by $\omega \sqrt{\frac{\rho }{F}}$. Give with the help of complex representation the expression of relative transmitted and reflected wave amplitude at the transition area at x=0, with the help of the given boundary condition.

Solution:

The boundary condition at x=0 is given by: ${{y}_{o}}={{y}_{1}}$  and $\frac{\partial {{y}_{o}}}{\partial x}=\frac{\partial {{y}_{1}}}{\partial x}$.

Please see figure below:

 In this figure, Y_i, Y_r, and Y_t are the complex representation of incoming, reflected and transmitted wave respectively and can be given as follow:

${{Y}_{i}}(x,\omega )={{A}_{oi}}{{e}^{i{{k}_{i}}x}}$; ${{Y}_{r}}(x,\omega )={{A}_{or}}{{e}^{-i{{k}_{r}}x}}$ ; and ${{Y}_{t}}(x,\omega )={{A}_{ot}}{{e}^{i{{k}_{t}}x}}$.

At $x=0$ : ${{y}_{o}}={{y}_{1}}$ → ${{Y}_{i}}+{{Y}_{r}}={{Y}_{t}}$ →  ${{A}_{oi}}{{e}^{i{{k}_{i}}x}}+{{A}_{or}}{{e}^{-i{{k}_{r}}x}}={{A}_{ot}}{{e}^{i{{k}_{t}}x}}$ for  x=0 : ${{A}_{oi}}+{{A}_{or}}={{A}_{ot}}$

For the second boundary condition ($\frac{\partial {{y}_{o}}}{\partial x}=\frac{\partial {{y}_{1}}}{\partial x}$ ) at x=0: $\frac{\partial }{\partial x}\left( {{Y}_{i}}+{{Y}_{r}} \right)=\frac{\partial }{\partial x}\left( {{Y}_{t}} \right)$ →

${{A}_{oi}}i{{k}_{i}}{{e}^{i{{k}_{i}}x}}-{{A}_{or}}i{{k}_{r}}{{e}^{-i{{k}_{r}}x}}={{A}_{ot}}i{{k}_{t}}{{e}^{-i{{k}_{t}}x}}$ and at x=0 : ${{A}_{oi}}{{k}_{i}}-{{A}_{or}}{{k}_{r}}={{A}_{ot}}{{k}_{t}}$  since ${{A}_{ot}}={{A}_{oi}}+{{A}_{or}}$ we obtain: ${{A}_{oi}}{{k}_{i}}-{{A}_{or}}{{k}_{r}}=\left( {{A}_{oi}}+{{A}_{or}} \right){{k}_{t}}$ . It’s given that: $k=\omega \sqrt{\frac{\rho }{F}}$ ; equivalently: ${{k}_{i}}=\omega \sqrt{\frac{{{\rho }_{o}}}{F}}$ ; ${{k}_{r}}=\omega \sqrt{\frac{{{\rho }_{o}}}{F}}$ and ${{k}_{t}}=\omega \sqrt{\frac{{{\rho }_{1}}}{F}}$  → k_i = k_r.

            Therefore: $\left( {{A}_{oi}}-{{A}_{or}} \right)\omega \sqrt{\frac{{{\rho }_{o}}}{F}}=\left( {{A}_{oi}}+{{A}_{or}} \right)\omega \sqrt{\frac{{{\rho }_{o}}}{F}}$  and we obtain the reflection coefficient r:

            $r=\frac{{{A}_{or}}}{{{A}_{oi}}}=\frac{\sqrt{{{\rho }_{o}}}-\sqrt{{{\rho }_{1}}}}{\sqrt{{{\rho }_{o}}}+\sqrt{{{\rho }_{1}}}}$ .

Since ${{A}_{ot}}={{A}_{oi}}+{{A}_{or}}$ → ${{A}_{or}}={{A}_{ot}}-{{A}_{oi}}$; substituting this equation for A_or into ${{A}_{oi}}{{k}_{i}}-{{A}_{or}}{{k}_{r}}={{A}_{ot}}{{k}_{t}}$  we then obtain: ${{A}_{oi}}{{k}_{i}}-\left( {{A}_{ot}}-{{A}_{oi}} \right){{k}_{i}}={{A}_{ot}}{{k}_{t}}$ → ${{A}_{oi}}-{{A}_{ot}}{{k}_{i}}+{{A}_{oi}}{{k}_{i}}={{A}_{ot}}{{k}_{t}}$  we simplify further this equation, substitute the expression for k_i and k_t , and then we obtain: $2{{A}_{oi}}\omega \sqrt{\frac{{{\rho }_{o}}}{F}}={{A}_{ot}}\left( \sqrt{{{\rho }_{o}}}+\sqrt{{{\rho }_{1}}} \right)\frac{\omega }{\sqrt{F}}$  

Thus, the transmission coefficient t is: $t=\frac{{{A}_{ot}}}{{{A}_{oi}}}=\frac{2\sqrt{{{\rho }_{o}}}}{\sqrt{{{\rho }_{o}}}+\sqrt{{{\rho }_{1}}}}$.

2.      A transverse force F_tr(t) is working on the begin point O of a long rope, such that a wave propagates in the positive x-direction. The disturbance in the y-direction is given by: $y(x,t)=Ag(x-ct)$ with c the propagation velocity of the wave. Furthermore, it is also given that$\left| g \right|<1$, the mass of the rope per unit length is ρ, and the tension is F_s.

a.1. Give an expression for the propagation velocity v_tr(x,t).

a.2. The resultant of the horizontal tension F_h and the driving force F_tr at the begin point of the rope is in balance with the force F_s that works on the rope. Give a sketch of the equilibrium of forces and define the relationship between F_tr and F_s.

b. The power (as a function of time t) that works on the rope is given by: $p(t)={{F}_{tr}}(t){{v}_{tr}}(t)$ . Define p(t) in term of function g(u) with $u=x-ct$ . Hint: use chain rule for instance, $\frac{\partial y}{\partial x}=\frac{dy}{du}\frac{\partial u}{\partial x}=A{g}'(u)\frac{\partial u}{\partial x}$ .

c. Show that when there’s only transverse wave, the ratio ${}^{{{F}_{tr}}(t)}/{}_{{{v}_{tr}}(t)}$ does not depend on time and is equal to the ‘characteristic resistant’ $Z=\sqrt{\rho F}$ .

d. Give the kinetic energy dE_k(x,t) and potential energy dE_p(x,t) of an infinitesimal part of the rope dx in terms of g(u).

e. If the total energy works on the infinitesimal part of the rope dx is given by: $dE(x,t)=d{{E}_{k}}(x,t)+d{{E}_{p}}(x,t)$; show that for the energy density at x=0: $p(t)={{{}^{cdE(x,t)}/{}_{dx}}_{x=0}}$.

Solution:

a.1. ${{v}_{tr}}(x,t)=\frac{\partial y}{\partial t}=-Ac\frac{\partial g(x-ct)}{\partial t}$  with: $u=x-ct$  → ${{v}_{tr}}=A\frac{dg(u)}{du}\frac{\partial u}{\partial t}=-Ac\frac{\partial g(u)}{\partial u}$

a.2. The x-component:

 ${{F}_{s}}\cos \alpha -{{F}_{h}}=0$ → ${{F}_{s}}\cos \alpha ={{F}_{h}}$

            The y-component:

 ${{F}_{tr}}+{{F}_{s}}\sin \alpha =0$ → ${{F}_{tr}}=-{{F}_{s}}\sin \alpha$

       

For small angle α: $\sin \alpha \approx \alpha $ → ${{F}_{tr}}\approx -{{F}_{s}}\alpha $ ; since $\tan \alpha =\frac{\partial y}{\partial x}\approx \alpha $ ; we then obtain:   ${{F}_{tr}}\approx -{{F}_{s}}\frac{\partial y}{\partial x}=-{{F}_{s}}A\frac{\partial g(u)}{\partial u}\frac{\partial u}{\partial x}=-{{F}_{s}}A\frac{\partial g(u)}{\partial u}$.        

            b. $p(t)={{F}_{tr}}(t){{v}_{tr}}(t)=-{{F}_{s}}A\frac{\partial g(u)}{\partial u}\cdot -Ac\frac{\partial g(u)}{\partial u}$ therefore, the power is given by:

            $p(t)={{F}_{s}}{{A}^{2}}c{{\left( \frac{\partial g(u)}{\partial u} \right)}^{2}}$

c. $\frac{{{F}_{tr}}(t)}{{{v}_{tr}}(t)}=\frac{-{{F}_{s}}A\frac{\partial g(u)}{\partial u}}{-Ac\frac{\partial g(u)}{\partial u}}=\frac{{{F}_{s}}}{c}$  from the 1-dimensional wave equation: $\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=\frac{\rho }{{{F}_{s}}}\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$  :

    → $\frac{1}{c}=\sqrt{\frac{\rho }{{{F}_{s}}}}$ → $c=\sqrt{\frac{{{F}_{s}}}{\rho }}$  therefore: $\frac{{{F}_{tr}}(t)}{{{v}_{tr}}(t)}=\frac{{{F}_{s}}}{c}=\frac{{{F}_{s}}}{\sqrt{{}^{{{F}_{s}}}/{}_{\rho }}}=\sqrt{\rho {{F}_{s}}}$.

d. The kinetic energy is given by: $d{{E}_{k}}(x,t)=\frac{1}{2}\rho dx{{(\frac{\partial y}{\partial t})}^{2}}=\frac{1}{2}\rho dx{{(-Ac\frac{dg(u)}{du})}^{2}}$

    → $d{{E}_{k}}(x,t)=\frac{1}{2}\rho {{A}^{2}}\frac{{{F}_{s}}}{\rho }{{\left( \frac{dg(u)}{du} \right)}^{2}}dx=\frac{1}{2}{{A}^{2}}{{F}_{s}}{{\left( \frac{dg(u)}{du} \right)}^{2}}dx$

    The potential energy is given by: $d{{E}_{p}}(x,t)={{F}_{s}}(ds-dx)={{F}_{s}}dx\left( \frac{ds}{dx}-1 \right)$

    From figure in a.2 : $\cos \alpha =\frac{dx}{ds}$  and trigonometry identity leads us to:

    $d{{E}_{p}}(x,t)={{F}_{s}}\left( \sqrt{{{\tan }^{2}}\alpha +1}-1 \right)dx$  Taylor expansion of $\sqrt{{{\tan }^{2}}\alpha +1}$  gives:

    $d{{E}_{p}}(x,t)\approx {{F}_{s}}\left( 1+{}^{1}/{}_{2}{{\tan }^{2}}\alpha -1 \right)dx=\frac{1}{2}{{F}_{s}}{{\tan }^{2}}\alpha dx$  therefore:

    $d{{E}_{p}}(x,t)=\frac{1}{2}{{F}_{s}}{{\left( \frac{dy}{dx} \right)}^{2}}dx=\frac{1}{2}{{F}_{s}}{{A}^{2}}{{\left( \frac{dg(u)}{du} \right)}^{2}}dx$  

    e. $dE(x,t)=d{{E}_{k}}(x,t)+d{{E}_{p}}(x,t)={{A}^{2}}{{F}_{s}}{{\left( \frac{dg(u)}{du} \right)}^{2}}dx$ →

    $c\frac{dE(x,t)}{dx}=c{{A}^{2}}{{F}_{s}}{{\left( \frac{dg(u)}{du} \right)}^{2}}=p(t)$  

     

 

6. Superposition Principle and Complex Representation

Let us say that ψ₁ and ψ₂ are each separate solutions of one-dimensional differential wave equation $\frac{{{\partial }^{2}}\psi }{\partial {{x}^{2}}}=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}\psi }{\partial {{t}^{2}}}$ (see also Equation (3.5)) . The superposition principle says that: the sum of these two wave-functions, i.e. $\left( {{\psi }_{1}}+{{\psi }_{2}} \right)$ , is also a solution of the wave equation. This principle can be proven as follows:

$\frac{{{\partial }^{2}}{{\psi }_{1}}}{\partial {{x}^{2}}}=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}{{\psi }_{1}}}{\partial {{t}^{2}}}$                                                                (6.1)

$\frac{{{\partial }^{2}}{{\psi }_{2}}}{\partial {{x}^{2}}}=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}{{\psi }_{2}}}{\partial {{t}^{2}}}$                                                               (6.2)

Adding Equation (6.1) and (6.2) yields:

$\frac{{{\partial }^{2}}{{\psi }_{1}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}{{\psi }_{2}}}{\partial {{x}^{2}}}=\frac{1}{{{v}^{2}}}\left( \frac{{{\partial }^{2}}{{\psi }_{1}}}{\partial {{t}^{2}}}+\frac{{{\partial }^{2}}{{\psi }_{2}}}{\partial {{t}^{2}}} \right)$                                              (6.3)

And then, by simplifying Equation (6.3) we obtain:

$\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\left( {{\psi }_{1}}+{{\psi }_{2}} \right)=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}}{\partial {{t}^{2}}}\left( {{\psi }_{1}}+{{\psi }_{2}} \right)$                                             (6.4)

 Equation (6.4) establishes that $\left( {{\psi }_{1}}+{{\psi }_{2}} \right)$ is definitely a solution.

This means that when two separate waves arrive at the same place in space wherein they overlap, both waves will add (or subtract from) one another without disrupting either wave. By taking the algebraic sum of the individual constituent waves at a certain location, the disturbance at each point in the region of overlap can be calculated. For instance, when the two constituent waves rise and fall in-step, reinforcing each other, they are said to be in-phase, i.e. their phase-angle difference is zero (see Figure 6.1. below).

Figure 6.1. The superposition of two sinusoidal in-phase waves with amplitudes 1.0 and 0.9.

In contrast to in-phase sinusoidal waves, there are also waves that tend to diminish each other, the so-called the out-of-phase sinusoidal waves when the phase-angle difference equals π (see Figure 6.2. below).

Figure 6.2. The superposition of two sinusoidal out-of-phase waves with amplitudes 1.0 and 0.9.

The complex number representation that describes a wave function offers an alternative description that is mathematically simpler to process. The complex number $\tilde{z}$ has the form:

$\tilde{z}=x+iy$                                                            (6.5)

Where $\sqrt{-1}=i$. Both x and y are real numbers, and they can also be expressed in terms of real and imaginary parts of $\tilde{z}$ respectively, i.e. Re $\left( {\tilde{z}} \right)$ and Im $\left( {\tilde{z}} \right)$.

The complex number $\tilde{z}$ in Equation (6.5) can be expressed graphically in the Argand diagram as given in Figure 6.3.

Figure 6.3. A representation of a complex number in an Argand diagram. In terms of x and y (a). In terms of polar coordinates r and θ  (b). Figure (c) illustrates the complex conjugate $\tilde{z}*$ of $\tilde{z}$. And figure (d) shows when θ is a constantly changing function of time, the arrow rotates at a rate ω.

In terms of polar coordinates (r,θ), the real and imaginary parts in Equation (6.5) can be expressed:

$x=r\cos \theta $                                                                 (6.6)

$y=r\sin \theta $                                                                  (6.7)

$\tilde{z}=x+iy=r(\cos \theta +i\sin \theta )$                                                  (6.8)

Where r denotes the distance from the center of Argand diagram to a point in complex plane (Figure 6.3.(b)), i.e. $r=\left| {\tilde{z}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ which is also defined as the modulus of complex number $\tilde{z}$. The angle θ is the argument of complex number $\tilde{z}$ which varies between 0 and 2π ($0\le \arg \tilde{z}\le 2\pi $). Therefore complex number $\tilde{z}$ in Equation (6.8) can be written:

$\tilde{z}=\left| {\tilde{z}} \right|[\cos (\arg \tilde{z})+i\sin (\arg \tilde{z})]$                (6.9)

Euler formula defines that a complex number (in term of polar coordinate) can be expressed in term of complex exponential function:

${{e}^{i\theta }}=\cos \theta +i\sin \theta $                                                     (6.10)

Equivalently, for an exponential function with negative power:

${{e}^{-i\theta }}=\cos \theta -i\sin \theta $                                                    (6.11)

From Euler formula, it can be shown that:

$\cos \theta =\frac{{{e}^{i\theta }}+{{e}^{-i\theta }}}{2}$                                                       (6.12)

And

$\sin \theta =\frac{{{e}^{i\theta }}-{{e}^{-i\theta }}}{2i}$                                                      (6.13)

Euler formula also allows us to rewrite the complex number $\tilde{z}$  in terms of polar coordinates (r,θ) as follows:

$\tilde{z}=r{{e}^{i\theta }}=r\cos \theta +ir\sin \theta $                                                (6.14)

For the complex conjugate $\tilde{z}*$ (see Figure 6.3.(c)) applies:

$\tilde{z}*=x-iy$                                                            (6.15)

$\tilde{z}*=r(\cos \theta -i\sin \theta )$                                                     (6.16)

And

$\tilde{z}*=r{{e}^{-i\theta }}$                                                           (6.17)

Notice that, in term of a complex number and its corresponding complex conjugate, the modulus of a complex number can be expressed as follows: $r=\left| {\tilde{z}} \right|\equiv \sqrt{\tilde{z}\tilde{z}*}$.

Addition and substraction of two complex numbers ${{\tilde{z}}_{1}}$ and ${{\tilde{z}}_{2}}$ are straight forward:

${{\tilde{z}}_{1}}\pm {{\tilde{z}}_{2}}=({{x}_{1}}+i{{y}_{1}})\pm ({{x}_{2}}+i{{y}_{2}})=({{x}_{1}}\pm {{x}_{2}})+i({{y}_{1}}\pm {{y}_{2}})$                       (6.18)

For multiplication and division, the following rules of thumb can be applied:

${{\tilde{z}}_{1}}{{\tilde{z}}_{2}}={{r}_{1}}{{r}_{2}}{{e}^{i({{\theta }_{1}}+{{\theta }_{2}})}}$                                                      (6.19)

$\frac{{{{\tilde{z}}}_{1}}}{{{{\tilde{z}}}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i({{\theta }_{1}}-{{\theta }_{2}})}}$                                                        (6.20)

Another important equation is:

${{e}^{z}}={{e}^{x+iy}}={{e}^{x}}{{e}^{iy}}={{e}^{x}}(\cos y+i\sin y)$                                         (6.21)

Therefore

$\left| {{e}^{z}} \right|={{e}^{x}}$                                                            (6.22)

And for $n\in \mathbb{Z}$ 

$\arg ({{e}^{z}})=y+2\pi n$                                                        (6.23)

For all ${{\tilde{z}}_{1}},{{\tilde{z}}_{2}}\in \mathbb{Z}$  the following equation is also valid:

${{e}^{{{{\tilde{z}}}_{1}}}}{{e}^{{{{\tilde{z}}}_{2}}}}={{e}^{{{{\tilde{z}}}_{1}}+{{{\tilde{z}}}_{2}}}}$ .                                                              (6.24)