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

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