1.
A
tension F is working on a rope, the
mass per unit length of the rope is ρo
[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 $\frac{\partial {{y}_{o}}}{\partial
x}=\frac{\partial {{y}_{1}}}{\partial x}$.
In this figure, Yi, Yr, and Yt 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 ${{k}_{t}}=\omega \sqrt{\frac{{{\rho
}_{1}}}{F}}$ → ki = kr.
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 Aor 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 ki and kt , 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 Ftr(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 Fs.
a.1. Give an
expression for the propagation velocity vtr(x,t).
a.2. The resultant of
the horizontal tension Fh
and the driving force Ftr
at the begin point of the rope is in balance with the force Fs that works on the rope.
Give a sketch of the equilibrium of forces and define the relationship between Ftr and Fs.
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 dEk(x,t) and
potential energy dEp(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. $dE(x,t)=d{{E}_{k}}(x,t)+d{{E}_{p}}(x,t)={{A}^{2}}{{F}_{s}}{{\left(
\frac{dg(u)}{du} \right)}^{2}}dx$ →
$c\frac{dE(x,t)}{dx}=c{{A}^{2}}{{F}_{s}}{{\left( \frac{dg(u)}{du}
\right)}^{2}}=p(t)$
Posts
