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\begin{document}
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\heading{Physics 3 for physics - Quiz 2006/7 - solution}
\auname{Alexander Stotland}
{\bf Question 1:}
The mass movement along the $y$ axis produces a wave in the string propagating
in the $+x$ direction.
The force applied by the string to the mass is
\[
F=T\pd{y}{x}
\]
For a wave propagating in the $+x$ direction $y(x,t)=A\eexp{\imath(\omega t-kx)}$ the force can be written as
\[
F=-Z\pd{y}{t}
\]
where $y$ is the transversal displacement of the string, and
$Z=\sqrt{\rho T}=5\ \mbox{kg/sec}$ is the impedance of the string. It can be
seen that this force acts on the mass as a friction force which is proportional
to the velocity of the mass.
\Dn
The solution of
\[
\ppd{y}{t} + \Gamma \pd{y}{t} +\omega_0^2 y = 0
\]
is
\[
y(t)=A\eexp{-\frac{\Gamma}{2}t}\cos(\omega t + \phi)
\]
where $\omega^2=\omega_0^2-\left(\frac{\Gamma}{2}\right)^2$.
In our case $\Gamma = \frac{Z}{M}=2.5\ \mbox{sec}^{-1}$ and
$\omega=\sqrt{\frac{k}{M}-\left(\frac{\Gamma}{2}\right)^2}=
\sqrt{\frac{32}{2}-\left(\frac{2.5}{2}\right)^2}\simeq 3.8\ \mbox{sec}^{-1}$.
So that the period is: $T=\frac{2\pi}{\omega} \simeq 1.65 \mbox{sec}$.
The amplitude lowers to the $1/e$ of its initial value for
$\eexp{-\frac{\Gamma}{2}t}=\eexp{-1}$. In our case it happens at
$t=\frac{2}{\Gamma}=\frac{2M}{Z}=0.8\;\mbox{sec}$.
The phase velocity in the string equals to the group velocity.
\[
v = \sqrt{\frac{T}{\rho}} = 20\;\mbox{m/s}
\]
So the pulse travels $z = vt = 16\;\mbox{m}$
Since the force $F=-Z\pd{y}{t}$ decays with time, the amplitude of the
produced wave at $z=0$ decays, so that at larger $z$ the amplitude is larger
(since it was generated at earlier times).
We are interested at the times $t \lesssim T/2$. In the first quarter of
the period the string is above the $x$ axis and and it starts moving from $A$
and in the second quarter the
string is below the $x$ axis and the largest amplitude is $A/e$.
\includegraphics[width=0.7\hsize]{quiz_99_1_2007a_A_s_p0}
In order to find the the propagating wave $y(x,t)$ in the string, first of all
we write the boundary conditions
%
\begin{eqnarray*}
y(x=0,t)=B\eexp{-\frac{\Gamma}{2}t}\cos(\omega t + \phi)
\end{eqnarray*}
%
At $t=0$ the mass is at $A=0.05$:
%
\begin{eqnarray*}
y(x=0,t=0)=B\eexp{-\frac{\Gamma}{2}t}\cos(\omega t + \phi)|_{t=0} = B\cos{\phi} = A
\end{eqnarray*}
%
and its velocity is zero:
%
\begin{eqnarray*}
\dot y(x=0,t=0)=B\eexp{-\frac{\Gamma}{2}t}
\left(-\frac{\Gamma}{2}\cos(\omega t + \phi) - \omega\sin(\omega t + \phi) \right)|_{t=0}
= B \left(-\frac{\Gamma}{2}\cos\phi - \omega\sin\phi \right) = 0
\end{eqnarray*}
%
So that
%
\begin{eqnarray*}
\tan\phi &=& -\frac{\Gamma}{2\omega} \\
\cos^2\phi &=& \frac{1}{\tan^2\phi+1}
\end{eqnarray*}
%
In non-dispersive medium the wave should satisfy the
d'Alembert solution of the wave equation $y(x,t) = y(x-vt)$ for the wave
propagating in the positive direction. Or, in other words, if $t'=t-\frac{x}{v}$
then $y(x,t)=y(0,t')$. Therefore,
\[
y(x,t)=A\sqrt{1+\left(\frac{\Gamma}{2\omega}\right)^2}\ \eexp{-\frac{\Gamma}{2}\left(t-\frac{x}{v}\right)}\cos\left(\omega t - kx - \arctan\left(\frac{\Gamma}{2\omega}\right)\right)
\]
where $\omega = v k$. Substituting the numerical values into the expression we get
\[
y(x,t)\simeq 0.053\ \eexp{-1.25\left(t-\frac{x}{20}\right)}\cos(3.8 t - 0.21x - 0.31)
\]
\newpage
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{\bf Question 2:}
\Dn
The unstretched springs have a length $L$.
After stretching the springs the tension in the spring {\bf A} is
\[
T_A = k_A (2L-L) = k_A L
\]
and the tension in the spring {\bf B} is
\[
T_B = k_B (3L-L) = 2k_B L
\]
Since the tensions should be equal at the attachment point $k_A=2k_B$.
\Dn
{\bf Transversal vibrations}
After stretching the springs the mass densities are:
\[
\rho_A=\frac{M}{2L}, \;\;\;
\rho_B=\frac{M}{3L}
\]
So that $\rho_A = \frac32 \rho_B$. The impedance for transversal vibrations is
$Z=\sqrt{\rho T}$. Then
\[
Z_A=\sqrt{\rho_A T}=\sqrt{\frac32 \rho_B T}=\sqrt{\frac32} Z_B
\]
The reflection coefficient is:
\[
R = \frac{Z_A-Z_B}{Z_A+Z_B}=\frac{\sqrt{\frac32}-1}{\sqrt{\frac32}+1}\simeq
0.101
\]
The trasmission coefficient is:
\[
T=1+R\simeq 1.101
\]
{\bf Longitudinal vibrations}
For longitudinal vibrations the impedance is $Z=\sqrt{K M}$.
\[
Z=\sqrt{K M} = \sqrt{(K a)\times(M/a)} = \sqrt{\rho T}
\]
where $a$ is the length of a spring of mass $M$ and the constant $K$. For the two springs $T$ is equal, but $\rho$ is different (like in the previous case).
Therefore, the reflection and transmission coefficients are the same for longitudinal and transversal oscillations.
\[
R = \frac{Z_A-Z_B}{Z_A+Z_B}= \simeq 0.101, \;\;\;
T=1+R\simeq 1.101
\]
{\em Comment:}
The reflection coefficient (and the impedance) can be derived in the following way:
Write the dispersion relation for a chain of masses $M$ connected by springs $K$
(${\omega = 2\sqrt{\frac{K}{M}}\sin\left(\frac12 ka\right)}$). Now suppose there two such chains connected at $x=0$: one chain with the coefficient $K_1$ and the second one with $K_2$. Assume a harmonic wave coming from the right, write the equation of motion for the mass connecting the chains and solve it in the continuous limit $a\to 0$. This leads to the formula for the reflection coefficient $R$.
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