Standing Waves Part II
Problem 1
An ideal taut string of length \( l \) and mass \( m \) is attached to a fixed point at one end and to a massive ring of mass \( M_R \) at the other end as shown below. The ring is free to move on a horizontal frictionless rod which is perpendicular to the string in its equiibrium position. The tension in the string, \( T \), can be considered constant at all times and gravity can be ignored.
 Starting with the general form of the normal mode solutions of the classical wave equation for a taut string, derive an equation which, when solved, gives the angular frequency \( \omega \) of the lowest mode of the system. YOU ARE NOT ASKED TO SOLVE THIS EQUATION.
 Determine the angular frequency of the lowest normal mode for
 \( M_R = 0 \)
 \( M_R = \infty \).
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\begin{eqnarray} \nonumber \omega &=& \frac{T}{M_R}\sqrt{\frac{m}{Tl}}\cot \left( \omega l \sqrt{\frac{m}{Tl}}\right) \\ \nonumber &=& \frac{1}{M_R}\sqrt{\frac{Tm}{l}}\cot \left( \omega \sqrt{\frac{ml}{T}}\right) \end{eqnarray}
This is a transcendental equation with no simple solution.

 For \( M_R = 0 \), we get \begin{eqnarray} \nonumber &&\cot \left( \frac{\omega}{v}l\right) = 0 \hspace{3mm} with \hspace{3mm} v = \sqrt{\frac{Tl}{m}} \\ \nonumber &\Rightarrow& \frac{\omega}{v}l = \frac{\pi}{2} \\ \nonumber &\Rightarrow& \omega = \frac{\pi}{2l}v = \frac{\pi}{2l}\sqrt{\frac{Tl}{m}} = \frac{\pi}{2}\sqrt{\frac{T}{ml}} \end{eqnarray} for the lowest normal mode. Intuitively, this makes sense since this situation corresponds to a string which is fixed at one end and free at the other, so the lowest normal mode has \( l = \lambda / 4 \) or \( \lambda = 4l \) and thus \( \omega = \dfrac{2 \pi v}{\lambda} = \dfrac{2 \pi v}{4l} = \dfrac{\pi}{2l}v \).
 For \( M_R = \infty \), we get \begin{eqnarray} \nonumber &&\tan \left( \frac{\omega}{v}l\right) = 0 \\ \nonumber &\Rightarrow& \frac{\omega}{v}l = \pi \\ \nonumber &\Rightarrow& \omega = \frac{\pi}{l}v = \frac{\pi}{l}\sqrt{\frac{Tl}{m}} = \pi\sqrt{\frac{T}{ml}} \end{eqnarray} for the lowest normal mode. Intuitively, this makes sense since this situation corresponds to a string which is fixed at both ends, so the lowest normal mode has \( l = \lambda / 2 \) or \( \lambda = 2l \) and thus \( \omega = \dfrac{2 \pi v}{\lambda} = \dfrac{2 \pi v}{2l} = \dfrac{\pi}{l}v \).
Problem 2
A system consists of two materials glued together at \(x = L\) and subject to longitudinal (sound) wave propagation. The end at \(x = 0\) is fixed and the end at \(x = 4L\) is free. A tungstenlead alloy is used for the segment \(0 \leq x \leq L\), with a Young’s modulus of \(Y = 7 \times 10^{10}\) N/m\(^2\) and density \(\rho = 14.4 \times 10^{3}\) kg/m\(^3\), while the segment \(L \leq x \leq 4L\) is comprised of a carbon fiber alloy that is as strong as the tungstenlead alloy with \(Y = 7 \times 10^{10}\) N/m\(^2\) but nine times less dense, with \(\rho = 1.6 \times 10^{3}\) kg/m\(^3\).
 Derive a condition for the normal mode frequency of longitudinal oscillation, \(\omega_n\), of the twometal system.
 Evaluate the frequencies for the first four modes, in terms of the length \(L\).
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The phase velocity \(v\) of longitudinal waves \( = \sqrt{\dfrac{Y}{\rho}}\)
\(\therefore \dfrac{v_2}{v_1} = 3\) and the essence of the problem is the boundary conditions at \(x\) = 0, \(L\) and \(4L\).
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 \( sin{\dfrac{\omega_n L}{v_1}} = \pm \dfrac{\sqrt{3}}{2} \) with \(v_1 = \sqrt{\dfrac{Y_1}{\rho_1}} \approx 2.2 \times 10^{3}\) m/s
 \( \omega_1 = {\dfrac{v_1}{L}}{\dfrac{\pi}{3}} \); \( \omega_2 = {\dfrac{v_1}{L}}{\dfrac{2 \pi}{3}} \); \( \omega_3 = {\dfrac{v_1}{L}}{\dfrac{4 \pi}{3}} \); \( \omega_4 = {\dfrac{v_1}{L}}{\dfrac{5 \pi}{3}} \)
[Note that just about any factor other than 3 for speed and length ratios would give an equation which could not be easily solved.]