Fall 2007 11 Mech

Three identical strings are connected to a ring of mass m that can slide frictionlessly along a vertical pole. Each string has tension $\tau$ and the linear mass density $\sigma$. In equilibrium, all strings are in the same horizontal plane. The motion of the strings is in the vertical z-direction. We can choose coordinates x1, x2, and x3 for the three strings, with $-\infty < x_i \leq 0$ and the ring position being xi = 0. When a plane wave of a given momentum k is incident on the ring from the first string, it creates transmitted waves down the other two strings and a reflected wave on the first string:

(1)
\begin{align} z_1=\hat{f}(k)\exp{(ikx_1-i\omega t)}+\hat{g}(k)\exp{(-ikx_1-i\omega t)} \end{align}
(2)
\begin{align} z_2=\hat{h}_A(k)\exp{(-ikx_2-i\omega t)},\;\; z_3=\hat{h}_B(k)\exp{(-ikx_3-i\omega t)} \end{align}

(a) Write the set of equations of motion for the problem. Define all coefficients, e.g., $c\equiv \omega/k$.

(b) Find the reflection coefficient $\hat{g}(k)/\hat{f}(k)$.

(c) Test the correctness of your formula by considering two limits, $k \rightarrow 0$ (a very long and slow pulse) and $k \rightarrow \infty$ (a very short and fast one).

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(3)
\begin{align} \frac{\partial^2 z_i}{\partial x_i^2} = \frac{1}{c^2} \frac{\partial^2 z_i}{\partial t^2}, \;\; c=\sqrt{\frac{\tau}{\sigma}} \end{align}

Next, if $Z \exp{(-i \omega t)}$ is the vertical coordinate of the ring, then Newton’s second law implies $F = -m\omega^2 Z\exp{(-i\omega t)}$. Here the force F on the ring is the sum of the vertical components of the tension in the three strings at xi = 0:

(4)
\begin{align} F=-\tau\sum_{i=1}^{3}\frac{\partial z_i}{\partial x_i}\bigg|_{x_i=0}=-i\tau ke^{-i\omega t}(\hat{f}-\hat{g}-\hat{h}_A-\hat{h}_B) \end{align}

(b) The continuity at the ring demands

(5)
\begin{align} Z = \hat{f}+\hat{g}=\hat{h}_A=\hat{h}_B \end{align}

Eliminating $\hat{h}_A$ and $\hat{h}_B$ from the Newton’s law for the ring, we readily obtain

(6)
\begin{align} \hat{g}(k)=-\left(\frac{k+iQ}{k+3iQ}\right)\hat{f}(k) \end{align}

where $Q\equiv\tau/mc^2$ has dimensions of inverse length. Substituting this into formulas for $\hat{h}_A$ and $\hat{h}_B$ , we have

(7)
\begin{align} \hat{h}_A(k) = \hat{h}_B(k) = \left(\frac{2iQ}{k+3iQ} \right)\hat{f}(k) \end{align}

(c) For a very long wavelength pulse, composed of plane waves for which |k| « Q, we have $\hat{g}(k)\simeq-\frac{1}{3}\hat{f}(k)$. Thus, the reflected pulse is inverted, and is reduced by a factor of 3 in amplitude. The other outgoing pulses have amplitudes $(2/3) \hat{f}$. This is consistent with the energy conservation: $1^2 = (1/3)^2 + (2/3)^2 + (2/3)^2$ (for $\omega \rightarrow 0$, the ring oscillates very slowly, and so it has no appreciable kinetic energy). Conversely, for a very short wavelength pulse, k » Q, we have perfect reflection with inversion, and no transmission. This is due to the inertia of the ring.