### Fall 2007

Fall 2007 1 Mech

Fall 2007 2 Mech

Fall 2007 3 EM

Fall 2007 4 EM

Fall 2007 5 QM

Fall 2007 6 QM

Fall 2007 7 SM

Fall 2007 8 SM

Fall 2007 9 Misc

Fall 2007 10 Misc

Fall 2007 11 Mech

Fall 2007 12 Mech

Fall 2007 13 EM

Fall 2007 14 EM

Fall 2007 15 QM

Fall 2007 16 QM

Fall 2007 17 SM

Fall 2007 18 SM

Fall 2007 19 Misc

Fall 2007 20 Misc

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 *x _{1}*,

*x*, and

_{2}*x*for the three strings, with $-\infty < x_i \leq 0$ and the ring position being

_{3}*x*= 0. When a plane wave of a given momentum

_{i}*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:

(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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Answer

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(a) The wave equations read

(3)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 *x _{i}* = 0:

(b) The continuity at the ring demands

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

(6)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)(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.

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