OSCILLATIONS AND WAVES BYU 6

Solution to Problem 6

Solution: Small Oscillations of a Ring on a Cord

1. Geometric Analysis

The cord has a fixed length $ 2l $. The ring is threaded on it, so the sum of distances from the ring to the two fixed ends is constant ($ 2l $). The locus of such a point is an ellipse with the fixed ends as foci.

Let the depth be $ y $. At equilibrium, the depth is $ h $. This corresponds to the semi-minor axis of the ellipse being $ b = h $. Since the string length is $ 2l $, the semi-major axis is $ a = l $.

2. Radius of Curvature

For small oscillations about the lowest point (the vertex of the minor axis), the motion is equivalent to a simple pendulum with length equal to the radius of curvature $ \rho $ of the ellipse at that point.

The radius of curvature at the vertex of the minor axis is given by:

$$ \rho = \frac{a^2}{b} = \frac{l^2}{h} $$

3. Period Calculation

Using the simple pendulum formula with effective length $ L_{eff} = \rho $:

$$ T = 2\pi \sqrt{\frac{\rho}{g}} = 2\pi \sqrt{\frac{l^2}{gh}} $$