OSCILLATIONS AND WAVES BYU 23

Solution – Problem 23

Problem 23 Solution

v R x (deformation)
1. Restoring Force

When the ball hits the wall, it deforms by a small distance \(x\). The contact area \(S\) is circular. From geometry (Chord theorem for small \(x\)): The radius of the contact circle \(r\) satisfies \(r^2 \approx 2Rx\). Area of contact \(S = \pi r^2 = 2\pi R x\).

The restoring force is generated by the excess pressure \(\Delta p\) acting on this contact area: \[ F = (\Delta p) S = (\Delta p)(2\pi R) x \] This is a linear restoring force \(F = kx\) where spring constant \(k = 2\pi R \Delta p\).

2. Motion Analysis

The motion during contact is Simple Harmonic Motion (SHM). Mass = \(m\). Stiffness \(k = 2\pi R \Delta p\). Angular frequency \(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2\pi R \Delta p}{m}}\).

3. Contact Time

The ball stays in contact for half an oscillation cycle (compression from 0 to max and restitution back to 0). \[ t = \frac{T}{2} = \frac{1}{2} \left( \frac{2\pi}{\omega} \right) = \frac{\pi}{\omega} \] \[ t = \pi \sqrt{\frac{m}{2\pi R \Delta p}} = \sqrt{\frac{\pi m}{2R \Delta p}} \]