OSCILLATIONS AND WAVES BYU 12

1. Elastic Collision Constraint

The collision is elastic, and the speeds are “unchanged but directions reversed”. Conservation of Momentum for reversed velocities: $$ M v_p – m v_b = M (-v_p) + m (v_b) $$ $$ 2 M v_p = 2 m v_b \implies M v_p = m v_b $$

2. Synchronization Condition

The process repeats indefinitely. The ball must return to the collision point exactly when the platform returns.

• Ball: Time of flight for a projectile going up and down with speed \( v_b \): \( t_b = \frac{2v_b}{g} \).
• Platform: Moves down distance \( H \) and returns. Since it starts from equilibrium with max velocity, this corresponds to half a period \( T/2 \). However, for generalized synchronization (repeats indefinitely), the ball’s airtime must be an odd integral multiple of the half-period (so they meet again at the center moving opposite directions).

$$ t_b = (2n + 1) \frac{T}{2} $$ $$ \frac{2v_b}{g} = \frac{(2n+1)T}{2} \implies v_b = \frac{(2n+1)gT}{4} $$

3. Solving for Mass

Platform velocity \( v_p \) is related to its displacement amplitude \( H \). Since it travels distance \( H \) from equilibrium to rest, \( H \) is the amplitude. $$ v_p = \omega H = \frac{2\pi}{T} H $$ Substitute \( v_p \) and \( v_b \) into the momentum equation \( m v_b = M v_p \): $$ m \left[ \frac{(2n+1)gT}{4} \right] = M \left[ \frac{2\pi H}{T} \right] $$