OSCILLATIONS AND WAVES BYU 14

Solution to Problem 14

Solution: Putty Falling in Oscillating Beaker

1. Height of the Beaker

When the putty falls, the plate (with the beaker) starts oscillating. The putty hits the plate after the plate has completed exactly one oscillation. This means the time of fall equals the time period of the plate. $$ t = T = 2\pi \sqrt{\frac{M}{k}} $$ The distance fallen $ h $ is:

            $$ h = \frac{1}{2} g t^2 = \frac{1}{2} g \left( 4\pi^2 \frac{M}{k} \right) = \frac{2\pi^2 Mg}{k} $$
        

2. Amplitude after Collision

Collision Details: The plate is at the bottom extremum of its path (velocity = 0) when the putty hits. The putty has velocity $ v_{putty} = gT $. Momentum Conservation (Inelastic): $$ (M+m) v_{new} = m v_{putty} \implies v_{new} = \frac{m}{M+m} (gT) $$

New Oscillation: The collision happens at the position corresponding to the old equilibrium + old amplitude. Coincidentally, this position is exactly the new equilibrium position for the combined mass $ M+m $. Therefore, the collision happens at the new equilibrium point, meaning the velocity $ v_{new} $ is the maximum velocity $ \omega_{new} A_{new} $.

$$ A_{new} = \frac{v_{new}}{\omega_{new}} = \frac{\frac{m g T}{M+m}}{\sqrt{\frac{k}{M+m}}} $$

Substituting $ T = 2\pi \sqrt{M/k} $:

$$ A_{new} = \frac{2\pi m g}{k} \sqrt{\frac{M}{M+m}} $$