OSCILLATIONS AND WAVES BYU 11

Solution to Problem 11

Solution: Oscillation of a Suspended Rope

1. System Setup

Equilibrium length in air is $ l_0 $. Mass $ M_0 = \lambda l_0 $. The spring force balances this weight: $ k \Delta_{eq} = \lambda l_0 g $.

2. Displacement Analysis

Let the rope be pulled down by a small distance $ x $.

Spring Force: The spring is stretched further. Upward force increases by $ kx $.
Weight: Since the rope moves down, a length $ x $ settles onto the heap on the floor. The suspended length decreases by $ x $. The suspended weight decreases by $ \lambda x g $.

3. Net Restoring Force

Let downward forces be positive. $$ F_{net} = W_{new} – F_{spring\_new} $$ $$ F_{net} = (\lambda l_0 g – \lambda x g) – (k \Delta_{eq} + k x) $$ Using the equilibrium condition $ \lambda l_0 g = k \Delta_{eq} $: $$ F_{net} = -\lambda g x – k x = – (k + \lambda g) x $$

Notice that both the spring stiffness and the loss of weight contribute to the restoring force, so they add up.

4. Angular Frequency

The effective spring constant is $ K_{eff} = k + \lambda g $. The oscillating mass is approximately $ M \approx \lambda l_0 $.

            $$ \omega = \sqrt{\frac{K_{eff}}{M}} = \sqrt{\frac{k + \lambda g}{\lambda l_0}} $$
        