Solution: Oscillations of a Rope Loop
Let the rope have length \( l \) and linear mass density \( \lambda \). Total mass \( m = \lambda l \). The thread connecting A and B is inextensible, so the total perimeter of the loop is constant.
When junction B is pulled down by a distance \( x \), junction A must move up by the same distance \( x \) (due to the inextensible thread connecting them over the pulley).
Initially, the system is balanced. After displacement \( x \):
- The left side of the rope loses a length \( x \).
- The right side of the rope gains a length \( x \).
The imbalance in rope length is \( 2x \). The weight of this excess portion provides the restoring force.
$$ F_{net} = -(\text{mass of length } 2x) g = -(\lambda \cdot 2x) g $$Imagine whole rope as a straight line. Then we can use the idea of net pulling force.
$$ m a = F_{net} $$ $$ (\lambda l) a = -2\lambda g x $$ $$ a = -\left( \frac{2g}{l} \right) x $$This is SHM with \( \omega^2 = 2g/l \).