Pendulums — Question 7

Question 7

A peg creates two half-cycles with different lengths

A simple pendulum has length $\ell$. A smooth peg on the vertical line a distance $3\ell/4$ below its original pivot intercepts the string only on one side of the swing, making that side's effective length $\ell/4$. Assume small angles about each pivot, a taut string, and no energy loss at interception.

1. Find the full oscillation period in terms of the original period $T_0=2\pi\sqrt{\ell/g}$.

2. Compare the small angular amplitudes on the long and short sides. Explain why they differ even though the bottom speed is continuous.

Question 7 - Solution

A peg creates two half-cycles with different lengths

(a) Derivation and calculation

The peg leaves a shortened swing radius$$\ell_s=\ell-3\ell/4=\ell/4,\quad T_s=2\pi\sqrt{\ell_s/g}=T_0/2.$$$$T=\frac{T_0}{2}+\frac{T_s}{2}\Rightarrow\boxed{T=\frac34T_0}.$$Each side contributes two quarter-swings about its own center.

(b) Result and interpretation

For small amplitudes, energy at the turning points gives$$mg\ell(1-\cos\theta_L)=mg\ell_s(1-\cos\theta_S).$$$$\tfrac12mg\ell\theta_L^2\simeq\tfrac12mg(\ell/4)\theta_S^2\Rightarrow\boxed{\theta_S\simeq2\theta_L}.$$Both angular amplitudes must be small, especially the larger short-side angle.

Physical check

The speed is continuous as the taut string meets the ideal peg at the bottom; the subsequent curvature changes.