Question
A simple pendulum, with a bob of mass m, oscillates in a vertical plane, with an angular amplitude θ0. The tension in its string when it passes through the mean position is 2mg. Neglecting the effect of air friction and the viscosity of air, the angular amplitude θ0 is :
Options
Solution
[Key Concept: At the lowest point T − mg = mv2/ℓ, and energy conservation links v2 to the amplitude.]
$$\begin{matrix} & & T - mg = \frac{{mv}^{2}}{\mathcal{l}} \\ & 2mg - mg = \frac{{mv}^{2}}{\mathcal{l}} & \text{(1)} \\ & mg\mathcal{l} = {mv}^{2} & \text{(1)} \\ & \text{ }h = \mathcal{l}(1 - cos\theta) & \text{(1)} \end{matrix}$$
Work done by gravity $= \Delta K$
$$mg\mathcal{l}(1 - cos\theta) = \frac{1}{2}mv^{2} - 0$$
From (1) :
$$\begin{matrix} & mg\mathcal{l}(1 - cos\theta) = \frac{1}{2}(mg\mathcal{l}) \\ & \frac{mg\mathcal{l}}{2} = mg\mathcal{l}cos\theta \\ & cos\theta = \frac{1}{2} \\ & \theta = 60^{\circ} \end{matrix}$$
Create a free account to view solution
View Solution Free