A simple pendulum having a bob of mass m undergoes small oscillations with amplitude theta_0. Find the tension in the string as a function of the angle made by the string with the vertical?

(Try not to use the energy conservation method)

Mar 19, 2018

$t = - m \left(l {\omega}^{2} + g \cos \theta\right)$

Explanation:

For any $\theta$ (bob angle with the vertical) we have

$m \vec{\alpha} = \vec{T} + \vec{P}$

where

$\vec{T} = t \left(\sin \theta , \cos \theta\right)$
$\vec{P} = \left(0 , - m g\right)$
$\vec{r} = l \left(\sin \theta , \cos \theta\right)$
vec alpha = ddot(vec r) = l(-Sin theta omega^2 + Cos theta dotomega, - (Cos theta omega^2 + Sin theta dotomega))

with $\omega = \dot{\theta}$

or

$\left\{\begin{matrix}l m \left(C o s \theta \dot{\omega} - S \in \theta {\omega}^{2}\right) = t S \in \theta \\ - l m \left(S \in \theta \dot{\omega} + C o s \theta {\omega}^{2}\right) = m g + t C o s \theta\end{matrix}\right.$

now solving for $\ddot{\theta} , t$ we obtain

$\ddot{\theta} = - \frac{m g}{l} \sin \theta , t = - m \left(l {\omega}^{2} + g \cos \theta\right)$

hence

$t = - m \left(l {\omega}^{2} + g \cos \theta\right)$