$7.1.22^*.$
Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge $Q$ is fixed directly under the equilibrium position of the ball at a distance $h$ from it. Length of the thread $l$, mass of the ball $m$, charge $q$.
where,
$$\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$$
and, it is known that
$$\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$$
as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then
$$\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$$
Modifying $(2)$,
$$\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$$
Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$
Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained
Graphically it can be seen that, with cosine and sine approximations applied,
$$\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$$
Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically
Since there are small oscillations, $\theta\rightarrow0$, but terms $\theta^2$ and $\theta^3$ tend to zero faster than $\theta$, so they are negligible respect to $\theta$.
$$\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$$
as $k = \frac{1}{4\pi\epsilon_0}$,
$$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$$
Finally,
$$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$$
valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$
BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba
