$7.1.22^*.$ 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 $q$.
### Solution
 Forces acting on pendulum.
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 $\sin{\theta}\approx\theta$
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where, $$\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)}\tag{2}$$ 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 $$\sin{(\beta +\theta )} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\label{E3}\tag{3}$$ Modifying $(2)$, $$\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}}\tag{4}$$ 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 $k = \frac{1}{4\pi\varepsilon_0}$, $$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\varepsilon_0mh^3l}\right]\theta = \ddot{\theta}$$ Finally, $$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\varepsilon_0mh^3l}}}$$ valid for the condition $\frac{qQ(h+l)}{4\pi\varepsilon_0h^3} < mg$
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### Statement
### Statement
$7.1.22^*.$ 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 $q$.
$7.1.22^*.$ 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 $q$.
### Solution
### Solution
 Forces acting on pendulum.
 Forces acting on pendulum.
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 $\sin{\theta}\approx\theta$
where, $$\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)}\tag{2}$$ 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 $$\sin{(\beta +\theta )} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\label{E3}\tag{3}$$ Modifying $(2)$, $$\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}}\tag{4}$$ 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
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
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 $k = \frac{1}{4\pi\varepsilon_0}$, $$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\varepsilon_0mh^3l}\right]\theta = \ddot{\theta}$$ Finally, $$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\varepsilon_0mh^3l}}}$$ valid for the condition $\frac{qQ(h+l)}{4\pi\varepsilon_0h^3} < mg$
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 $k = \frac{1}{4\pi\varepsilon_0}$, $$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\varepsilon_0mh^3l}\right]\theta = \ddot{\theta}$$ Finally, $$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\varepsilon_0mh^3l}}}$$ valid for the condition $\frac{qQ(h+l)}{4\pi\varepsilon_0h^3} < mg$
