<h3 id="back-link"><a href="/#7.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
      $7.1.22^*.$
      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 $q$.
    </p>
    <center>
    <figure>
      <img src="statement.png"
        loading="lazy" alt="Mountain landscape" width="150" />
      <figcaption>
        For the 7.1.22 problem
      </figcaption>
    </figure>
    </center>

    <h3>Solution</h3>
    <center>
      <figure>
      <img src="drawing.png"
        loading="lazy" alt="Mountain landscape" width="250" />
      <figcaption>
        Forces acting on pendulum.
      </figcaption>
      </figure>
    </center>
    <p>
      Applying Newton's Second Law:
      <div class="scroll-wrapper">
      $$mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} = m\frac{dv}{dt} \\; (1)$$
      </div>
      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 $\sin{\theta}\approx\theta$
      <div class="scroll-wrapper">
      $$mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$$
      </div>
      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          
      <div class="scroll-wrapper">
        $$\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \\; (5)$$
      </div>
      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
      <div class="scroll-wrapper">
        $$\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$$
      </div>
      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 $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$
    </p>


    <p style="text-align: right; font-style: italic; font-size: 14;">   
      BSc. Luis Daniel Fernández Quintana<br>
      Physics Department (FCNE)<br>
      Universidad de Oriente, Cuba<br>
    </p>



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