The solution at revision #1379 of , by astrosander. This is not the current version.
К задаче $7.1.22$
Savchenko Solutions

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

  <h4>Statement</h4>
  <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>
  <h4>Solution</h4>
  <center>
  <figure>
    <img src="drawing.png"
      loading="lazy" alt="Mountain landscape" width="250" />
    <figcaption>
      Forces acting on pendulum.
    </figcaption>
  </figure>
  </center>
          <p class="TxtSolutions">
      Applying Newton's Second Law:
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $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)$
    </p>
    <p class="TxtSolutions">
      where,
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \\; (2)$
    </p>
    <p class="TxtSolutions">   
      and, it is known that
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$
    </p>
    <p class="TxtSolutions">   
      as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \\; (3)$
    </p>
    <p class="TxtSolutions">   
      Modifying $(2)$,
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \\; (4)$
    </p>
    <p class="TxtSolutions">   
      Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$
    </p>
    <p class="TxtSolutions">   
      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
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \\; (5)$
    </p>
    <p class="TxtSolutions">   
      Graphically it can be seen that, with cosine and sine approximations applied,
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$
    </p>
    <p class="TxtSolutions">   
      Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$
    </p>
    <p class="TxtSolutions">   
      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 $\theta$.
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$
    </p>
    <p class="TxtSolutions">   
      as $k = \frac{1}{4\pi\epsilon_0}$,
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$
    </p>
    <p class="TxtSolutions">   
      Finally,
    </p>
    <p class="TxtSolutions" style="text-align: center;">   
      $\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$
    </p>
    <p class="TxtSolutions">   
      valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$
    </p>
    <p class="TxtSolutions" 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>

</article>