Added Luis's English solution of 5.6.22 and 7.1.17

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+ <header>
+ <h1>Решение задач из Савченко О.Я.</h1>
+ </header>
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+ <a href="../../#1">1.Кинематика</a>
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+ <h3>
+ Problem statement:
+ </h3>
+ <div class="task">
+ <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 $q$. <br>
+ </p>
+ <div class="StatementDiv">
+ <img class="Statement" src="statement.png" alt="9.1.14">
+ </div>
+ </div>
+ <hr>
+ <h3>
+ Solution:
+ </h3>
+
+ <div class="solutions" >
+
+ <img class="ScreenS" src="drawing.png" style="max-width: 250;">
+ <br>
+ <p class="TxtSolutions">
+ Applying Newton's Second Law:
+ </p>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions">
+ where,
+ </p>
+ <br>
+ <p class="TxtSolutions" style="text-align: center;">
+ $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$
+ </p>
+ <br>
+ <p class="TxtSolutions">
+ and, it is known that
+ </p>
+ <br>
+ <p class="TxtSolutions" style="text-align: center;">
+ $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$
+ </p>
+ <br>
+ <p class="TxtSolutions">
+ as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then
+ </p>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions">
+ Modifying $(2)$,
+ </p>
+ <br>
+ <p class="TxtSolutions" style="text-align: center;">
+ $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$
+ </p>
+ <br>
+ <p class="TxtSolutions">
+ Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$
+ </p>
+ <br>
+ <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>
+ <br>
+ <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>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions">
+ Graphically it can be seen that, with cosine and sine approximations applied,
+ </p>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions">
+ Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically
+ </p>
+ <br>
+ <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>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions" style="text-align: center;">
+ $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$
+ </p>
+ <br>
+ <p class="TxtSolutions">
+ as $k = \frac{1}{4\pi\epsilon_0}$,
+ </p>
+ <br>
+ <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>
+ <br>
+ <br>
+ <p class="TxtSolutions">
+ Finally,
+ </p>
+ <br>
+ <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>
+ <br>
+ <p class="TxtSolutions">
+ valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$
+ </p>
+ <br>
+ <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>
+
+ </div>
+
+ <h3>
+ Answer: $\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}$
+ </h3>
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