| <header> | | <header> |
| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <main> | | <main> |
| <article> | | <article> |
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
| | | |
| <h4>Statement</h4> | | <h4>Statement</h4> |
| <p> | | <p> |
| $7.1.22^*.$ | | $7.1.22^*.$ |
| 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 $q$. | | 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> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="Mountain landscape" width="150" /> | | loading="lazy" alt="Mountain landscape" width="150" /> |
| <figcaption> | | <figcaption> |
| For the 7.1.22 problem | | For the 7.1.22 problem |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <h4>Solution</h4> | | <h4>Solution</h4> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="drawing.png" | | <img src="drawing.png" |
| loading="lazy" alt="Mountain landscape" width="250" /> | | loading="lazy" alt="Mountain landscape" width="250" /> |
| <figcaption> | | <figcaption> |
| Forces acting on pendulum. | | Forces acting on pendulum. |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Applying Newton's Second Law: | | Applying Newton's Second Law: |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <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)} $$= m\frac{dv}{dt} \; (1)$ | | $mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} $$= m\frac{dv}{dt} \; (1)$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| where, | | where, |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$ | | $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| and, it is known that | | and, it is known that |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$ | | $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then | | as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$ | | $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Modifying $(2)$, | | Modifying $(2)$, |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$ | | $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ | | Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$ | | $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <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 | | 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> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$ | | $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Graphically it can be seen that, with cosine and sine approximations applied, | | Graphically it can be seen that, with cosine and sine approximations applied, |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$ | | $\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically | | Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$ | | $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <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 $\theta$. | | 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> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$ | | $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| as $k = \frac{1}{4\pi\epsilon_0}$, | | as $k = \frac{1}{4\pi\epsilon_0}$, |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$ | | $\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Finally, | | Finally, |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: center;"> | | <p class="TxtSolutions" style="text-align: center;"> |
| $\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$ | | $\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ | | valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ |
| </p> | | </p> |
| <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> | | <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
| | | |
| </article> | | </article> |
| </main> | | </main> |
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