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| | + | <h2>Solutions of Savchenko Physics Textbook</h2> |
| | + | <p class="author"> |
| | + | Aliaksandr Melnichenka <br/> |
| | + | October 2023 |
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| | + | <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 $q$. |
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| | + | For the 7.1.22 problem |
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| | + | <h3>Solution</h3> |
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| | + | <figcaption> |
| | + | Forces acting on pendulum. |
| | + | </figcaption> |
| | + | </figure> |
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| | + | <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> |
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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"> |
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| | − | <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: $\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}$ |
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