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правка #1379 предыдущая #1378 GitHub 454340f ← раньше позже →

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			<header>
			<h2>Solutions of Savchenko Physics Textbook</h2>
			<p class="author">
			Aliaksandr Melnichenka <br/>
			October 2023
			</p>
			</header>

			<main>
			<article>
			<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>

			<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>

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	</head>		</head>

	<body id="top">		<body id="top">
	<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>

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