Added Luis's English solution of 3.3.3

Luisito правка от 3 окт. 2024 г., 04:23

правка #9831 GitHub f766a22 позже →

en/3.3.3.md +93 −0

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		+	<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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		+	<p class="author">
		+	Solutions of Savchenko Problems in Physics <br>
		+	<i><b>knowledge must be free</b></i>
		+	</p>
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		+
		+	<h3 id="back-link"><a href="../../#10.1">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$3.3.3$
		+	A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations?
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So,
		+	$$x(t) = A~\sin{~\omega t}$$
		+	where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then,
		+	$$x_0 = A~\sin{~\omega t_1} \;(1)$$
		+	$$A = A~\sin{~\omega t_2}$$ or
		+	$$\sin{~\omega t_2} = 1$$
		+	Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$,
		+	$$t_1 = \frac{T}{4} - \Delta t \;(2)$$
		+	Putting (2) into (1) and separating $T$, it is obtained
		+	$$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$
		+	</p>
		+	<h4>Answer</h4>
		+	<p>
		+	$$T = 0.06 {\rm{s}}$$
		+	</p>
		+
		+
		+	<p 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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			<title>A plane capacitor is placed in a homogeneous magnetic field od induction B parallel to the plates. From point A, electrons enter perpendicularly to magnetic field direction. The tension (voltage) applied to the plates is V. What is the condition for which electrons pass through the capacitor?</title>
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			</p>
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			<h3 id="back-link"><a href="../../#10.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$3.3.3$
			A weight, oscillating freely on a spring, has moved from a distance of 0.5 cm from its equilibrium position to the largest one, equal to 1 cm, for a time of 0.01 s. What is the period of its oscillations?
			</p>

			<h3>Solution</h3>
			<p>
			For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So,
			$$x(t) = A~\sin{~\omega t}$$
			where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then,
			$$x_0 = A~\sin{~\omega t_1} \;(1)$$
			$$A = A~\sin{~\omega t_2}$$ or
			$$\sin{~\omega t_2} = 1$$
			Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$,
			$$t_1 = \frac{T}{4} - \Delta t \;(2)$$
			Putting (2) into (1) and separating $T$, it is obtained
			$$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$
			</p>
			<h4>Answer</h4>
			<p>
			$$T = 0.06 {\rm{s}}$$
			</p>


			<p 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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