fix minor issue and Added Luis's English solution of 12.1.4

Luisito правка от 4 окт. 2024 г., 12:27

правка #9844 GitHub 3c6f18e позже →

en/12.1.4.md +103 −0

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		+	Solutions of Savchenko Problems in Physics <br>
		+	<i><b>knowledge must be free</b></i>
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		+	<h3 id="back-link"><a href="../../#12.1">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$12.1.4$
		+	Two sinusoidal waves with the same polarization $E_1~\sin{[\omega(t-z/c)+\varphi_1]}$, $E_2~\sin{[\omega(t-z/c)+\varphi_2]}$ are superimposed on each other. What is the amplitude of the electric field strength of the resulting wave? What is the phase of this wave?
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	Since waves are superimposed,
		+	$$E_R = E_{w1} + E_{w2}$$
		+	$$E_R = E_1~\sin{[\omega(t-z/c)+\varphi_1]} + E_2~\sin{[\omega(t-z/c)+\varphi_2]}$$
		+	$$E_R = (E_1\cos{\varphi_1}+E_2\cos{\varphi_2})\sin{[\omega(t-z/c)]} + (E_1\sin{\varphi_1}+E_2\sin{\varphi_2})\cos{[\omega(t-z/c)]}$$
		+	Let's suppose that
		+	$$E_1\cos{\varphi_1}+E_2\cos{\varphi_2} = E \cos{\varphi} = E_x$$
		+	and
		+	$$E_1\sin{\varphi_1}+E_2\sin{\varphi_2} = E \sin{\varphi} = E_y$$
		+	and considering the trigonometric identity $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$,
		+	$$E_R = E \sin{[\omega(t-z/c)+\varphi]}$$
		+	As $E = \sqrt{{E_x}^2 + {E_y}^2}$ and taking in account that $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$
		+	</p>
		+	<h4>Answer 1</h4>
		+	<p>
		+	$$E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}$$
		+	</p>
		+	<p>
		+	Phase difference is
		+	$$\tan{\varphi} = \frac{E_y}{E_x} = \frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}$$
		+	Finally, the phase is,
		+	</p>
		+	<h4>Answer 2</h4>
		+	<p>
		+	$$\Phi = \omega(t-z/c) + \arctan{\frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}}$$
		+	</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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			<h3 id="back-link"><a href="../../#12.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$12.1.4$
			Two sinusoidal waves with the same polarization $E_1~\sin{[\omega(t-z/c)+\varphi_1]}$, $E_2~\sin{[\omega(t-z/c)+\varphi_2]}$ are superimposed on each other. What is the amplitude of the electric field strength of the resulting wave? What is the phase of this wave?
			</p>

			<h3>Solution</h3>
			<p>
			Since waves are superimposed,
			$$E_R = E_{w1} + E_{w2}$$
			$$E_R = E_1~\sin{[\omega(t-z/c)+\varphi_1]} + E_2~\sin{[\omega(t-z/c)+\varphi_2]}$$
			$$E_R = (E_1\cos{\varphi_1}+E_2\cos{\varphi_2})\sin{[\omega(t-z/c)]} + (E_1\sin{\varphi_1}+E_2\sin{\varphi_2})\cos{[\omega(t-z/c)]}$$
			Let's suppose that
			$$E_1\cos{\varphi_1}+E_2\cos{\varphi_2} = E \cos{\varphi} = E_x$$
			and
			$$E_1\sin{\varphi_1}+E_2\sin{\varphi_2} = E \sin{\varphi} = E_y$$
			and considering the trigonometric identity $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$,
			$$E_R = E \sin{[\omega(t-z/c)+\varphi]}$$
			As $E = \sqrt{{E_x}^2 + {E_y}^2}$ and taking in account that $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$
			</p>
			<h4>Answer 1</h4>
			<p>
			$$E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}$$
			</p>
			<p>
			Phase difference is
			$$\tan{\varphi} = \frac{E_y}{E_x} = \frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}$$
			Finally, the phase is,
			</p>
			<h4>Answer 2</h4>
			<p>
			$$\Phi = \omega(t-z/c) + \arctan{\frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}}$$
			</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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