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| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> |
| </div> | | </div> |
| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#12.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#12.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $12.1.4$ | | $12.1.4$ |
| 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? | | 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> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Since waves are superimposed, | | Since waves are superimposed, |
| $$E_R = E_{w1} + E_{w2}$$ | | $$E_R = E_{w1} + E_{w2}$$ |
| $$E_R = E_1~\sin{[\omega(t-z/c)+\varphi_1]} + E_2~\sin{[\omega(t-z/c)+\varphi_2]}$$ | | $$E_R = E_1~\sin{[\omega(t-z/c)+\varphi_1]} + E_2~\sin{[\omega(t-z/c)+\varphi_2]}$$ |
| $$E_R = (E_1\cos{\varphi_1}+E_2\cos{\varphi_2})\sin{[\omega(t-z/c)]} + (E_1\sin{\varphi_1}+E_2\sin{\varphi_2})\cos{[\omega(t-z/c)]}$$ | | $$E_R = (E_1\cos{\varphi_1}+E_2\cos{\varphi_2})\sin{[\omega(t-z/c)]} + (E_1\sin{\varphi_1}+E_2\sin{\varphi_2})\cos{[\omega(t-z/c)]}$$ |
| Let's suppose that | | Let's suppose that |
| $$E_1\cos{\varphi_1}+E_2\cos{\varphi_2} = E \cos{\varphi} = E_x$$ | | $$E_1\cos{\varphi_1}+E_2\cos{\varphi_2} = E \cos{\varphi} = E_x$$ |
| and | | and |
| $$E_1\sin{\varphi_1}+E_2\sin{\varphi_2} = E \sin{\varphi} = E_y$$ | | $$E_1\sin{\varphi_1}+E_2\sin{\varphi_2} = E \sin{\varphi} = E_y$$ |
| and considering the trigonometric identity $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$, | | and considering the trigonometric identity $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$, |
| $$E_R = E \sin{[\omega(t-z/c)+\varphi]}$$ | | $$E_R = E \sin{[\omega(t-z/c)+\varphi]}$$ |
| As $E = \sqrt{{E_x}^2 + {E_y}^2}$ and taking in account that $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$ | | 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> | | </p> |
| <h4>Answer 1</h4> | | <h4>Answer 1</h4> |
| <p> | | <p> |
| $$E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}$$ | | $$E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}$$ |
| </p> | | </p> |
| <p> | | <p> |
| Phase difference is | | Phase difference is |
| $$\tan{\varphi} = \frac{E_y}{E_x} = \frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}$$ | | $$\tan{\varphi} = \frac{E_y}{E_x} = \frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}$$ |
| Finally, the phase is, | | Finally, the phase is, |
