ทดสอบปลายภาค PY448

$E = 4 \cos(kx - \omega t + \frac{\pi}{6})$

3 kV/m

$\frac{\pi}{3}$

$E_1 = 3 \cos(ks_1 - \omega t + \frac{\pi}{5})$

$E_{total} = E_1 + E_2$

$k_1 = 3$ kV/m, $\omega_1 = \frac{\pi}{5}$ = $k_2 = 4$ kV/m, $\omega_2 = \frac{\pi}{6}$ $k_1 = 4$ kV/m, $\omega_1 = \frac{\pi}{6}$ = $k_2 = 3$ kV/m, $\omega_2 = \frac{\pi}{5}$ $k_1 = 3$ kV/m, $\omega_1 = \frac{\pi}{6}$ = $k_2 = 4$ kV/m, $\omega_2 = \frac{\pi}{5}$ $k_1 = 4$ kV/m, $\omega_1 = \frac{\pi}{5}$ = $k_2 = 3$ kV/m, $\omega_2 = \frac{\pi}{6}$

$E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{5})$ = $E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{6})$ $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{6})$ = $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{5})$ $E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{6})$ = $E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{5})$ $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{5})$ = $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{6})$

$\frac{\pi}{5}$ = $\frac{\pi}{6}$ $\frac{\pi}{6}$ = $\frac{\pi}{5}$ $\frac{\pi}{3}$ = $\frac{\pi}{4}$ $\frac{\pi}{4}$ = $\frac{\pi}{3}$

$E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{5})$ = $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{6})$ $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{6})$ = $E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{5})$ $E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{6})$ = $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{5})$ $E_1 = 4\cos(ks_1 - \omega t + \frac{\pi}{5})$ = $E_1 = 3\cos(ks_1 - \omega t + \frac{\pi}{6})$

$E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{6})$ = $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{5})$ $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{5})$ = $E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{6})$ $E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{5})$ = $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{6})$ $E_2 = 3\cos(ks_2 - \omega t + \frac{\pi}{6})$ = $E_2 = 4\cos(ks_2 - \omega t + \frac{\pi}{5})$

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