Question

Consider the three waves represented by y1=3sin (kx-wt), y2=3sin (kx-wt+2pi/3), y3=3sin (kx-wt+4pi/3)
then the amplitude of resultant of waves at x=0

Answer 1

Answer on Question #37378

Physics - Mechanics | Kinematics | Dynamics

Question:

Consider the three waves represented by $y_1 = 3\sin(kx - wt)$, $y_2 = 3\sin(kx - wt + 2pi/3)$, $y_3 = 3\sin(kx - wt + 4pi/3)$ then the amplitude of resultant of waves at $x = 0$

Solution:

At $x = 0$ one has

$$\begin{aligned} y(t) &= 3\left(\sin(-x) + \sin\left(\frac{2\pi}{3} - x\right) + \sin\left(\frac{4\pi}{3} - x\right)\right) \\ &= -3\left(\sin x + \sin x \cos \frac{2\pi}{3} - \cos x \sin \frac{2\pi}{3} + \sin x \cos \frac{4\pi}{3} - \cos x \sin \frac{4\pi}{3}\right) \\ &= -3\left(\sin x - \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x\right) \\ &= -3(\sin x - \sin x) = 0. \end{aligned}$$

Answer:

0.