Question 126: Waves

Question

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_1(x, t) = 4 \sin (kx - \omega t)$ and $y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3})$, are: (Take the angular frequency of initial waves same as $\omega$)

Accepted Answer

To determine the amplitude and phase of the wave formed by the superposition of the two harmonic traveling waves, $ y_1(x, t) = 4 \sin (kx - \omega t) $ and $ y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3}) $, we use the following steps:
First, calculate the resultant amplitude $ A $ using the formula for the resultant of two waves with different phases:
$ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} $
where:
$ A_1 = 4 $ (amplitude of the first wave)
$ A_2 = 2 $ (amplitude of the second wave)
$ \phi = 120^\circ $ (phase difference, given by $\frac{2\pi}{3}$).
Substituting the values, we get:
$ A = \sqrt{4^2 + 2^2 + 2 	imes 4 	imes 2 	imes \cos(120^\circ)} $
$ \cos(120^\circ) = -\frac{1}{2} $
Therefore:
$ A = \sqrt{16 + 4 + 16 	imes (-\frac{1}{2})} $
$ A = \sqrt{16 + 4 - 8} $
$ A = \sqrt{12} = 2\sqrt{3} $
Next, compute the phase $ \phi $ of the resultant wave:
$ 	an \phi = \frac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)} $
$ 	an \phi = \frac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)} $
$ \sin(120^\circ) = \frac{\sqrt{3}}{2} \quad 	ext{and} \quad \cos(120^\circ) = -\frac{1}{2} $
Substituting these values, we get:
$ 	an \phi = \frac{2 	imes \frac{\sqrt{3}}{2}}{4 + 2 	imes -\frac{1}{2}} $
$ 	an \phi = \frac{\sqrt{3}}{3} $
This simplifies to:
$ \phi = \frac{\pi}{6} $
Thus, the amplitude and phase of the resultant wave are $ 2\sqrt{3} $ and $ \frac{\pi}{6} $ respectively.

Answer

$\left[\sqrt{3}, \frac{\pi}{6}\right]$

Answer

$\left[2\sqrt{3}, \frac{\pi}{6}\right]$

Answer

$\left[6, \frac{2\pi}{3}\right]$

Answer

$\left[6, \frac{\pi}{3}\right]$

Waves

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y_1(x, t) = 4 \sin (kx - \omega t) and y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3}), are: (Take the angular frequency of initial waves same as \omega)

Choose the correct answer: