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 $)

• A. \( [\sqrt{3}, \frac{\pi}{6}] \)
• B. \( [6, \frac{\pi}{3}] \)
• C. \( [2\sqrt{3}, \frac{\pi}{6}] \)
• D. \( [6, \frac{2\pi}{3}] \)

Hint:

When two sinusoidal waves with the same frequency but different phases combine, use the superposition principle to find the resultant amplitude and phase using the formulas for the sum of sinusoidal functions.

Answer 1

The Correct Option is C

To determine the amplitude and phase of a resultant wave formed by the superposition of two harmonic traveling waves, we first analyze the individual waves:

• Wave 1: \( y_1(x, t) = 4 \sin(kx - \omega t) \)
• Wave 2: \( y_2(x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3}) \)

The resultant wave \( y(x, t) \) from their superposition is given by:

\(y(x, t) = A \sin(kx - \omega t + \phi)\)

Where \( A \) represents the resultant amplitude and \( \phi \) the resultant phase. The resultant amplitude \( A \) is calculated using the formula for the superposition of two sinusoidal functions:

• \( A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos(\delta)} \)

With the following values:

• \( a_1 = 4 \): Amplitude of the first wave.
• \( a_2 = 2 \): Amplitude of the second wave.
• \( \delta = \frac{2\pi}{3} \): Phase difference between the two waves.

Substituting these values yields:

\(\qquad A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos\left(\frac{2\pi}{3}\right)}\)

Given that \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \), the equation becomes:

\(\qquad A = \sqrt{16 + 4 + 16 \times (-\frac{1}{2})} = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3}\)

The amplitude of the resultant wave is \( A = 2\sqrt{3} \).

The phase \( \phi \) of the resultant wave is determined by the formula:

• \( \tan\phi = \frac{a_2 \sin\delta}{a_1 + a_2 \cos\delta} \)

Substituting the known values:

\(\qquad \tan\phi = \frac{2 \times \sin(\frac{2\pi}{3})}{4 + 2 \times \cos(\frac{2\pi}{3})}\)

Using \( \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} \) and \( \cos(\frac{2\pi}{3}) = -\frac{1}{2} \):

\(\qquad \tan\phi = \frac{2 \times \frac{\sqrt{3}}{2}}{4 + 2 \times (-\frac{1}{2})} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3}\)

From this, we find \( \phi = \frac{\pi}{6} \).

Consequently, the amplitude and phase of the resultant wave are \( [2\sqrt{3}, \frac{\pi}{6}] \).

Correct Answer: \( [2\sqrt{3}, \frac{\pi}{6}] \)