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 solve the problem of finding the amplitude and phase of the resultant wave formed by the superposition of two harmonic traveling waves, we need to analyze the given waves:

• The first wave is \( y_1(x, t) = 4 \sin(kx - \omega t) \)
• The second wave is \( y_2(x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3}) \)

When two waves superimpose, they add up to create a resultant wave. The equation for the resultant wave \( y(x, t) \) can be written as:

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

Where \( A \) is the amplitude and \( \phi \) is the phase of the resultant wave. The resultant amplitude can be found using the formula for the superposition of two sinusoidal functions:

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

Where:

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

By substituting the values, we get:

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

Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), substitute in the equation:

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

Thus, the amplitude \(A = 2\sqrt{3}\).

Next, we find the phase \( \phi \) using the formula:

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

Substituting the values, we get:

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

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

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

Thus, \(\phi = \frac{\pi}{6}\).

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

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

Answer 2

The superposition of two waves with the same frequency and different phases gives a resultant wave. The amplitude and phase of the resultant wave can be calculated using the principle of superposition. The waves are given as: 
- \( y_1 (x, t) = 4 \sin(kx - \omega t) \) 
- \( y_2 (x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3}) \)

To find the resultant amplitude, we use the formula for the amplitude of the sum of two sinusoidal waves: 
\( A_{\text{resultant}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi_2 - \phi_1)} \)
Where:
- \( A_1 = 4 \) is the amplitude of \( y_1 \),
- \( A_2 = 2 \) is the amplitude of \( y_2 \),
- \( \phi_1 = 0 \) is the phase of \( y_1 \),
- \( \phi_2 = \frac{2\pi}{3} \) is the phase of \( y_2 \).

Now, calculating the resultant amplitude: 
\( A_{\text{resultant}} = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos\left( \frac{2\pi}{3} \right)} \) 
\( A_{\text{resultant}} = \sqrt{16 + 4 + 2 \times 4 \times 2 \times \left( -\frac{1}{2} \right)} \) 
\( A_{\text{resultant}} = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3} \)

So, the amplitude of the resultant wave is \( 2\sqrt{3} \).

To find the phase of the resultant wave, we use the formula for the phase of the sum of two waves: 
\( \phi_{\text{resultant}} = \tan^{-1}\left( \frac{A_2 \sin(\phi_2) + A_1 \sin(\phi_1)}{A_2 \cos(\phi_2) + A_1 \cos(\phi_1)} \right) \)

Substituting the values: 
\( \phi_{\text{resultant}} = \tan^{-1}\left( \frac{2 \sin\left( \frac{2\pi}{3} \right) + 4 \sin(0)}{2 \cos\left( \frac{2\pi}{3} \right) + 4 \cos(0)} \right) \) 
\( \phi_{\text{resultant}} = \tan^{-1}\left( \frac{2 \times \frac{\sqrt{3}}{2}}{2 \times -\frac{1}{2} + 4} \right) \) 
\( \phi_{\text{resultant}} = \tan^{-1}\left( \frac{\sqrt{3}}{-1 + 4} \right) = \tan^{-1}\left( \frac{\sqrt{3}}{3} \right) = \frac{\pi}{6} \)

Thus, the phase of the resultant wave is \( \frac{\pi}{6} \). 
Therefore, the amplitude and phase of the resultant wave are \( 2\sqrt{3} \) and \( \frac{\pi}{6} \), respectively.

The correct answer is Option (3).