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 $)
Show Hint
- \( [\sqrt{3}, \frac{\pi}{6}] \)
- \( [6, \frac{\pi}{3}] \)
- \( [2\sqrt{3}, \frac{\pi}{6}] \)
- \( [6, \frac{2\pi}{3}] \)
The Correct Option is C
Approach Solution - 1
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: \( [2\sqrt{3}, \frac{\pi}{6}] \)
Approach Solution -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).
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