Question

The amplitude and phase of the wave when two travelling waves given as $ y_1(x, t) = 4 \sin(\omega t - kx) $ and $ y_2(x, t) = 2 \sin\left(\omega t - kx + \frac{2\pi}{3}\right) $ are superimposed.

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

Hint:

When two waves of the same frequency interfere, their resultant amplitude is calculated using the formula \( A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)} \), and the phase is determined by \( \tan(\phi) = \frac{A_2 \sin(\Delta \phi)}{A_1 + A_2 \cos(\Delta \phi)} \).

Answer 1

The Correct Option is C

The input provides two traveling waves: \(y_1(x, t) = 4 \sin(\omega t - kx)\) and \(y_2(x, t) = 2 \sin\left(\omega t - kx + \frac{2\pi}{3}\right)\). The resultant wave, \(y(x, t)\), is the sum of these two waves. We will determine its amplitude and phase using the principle of superposition. The general form of the sum of two sine waves with the same frequency is \(y(x, t) = A \sin(\omega t - kx + \phi)\), where \(A\) is the resultant amplitude and \(\phi\) is the phase.

Step 1: Resultant Amplitude
The resultant amplitude \(A\) is calculated using the formula: \[A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)}\] Given \(A_1 = 4\), \(A_2 = 2\), and the phase difference \(\Delta \phi = \frac{2\pi}{3}\). Substituting these values: \[A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \cos\left(\frac{2\pi}{3}\right)}\] Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\): \[A = \sqrt{16 + 4 + 2 \times 4 \times 2 \times \left(-\frac{1}{2}\right)} = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3}\] The resultant amplitude is \(2\sqrt{3}\).

Step 2: Phase of the Resultant Wave
The phase \(\phi\) of the resultant wave is determined by: \[ \tan(\phi) = \frac{A_2 \sin(\Delta \phi)}{A_1 + A_2 \cos(\Delta \phi)} \] Substituting the known values: \[ \tan(\phi) = \frac{2 \sin\left(\frac{2\pi}{3}\right)}{4 + 2 \cos\left(\frac{2\pi}{3}\right)} \] Using \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\): \[ \tan(\phi) = \frac{2 \times \frac{\sqrt{3}}{2}}{4 + 2 \times \left(-\frac{1}{2}\right)} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3} \] Therefore, \(\phi = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\).

Final Answer:
The resultant wave has an amplitude of \(2\sqrt{3}\) and a phase of \(\frac{\pi}{6}\). The correct answer is \( \boxed{(3)} \).