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

We are given two travelling 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.
To find the amplitude and phase, we use the principle of superposition.
The general form for the sum of two sine waves of 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 given by: \[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\Delta \phi)} \] where \( A_1 = 4 \), \( A_2 = 2 \), and \( \Delta \phi = \frac{2\pi}{3} \) is the phase difference between the two waves. Substituting the values: \[ A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \cos\left(\frac{2\pi}{3}\right)} \] We know that \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \), so: \[ A = \sqrt{16 + 4 + 2 \times 4 \times 2 \times \left(-\frac{1}{2}\right)} = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3} \] Thus, the resultant amplitude is \( 2\sqrt{3} \).
Step 2: Phase of the Resultant Wave The phase \( \phi \) of the resultant wave is given by: \[ \tan(\phi) = \frac{A_2 \sin(\Delta \phi)}{A_1 + A_2 \cos(\Delta \phi)} \] Substituting the known values: \[ \tan(\phi) = \frac{2 \sin\left(\frac{2\pi}{3}\right)}{4 + 2 \cos\left(\frac{2\pi}{3}\right)} \] We know that \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \), so: \[ \tan(\phi) = \frac{2 \times \frac{\sqrt{3}}{2}}{4 + 2 \times \left(-\frac{1}{2}\right)} = \frac{\sqrt{3}}{4 - 1} = \frac{\sqrt{3}}{3} \] Thus, \( \phi = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \).
Final Answer:
The amplitude of the resultant wave is \( 2\sqrt{3} \) and the phase is \( \frac{\pi}{6} \).
Hence, the correct answer is \( \boxed{(3)} \).