Question

Two waves having equation $x_1=a \sin(\omega t-kx+\phi_1),$ $ x_2=a \sin (\omega t-kx +\phi_2)$ If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is

• A. $\frac{\pi}{6}$
• B. $\frac{2\pi}{3}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{3}$

Answer 1

The Correct Option is B

To determine the phase difference between the two waves given by the equations:

x_1 = a \sin(\omega t - kx + \phi_1), x_2 = a \sin(\omega t - kx + \phi_2)

We use the principle of superposition which states that when two waves interfere, their amplitudes add up vectorially. According to the conditions given, the amplitude of the resultant wave remains equal to the amplitude of the individual waves, a.

The formula for the amplitude of the resultant wave when two waves interfere is:

R = \sqrt{a^2 + b^2 + 2ab \cos(\phi_2 - \phi_1)}

Since the amplitudes of both waves are equal, a = b, the equation simplifies to:

R = \sqrt{2a^2(1 + \cos(\phi_2 - \phi_1))}

Given that R = a, we equate and simplify:

a = \sqrt{2a^2(1 + \cos(\phi_2 - \phi_1))}

Squaring both sides gives:

a^2 = 2a^2(1 + \cos(\phi_2 - \phi_1))

1 = 2(1 + \cos(\phi_2 - \phi_1))

1 = 2 + 2\cos(\phi_2 - \phi_1)

-1 = 2\cos(\phi_2 - \phi_1)

\cos(\phi_2 - \phi_1) = -\frac{1}{2}

The cosine of \pm \frac{2\pi}{3} is -\frac{1}{2}. Thus, the phase difference, (\phi_2 - \phi_1) must be:

\frac{2\pi}{3}

Therefore, the correct answer is:

\frac{2\pi}{3}