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Two waves having equation $x_1 = a\sin(\omega t + \phi_1)$ and $x_2 = a\sin(\omega t + \phi_2)$. If in the resultant wave the frequency and amplitude remain equal to those of the superimposing waves, then phase difference between them : -
A
$\dfrac{\pi}{6}$
B
$\dfrac{2\pi}{3}$
C
$\dfrac{\pi}{4}$
D
$\dfrac{\pi}{3}$
Detailed Solution
Resultant amplitude of two waves of equal amplitude a with phase difference $\phi$:
$A^2 = a^2 + a^2 + 2a^2\cos\phi$
Given A = a: $a^2 = 2a^2 + 2a^2\cos\phi \Rightarrow \cos\phi = -\dfrac{1}{2}$
$\phi = \dfrac{2\pi}{3}$ (i.e. $120^\circ$)
The frequency of the resultant is the same, $\omega$, as the frequencies of both waves are equal.
$A^2 = a^2 + a^2 + 2a^2\cos\phi$
Given A = a: $a^2 = 2a^2 + 2a^2\cos\phi \Rightarrow \cos\phi = -\dfrac{1}{2}$
$\phi = \dfrac{2\pi}{3}$ (i.e. $120^\circ$)
The frequency of the resultant is the same, $\omega$, as the frequencies of both waves are equal.