Two waves have equations $x_1=a \sin \left(\omega t+\phi_1\right)$ and $x_2=a \sin \left(\omega t+\phi_2\right)$. If in the resultant wave the frequency and amplitude remain equal to amplitude of superimposing waves, the phase difference between them is ........
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(b)
$x_1=a \sin \left(\omega t+\phi_1\right)$
$x_2=a \sin \left(\omega t+\phi_2\right)$
$x^{\prime}=x_1+x_2$
$=a\left[\sin \left(\omega t+\phi_1\right)+\sin \left(\omega t+\phi_2\right)\right]$
$=2 a \sin \left(\omega t+\frac{\phi_1+\phi_2}{2}\right) \cos \left(\frac{\phi_1-\phi_2}{2}\right)$
Now as given in question
$2 a \cos \frac{\phi_1-\phi_2}{2}=a$
$\cos \left(\frac{\phi_1-\phi_2}{2}\right)=\frac{1}{2}$
$\frac{\phi_1-\phi_2}{2}=\frac{\pi}{3}$
$\phi_1-\phi_2=\frac{2 \pi}{3}$
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