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(A) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi = \phi_1 - \phi_2$ is given by the formula:$A^

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$\cos ^{-1}\left(-\frac{1}{2}\right)$

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(A) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi = \phi_1 - \phi_2$ is given by the formula:$A^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi$Given that $a_1 = a_2 = a$ and the resultant amplitude $A = a$,we substitute these into the equation:$a^2 = a^2 + a^2 + 2(a)(a) \cos \phi$$a^2 = 2a^2 + 2a^2 \cos \phi$Subtracting $2a^2$ from both sides:$-a^2 = 2a^2 \cos \phi$Dividing by $2a^2$:$\cos \phi = -\frac{1}{2}$Therefore,the phase difference is:$\phi = \phi_1 - \phi_2 = \cos ^{-1}\left(-\frac{1}{2}\right)$

The equations of two waves are given as
$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}$
If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ is

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The equations of two waves are given as$\begin{aligned}& y_1=a \sin \left(\omega t+\phi_1\right) \\& y_2=a \sin \left(\omega t+\phi_2\right)\end{aligned}$If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ is