Two waves having an equal amplitude A and the same frequency...

Let the two waves be $y_1 = A \sin(\omega t - kx)$ and $y_2 = A \sin(\omega t - kx + \phi)$.

By the principle of superposition, the amplitude of the resultant wave is given by $A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}$.

Substituting $A_1 = A_2 = A$, we get $A_R = \sqrt{A^2 + A^2 + 2A^2\cos\phi} = \sqrt{2A^2(1+\cos\phi)}$.

Using the trigonometric identity $1+\cos\phi = 2\cos^2$\left(\dfrac{\phi}{2}\right), we get $A_R = 2A\left|\cos\left(\dfrac{\phi}{2}\right)\right|.<br><br>Since the phase difference$\phi$is not specified, it can take any value.<br><br>The maximum amplitude occurs when$\cos\left(\dfrac{\phi}{2}\right) = \pm$1$, giving A_\max = 2A.

The minimum amplitude occurs when $\cos\left(\dfrac{\phi}{2}\right) =$0$, giving$A_\min = 0$.<br><br>Thus, the amplitude of the resultant wave lies between$0$and$2A$.$