When two sound waves having amplitude A, angular frequency \(\omega\) and a phase difference of \(\pi / 2\) superposes, the maximum amplitude and angular frequency of the resultant wave is \(\ldots \ldots \ldots \ldots\) (A) \(\sqrt{2} \mathrm{~A}, \omega\) (B) \((\mathrm{A} / \sqrt{2}),(\omega / 2)\) (C) \((\mathrm{A} / \sqrt{2}), \omega\) (D) \(\sqrt{2} \mathrm{~A},(\omega / 2)\)

The general equation of a wave is given by: \(y(t) = A \cos(\omega t + \phi)\) Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase. For the two waves, the equations are as follows: - Wave 1: \(y_1(t) = A \cos(\omega t)\) - Wave 2: \(y_2(t) = A \cos(\omega t + \frac{\pi}{2})\)

Using the principle of superposition, we add the two waves to get the resultant wave: \(y_R(t) = y_1(t) + y_2(t) = A \cos(\omega t) + A \cos(\omega t + \frac{\pi}{2})\)

We can use the cosine of the sum formula to simplify the second term in the equation: \(\cos(\omega t + \frac{\pi}{2}) = \cos(\omega t)\cos(\frac{\pi}{2}) - \sin(\omega t)\sin(\frac{\pi}{2})\) Since \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\), the above equation simplifies to: \(\cos(\omega t + \frac{\pi}{2}) = -\sin(\omega t)\)

Now, we can substitute this expression back into the equation for the resultant wave: \(y_R(t) = A \cos(\omega t) - A \sin(\omega t)\)

We can rewrite the equation as follows: \(y_R(t) = A (\cos(\omega t) - \sin(\omega t))\) The maximum amplitude of the resulting wave can be found by finding the maximum value of \((\cos(\omega t) - \sin(\omega t))\). By applying the Pythagorean identity, i.e., \(\sin^2 x + \cos^2 x = 1\), the sum of the squares of sine and cosine is equal to 1: \((\cos(\omega t))^2 + (\sin(\omega t))^2 = 1\) Therefore, the maximum value of \((\cos(\omega t) - \sin(\omega t))\) occurs when the square of it is equal to the square of the maximum amplitude: \((\cos(\omega t) - \sin(\omega t))^2 = A^2_{max}\) By squaring and expanding the left side, we get: \(\cos^2(\omega t) - 2\cos(\omega t)\sin(\omega t) + \sin^2(\omega t) = A^2_{max}\) Since \(\cos^2(\omega t) + \sin^2(\omega t) = 1\), the equation simplifies to: \(1 - 2\cos(\omega t)\sin(\omega t) = A^2_{max}\) Using the double-angle formula for sine, i.e., \(2\sin x \cos x = \sin 2x\), we get: \(1 - \sin(2\omega t) = A^2_{max}\) The maximum value of \(\sin(2\omega t)\) is 1. Therefore, the maximum value of \(A^2_{max}\) is: \(A^2_{max} = 1 - 1 = 0\) So, the maximum amplitude of the resultant wave is \(A_{max} = \sqrt{2}A\). Thus, the maximum amplitude and angular frequency of the resultant wave is \(\boxed{\sqrt{2}A, \omega}\), which corresponds to option (A).