The expression can be written as \(y = P \sin \left( \frac{2\pi Q}{\lambda}t - \frac{2\pi}{\lambda}x \right)\). The angular frequency is \(\omega = \frac{2\pi Q}{\lambda}\). Thus, the frequency is \(f = \frac{\omega}{2\pi} = \frac{Q}{\lambda}\).

Since reducing the tension in B decreases its frequency to 630 Hz (unison with A), B's original frequency must have been higher than A's. Therefore, \(f_B = 630 + 3 = 633\text{ Hz}\).

The linear frequencies are \(f_1 = \frac{1020\pi}{2\pi} = 510\text{ Hz}\) and \(f_2 = \frac{1004\pi}{2\pi} = 502\text{ Hz}\). The beat frequency is \(f_b = |f_1 - f_2| = 8\text{ Hz}\).

Wave velocity is given by \(v = \frac{\omega}{k}\). From the wave equation, \(\omega = 4\pi\) and \(k = 0.4\pi\), which gives \(v = \frac{4\pi}{0.4\pi} = 10\text{ m/s}\).

The wave equation can be rewritten as \(y = P \sin \left(\frac{2\pi Qt}{\lambda} - \frac{2\pi x}{\lambda}\right)\). Here, angular frequency \(omega = \frac{2\pi Q}{\lambda}\). Thus, linear frequency is \(f = \frac{\omega}{2\pi} = \frac{Q}{\lambda}\).

Beats = \(|f_A - f_B| = 3\text{ Hz}\). Since reducing tension in B decreases its frequency to unison (630 Hz), B's initial frequency must have been higher than 630 Hz. Thus, \(f_B = 633\text{ Hz}\).

From \(\omega_1 = 1020\pi\) and \(\omega_2 = 1004\pi\), we get $ f_1 = 510\text{ Hz} $ and \( f_2 = 502\text{ Hz}\). The beat frequency is \(|f_1 - f_2| = 8\text{ Hz}\).

A progressive wave along the y-axis must have a phase containing the term \((ky \pm \omega t)\). Thus, \(x = A sin(ky - \omega t)\) represents a transverse wave travelling along the y-axis.

Wavelength \(\lambda = \frac{v}{f} = \frac{330}{500} = 0.66\text{ m}\). Path difference \(\Delta x = \frac{\lambda}{2\pi} \Delta \phi = \frac{0.66}{2\pi} \left(\frac{\pi}{6}\right) = 0.055\text{ m}\).