The equation of a progressive wave is $y = 8 \sin \left[ \pi \left( \frac{t}{10} - \frac{x}{4} \right) + \frac{\pi}{3} \right]$. The wavelength of the wave is .... $m$.

$A$ particle performs $S.H.M.$ of amplitude $A$ and the wave has a wavelength $\lambda$. If $V$ is the wave velocity and $\nu$ is the maximum particle velocity,then they are related as:

The displacement of two sinusoidal waves is given by the equations:
$y_1 = 8 \sin(20x - 30t)$
$y_2 = 8 \sin(25x - 40t)$
Then the phase difference between the waves at time $t = 2 \ s$ and distance $x = 5 \ cm$ will be:

Two waves are given by $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \cos(\omega t - kx)$. The phase difference between the two waves is

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

In the given progressive wave equation,what is the maximum velocity of the particle? $Y = 0.5 \sin(10\pi t - 5x) \text{ cm}$