Two waves are represented by the equations $y_1 = A \sin (\omega t + kx + 0.57) \ m$ and $y_2 = A \cos (\omega t + kx) \ m$,where $x$ is in metre and $t$ is in second. What is the phase difference between them?

$A$ simple harmonic progressive wave is represented as $y = 0.03 \sin \pi (2 t - 0.01 x) \ m$. At a given instant of time,the phase difference between two particles $25 \ m$ apart is

The equation of a wave is $y = 2 \sin \pi (0.5x - 200t)$,where $x$ and $y$ are expressed in $cm$ and $t$ in $sec$. The wave velocity is ...... $cm/sec$.

The phase difference between the following two waves $y_1$ and $y_2$ is:
$y_1 = a \sin(\omega t - kx)$
$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$

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

The amplitude of a wave disturbance propagating in the positive $X-$ direction is given by $y = 1/(1 + x^2)$ at time $t = 0$ and by $y = 1/[1 + (x - 1)^2]$ at $t = 2$ seconds,where $x$ and $y$ are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is ..... $ms^{-1}$.