If the equation of a transverse wave is $y = 5\sin 2\pi \left[ \frac{t}{0.04} - \frac{x}{40} \right]$,where distance is in $cm$ and time is in seconds,then the wavelength of the wave is .... $cm$.

If a given wave has a propagation constant of $\frac{5 \pi}{7} \, rad/m$,then the phase difference between two particles having a distance difference of $\frac{49}{22} \, m$ is ..... $rad.$ (Take $\pi = \frac{22}{7}$)

$A$ wave travelling in the positive $x-$direction with amplitude $A = 0.2\;m$ has a velocity of $v = 360\;m/s.$ If the wavelength $\lambda = 60\;m,$ then the correct expression for the wave is:

Two waves are represented by: $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$. Then the phase difference between them is

The wave function of a pulse is given by $y = \frac{5}{(4x + 6t)^2}$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The velocity of the pulse is ......... $m/s$.

$A$ wave motion has the function $y = a_0 \sin(\omega t - kx)$. The graph in the figure shows how the displacement $y$ at a fixed point varies with time $t$. Which one of the labelled points shows a displacement equal to that at the position $x = \frac{\pi}{2k}$ at time $t = 0$?