The distance between two consecutive points with a phase difference of $60^{\circ}$ in a wave of frequency $500 \text{ Hz}$ is $0.6 \text{ m}$. The velocity with which the wave is travelling is: (in $\text{ km/s}$)

$A$ sound wave of frequency $200 Hz$ is travelling in air. The speed of sound in the air is $340 m s^{-1}$. What is the phase difference at a given instant between two points separated by a distance of $85 cm$ along the direction of propagation?

$A$ transverse wave is described by the equation $y = y_0 \sin 2 \pi \left( \nu t - \frac{x}{\lambda} \right)$. The maximum particle velocity is equal to four times the wave velocity if $\lambda =$

$A$ wave travelling along the $x$-axis is described by the equation $y(x, t) = 0.005 \cos(\alpha x - \beta t)$. If the wavelength and the time period of the wave are $0.08 \ m$ and $2.0 \ s$,respectively,then $\alpha$ and $\beta$ in appropriate units are:

The equation of a transverse wave is given by $y=0.05 \sin \pi(2 t-0.02 x)$,where $x, y$ are in metre and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively

The equation of a transverse wave travelling on a rope is given by $y = 10\sin \pi (0.01x - 2.00t)$ where $y$ and $x$ are in $cm$ and $t$ is in $seconds$. The maximum transverse speed of a particle in the rope is about .... $cm/s$.