The distance between two consecutive points with a phase difference of $60^{\circ}$ in a wave of frequency $500\,Hz$ is $6.0\,m$. The velocity with which the wave is traveling is $.........\,km/s$.

The equation of a wave is $y = 60 \sin (1200 t - 6 x)$,where '$y$' is in micron,'$t$' is in second,and '$x$' is in metre. The ratio of the maximum particle velocity to the wave velocity of wave propagation is:

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t) = e^{-(ax^2 + bt^2 + 2\sqrt{ab}xt)}$. This represents a:

$A$ longitudinal wave is represented by $x=x_0 \sin 2 \pi(n t-x / \lambda)$. The maximum particle velocity will be four times the wave velocity if:

$A$ wave travelling along a string is described by,
$y(x, t) = 0.005 \sin (80.0 x - 3.0 t)$
In which the numerical constants are in $SI$ units ($0.005 \, m, 80.0 \, rad \, m^{-1},$ and $3.0 \, rad \, s^{-1}$). Calculate
$(a)$ the amplitude,
$(b)$ the wavelength,and
$(c)$ the period and frequency of the wave.
Also,calculate the displacement $y$ of the wave at a distance $x = 30.0 \, cm$ and time $t = 20 \, s$.

For a wave,the path difference is $x$. The corresponding phase difference is ...