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 $.........\,km/s$.

$A$ wave is represented by the equation $y = 7 \sin(7\pi t - 0.04\pi x + \frac{\pi}{3})$,where $x$ is in metres and $t$ is in seconds. The speed of the wave is:

The equations of two simple harmonic waves are given by $Y_1 = 2 \sin 8 \pi \left(\frac{t}{0.2} - \frac{x}{2}\right) \text{ m}$ and $Y_2 = 4 \sin 8 \pi \left(\frac{t}{0.16} - \frac{x}{1.6}\right) \text{ m}$. Then both waves have:

Two points on a travelling wave having frequency $500\, Hz$ and velocity $300\, m/s$ are $60^{\circ}$ out of phase,then the minimum distance between the two points is ..... $m$.

The displacement of a wave is expressed as $x(t) = 5 \cos \left(628 t + \frac{\pi}{2}\right) \text{ m}$. The wavelength of the wave when its velocity is $300 \text{ m/s}$ is: (in $\text{ m}$)

$A$ person is producing a wave in a string by moving his hand first up and then down. If the frequency is $\frac{1}{8} \, Hz$,then find out the time taken by a particle which is at a distance of $9 \, m$ from the source to move to the lower extreme for the first time in seconds. (Given $\lambda = 24 \, m$)