$A$ transverse wave is described by the equation $y = A \sin [2\pi (f t - x/\lambda ) ]$. The maximum particle velocity is equal to four times the wave velocity if:

The distance between two consecutive points with a phase difference of $60^{\circ}$ in a wave of frequency $n$ is $x$ meters. The velocity with which the wave is traveling is:

$A$ transverse wave in a medium is given by $y = A \sin 2(\omega t - kx)$. It is found that the magnitude of the maximum velocity of particles in the medium is equal to the magnitude of the wave velocity. What is the value of $A$?

Two points are located at a distance of $10\; m$ and $15 \;m$ from the source of oscillation. The period of oscillation is $0.05 \;s$ and the velocity of the wave is $300 \;m/s$. What is the phase difference between the oscillations of the two points?

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 $\text{ m}$)

If a travelling wave is given by $y(x, t) = 0.5 \sin (70.1 x - 10 \pi t)$,where $x$ and $y$ are in metre,the time $t$ is in second,then the frequency of the wave is (in $Hz$)