Two waves are given by $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \cos(\omega t - kx)$. The phase difference between the two waves is

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 $y$ of a wave travelling in the $x$-direction is given by $y = 10^{-4} \sin(600t - 2x + \frac{\pi}{3}) \, \text{m}$,where $x$ is expressed in metres and $t$ in seconds. The speed of the wave in $\text{m/s}$ is:

$Assertion :$ Speed of wave $= \frac{\text{wavelength}}{\text{time period}}$
$Reason :$ Wavelength is the distance between two nearest particles in phase.

An equation of a simple harmonic progressive wave is given by $y=A \sin (100 \pi t-3 x)$. The distance between two particles having a phase difference of $\frac{\pi}{18}$ in meters is:

The equation of a simple harmonic wave is given by $y = 6 \sin 2 \pi (2t - 0.1x)$,where $y$ and $x$ are in $mm$ and $t$ is in seconds. The phase difference between two particles $2 \ mm$ apart at any instant is (in $^{\circ}$)