The phase difference between the following two waves $y_1$ and $y_2$ is:
$y_1 = a \sin(\omega t - kx)$
$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$

Two waves are represented by the equations, $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$, where $x$ is in metre and $t$ is in second. What is the phase difference between them? $(\pi = 3.14)$

The displacement equations of two waves are given as ${y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)$ and ${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$. What is the ratio of their amplitudes?

The relation between frequency '$n$',wavelength '$\lambda$',and velocity of propagation '$v$' of a wave is:

$A$ 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 $t = 20 \, s$.

What will be the change in phase of a wave due to reflection from a free support?