Relation Between Phase Difference And Path Difference - Practice Questions & MCQ

The general equation of sine wave is given as : 

$
y=A \sin (\omega t-k x+\phi)
$

Here $\omega$, is the angular frequency i.e,

$
\omega=\frac{2 \pi}{T}=2 \pi f_{\text {It defines how many cycles of the oscillations are there. }}
$

and $\phi=$ phase angle.
Phase:
The quantity which expresses at any instant, the displacement of the particle and its direction of motion is called the phase of the particle.

If two particles of the medium, at any instant are in the same state of motion ( parameters such as particle's displacement, velocity, and acceleration are same) then they are said to be in the same phase.

The phase of the wave is the quantity inside the brackets of the sin-function, and it is an angle measured either in degrees or radians.

$
\phi=\left(\frac{2 \pi}{T} t-\frac{2 \pi}{\lambda} x\right)
$

At a particular time t , The phase difference of the wave between point $\mathrm{A}\left(x_1\right)$ and point $\mathrm{B}\left({ }^{x_2}\right)$ is given by

$
\begin{aligned}
\phi_1-\phi_2 & =\left(\frac{2 \pi}{\lambda} x_2-\frac{2 \pi}{\lambda} x_1\right) \\
\phi_1-\phi_2 & =\frac{2 \pi}{\lambda}\left(x_2-x_1\right)
\end{aligned}
$

The important result here is that the two waves can be:
(1). In phase if $x_2-x_1=n \lambda_{\text {, }}$ i.e the particles corresponding to positions $x_1$ and $x_2$ are in the same state of motion.

2) Out of phase if

$
x_2-x_1=\left(n+\frac{1}{2}\right) \lambda
$

$x_{2 \text { is moving downwards but symmetrically. }}$