Two particles are in S.H.M. along parallel straight lines with the same amplitude and time period. If they cross each other in opposite directions at the midpoint of mean and extreme positions. Phase difference between them is:
A. \[{{30}^{0}}\]
B. \[{{120}^{0}}\]
C. \[{{150}^{0}}\]
D. \[{{180}^{0}}\]

Hint: To solve the above given question where two particles are in simple harmonic motion and we have to find the phase difference, we will first find out what exactly simple harmonic motion mean and then we will further use the equation of simple harmonic motion to solve our question.

Complete step-by-step solution:
In the question it is given that two particles are in simple harmonic motion that is S.H.M. moving along a straight line, these two particles have the same amplitude and same time period. They cross each other in opposite directions at the mid-point of mean and extreme position. And we have to find the angle of phase difference between them.
So, we will first discuss what is simple harmonic motion and we will further proceed to our solution
Simple Harmonic Motion or SHM can be defined as any movement in which the restoring force of a particle is directly proportional to the displacement of the given particle from its mean position. The direction of the restoring force of the particles is always towards the mean position.
It can also be described as any oscillatory motion of a particle in which the acceleration of the given particle at any position is directly proportional to the displacement from the mean position. It is a special case of oscillatory motion.
All the Simple Harmonic Motions are oscillatory and are also periodic but not all oscillatory motions are Simple harmonic motion. Oscillatory motion is often also known as the harmonic motion of all the oscillatory motions where the most important one is simple harmonic motion (SHM).
Now, write the equation of simple harmonic wave which is as follows:
\[y=A\sin (\omega t+\phi )\]
Where \[y=\dfrac{A}{2}\]
So, we can clearly say that
\[A\sin (\omega t+\phi )=\dfrac{A}{2}\]
Which means
\[\delta =\omega t+\phi =\dfrac{\pi }{6}or\dfrac{5\pi }{6}\]
Also, the phase difference of the given two particles when they are crossing each other at the mid-point that is \[y=\dfrac{A}{2}\]in opposite direction are –
\[\delta ={{\delta }_{1}}-{{\delta }_{2}}\]
\[=\dfrac{5\pi }{6}-\dfrac{\pi }{6}\]
\[=\dfrac{2\pi }{3}\]
Now, we know that the value of \[\pi \] is \[{{180}^{o}}\]
So, the value of \[\dfrac{2\pi }{3}\]will be equal to \[{{120}^{o}}\]
Hence, we can say that option(B) is the correct answer.

Note: When a body is having Simple harmonic motion by oscillating to and fro then its angular displacement should be small other SHM is not be done and when body is executing SHM the acceleration of that body is always directly proportional to negative of displacement.