Question

Two particles execute simple harmonic motions of the same amplitude and frequency along the same straight line. They cross one another when going opposite directions. The phase difference between them when their displacements are one half of their amplitude is then,
A) ${60^ \circ }$
B) ${30^ \circ }$
C) ${120^ \circ }$
D) ${150^ \circ }$

Answer 1

Hint:Two particles tend to execute simple harmonic motion of same amplitude and same frequency. Hence, the equation of simple harmonic motion of both of the particles are the same. It is given that; the both the particles have displacement equal to half of the amplitude of the wave. Hence, both the particles have the same displacement. By substituting the displacement value in the equation simple harmonic motion, the phase values can be obtained. Using the two-phase values, the phase difference between the waves can be calculated.

Useful formula:
The equation of simple harmonic motion (SHM) is given by,
$y = A\sin \left( {\omega t + \phi } \right)$
Where, $y$ is the displacement of the particle, $A$ is the amplitude of the wave and $\left( {\omega t + \phi } \right)$ is the phase angle of the wave.

Complete step by step answer:
Given data:
The displacement of the wave is equal to one half of the amplitude of the wave.
Hence, $y = \dfrac{A}{2}$

Since, both of the particles are executing the same simple harmonic motion (SHM) with same amplitude and same frequency, then the equation of the SHM for both the particles is the same. It is given by,
$y = A\sin \left( {\omega t + \phi } \right)\;......................................\left( 1 \right)$

And it is given that the displacement of both the particles is equal to one half of the amplitude of the waves.
Hence, the displacement of the particles $y = \dfrac{A}{2}$
Substituting the displacement value in the equation (1), we get
$
  \dfrac{A}{2} = A\sin \left( {\omega t + \phi } \right) \\
  \sin \left( {\omega t + \phi } \right) = \dfrac{A}{{2A}} \\
  \sin \left( {\omega t + \phi } \right) = \dfrac{1}{2} \\
 $
Taking the term $\sin $ to the RHS of the equation,
$\omega t + \phi = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Since, ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {30^ \circ }\left( {or} \right){150^ \circ }$
Hence,
$\omega t + \phi = {30^ \circ }$ (or) $\omega t + \phi = {150^ \circ }$
Thus, the phase angle of the first wave is ${30^ \circ }$ and the phase angle of the second wave is ${150^ \circ }$.
So, the phase difference between the two particles is ${150^ \circ } - {30^ \circ } = {120^ \circ }$

Hence, the option (C) is correct.

Note:
Even though the two particles execute the same simple harmonic motion with same amplitude and same frequency but they are executing in opposite directions. The phase difference between the two particle waves is ${120^ \circ }$, it is clear that the two particles are moving in different directions.