Question

Two particles executing SHM of same frequency, meet at $x = + {A \mathord{\left/{\vphantom {A 2}} \right.} 2}$, while moving in the opposite directions. Calculate phase difference between the particles.

Answer 1

Hint
From the wave equations of both the particles having a phase difference $y$, by substituting $x = + {A \mathord{\left/{\vphantom {A 2}} \right.} 2}$ in one of them, we can find the value of the phase angle $\theta$. Then by substituting that in the other equation we get the phase difference.
In this solution we will be using the following formula,
$\Rightarrow x = A\sin \left( {\omega t + \varphi } \right)$
Where $x$ is the position and $A$ is the amplitude of the wave.
$\omega$ is the angular frequency, $t$ is the time and $\varphi$ is the phase difference.

Complete step by step answer
In the question we are given that 2 particles are executing SHM of the same frequency but having a phase difference.
So we can write the wave equation of the 2 particles in the SHM as,
$\Rightarrow {x_1} = A\sin \left( {\omega t} \right)$ for the first particle
$\Rightarrow {x_2} = A\sin \left( {\omega t + \varphi } \right)$ for the second particle
Now, it is said that the two particles meet at the position given by $x = + {A \mathord{\left/ {\vphantom {A 2}} \right.} 2}$
So we can write that ${x_1} = {x_2} = + {A \mathord{\left/ {\vphantom {A 2}} \right.} 2}$
So we can substitute this value in the first equation and we get,
$\Rightarrow {A \mathord{\left/ {\vphantom {A 2}} \right.} 2} = A\sin \left( {\omega t} \right)$
From here we can cancel the amplitude on both the sides of the equation.
So we get
$\Rightarrow \sin \left( {\omega t} \right) = \dfrac{1}{2}$
Now taking sine inverse on both the sides of the equation we have,
$\Rightarrow \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Now ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ has 2 values which are, $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$
Therefore we get the value of $\omega t$ as
$\Rightarrow \omega t = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$
Now the 2 particles are moving in opposite directions. So if we take the value of $\omega t$ of the first particle as $\dfrac{\pi }{6}$ and the second particle as $\dfrac{{5\pi }}{6}$.
So we can write the wave equation of the second particle as,
$\Rightarrow {x_2} = A\sin \left( {\dfrac{{5\pi }}{6} + \varphi } \right)$
Now substituting the value of ${x_2} = + {A \mathord{\left/ {\vphantom {A 2}} \right.} 2}$ we get
$\Rightarrow \dfrac{A}{2} = A\sin \left( {\dfrac{{5\pi }}{6} + \varphi } \right)$
Cancelling $A$ on both sides of the equation we get
$\Rightarrow \dfrac{1}{2} = \sin \left( {\dfrac{{5\pi }}{6} + \varphi } \right)$
Therefore, taking sine inverse on both sides we have,
$\Rightarrow \dfrac{{5\pi }}{6} + \varphi = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Hence we can write from the above equation,
$\Rightarrow \dfrac{{5\pi }}{6} + \varphi = \dfrac{\pi }{6}$
Therefore to calculate the value of the phase difference we take $\dfrac{{5\pi }}{6}$ to the RHS and get
$\Rightarrow \varphi = \dfrac{\pi }{6} - \dfrac{{5\pi }}{6}$
On doing the calculation we get,
$\Rightarrow \varphi = - \dfrac{{2\pi }}{3}$
So the phase difference between the particles is $\varphi = - \dfrac{{2\pi }}{3}$ where the negative sign indicates the particles moving in opposite sides.

Note
The difference between the phases of two periodic signals is called the phase difference between those two signals. When this difference is equal to zero, then the two signals are said to be in phase, else the signals are said to be out of phase.