Question

Two particles are executing simple harmonic motion of the same amplitude$\;A$ and frequency $\omega$ along the x-axis. Their mean position is separated by distance ${X_0}{\text{ }}\left( {{X_0} > A} \right).$ If maximum separation between them is $\left( {{X_0} + A} \right)$, the phase difference between their motion is:
A) $\dfrac{\pi }{2}$
B) $\dfrac{\pi }{3}$
C) $\dfrac{\pi }{4}$
D) $\dfrac{\pi }{6}$

Answer 1

As SHM is an oscillatory motion of a system where the restoring force is proportional to the displacement, in order to find the phase difference, calculate the difference in the displacements of the two particles and compare the value with the maximum separation between the two points. From this relation find the value of the phase difference.

Complete step by step solution:
It is said that the two particles are executing simple harmonic motion of the same amplitude and frequency along the x-axis.
Let’s define the data given in the question
Amplitude of SHM=$\;A$
Frequency of SHM= $\omega$
Separation of mean position of the two particles=${X_0}$
Maximum separation between the two particles= $\left( {{X_0} + A} \right)$
The displacement of the particle which possess a simple harmonic motion is given by,
$x = A\sin (\omega t + \phi )$
Where, $t$ is the time period
$\phi$ is the phase shift
Displacement of the particle 1,${x_1} = A\sin (\omega t + {\phi _1})$
Displacement of the particle 2,${x_2} = A\sin (\omega t + {\phi _2})$
We are calculating the difference between the displacements of the two particles along the x- axis.
$\Rightarrow$ ${x_1} - {x_2} = A\sin (\omega t + {\phi _1}) - A\sin (\omega t + {\phi _2})$
\Rightarrow {x_1} - {x_2} = A\left\\{ {\sin (\omega t + {\phi _1}) - \sin (\omega t + {\phi _2})} \right\\}
Applying trigonometric conversions we get,
\Rightarrow {x_1} - {x_2} = A\left\\{ {2\sin \left( {\dfrac{{\omega t + {\phi _1} + \omega t + {\phi _2}}}{2}} \right)} \right\\}\sin \left\\{ {\dfrac{{\omega t + {\phi _1} - \omega t - {\phi _2}}}{2}} \right\\}
Cancelling the terms, we get,
\Rightarrow {x_1} - {x_2} = 2A\sin \left\\{ {\omega t + \dfrac{{{\phi _1} + {\phi _2}}}{2}} \right\\}\sin \left\\{ {\dfrac{{{\phi _2} - {\phi _1}}}{2}} \right\\}
\Rightarrow {x_1} - {x_2} = 2A\sin \left\\{ {\dfrac{{{\phi _2} - {\phi _1}}}{2}} \right\\}\sin \left\\{ {\omega t + \dfrac{{{\phi _2} + {\phi _1}}}{2}} \right\\}
The resultant motion can be treated as a simple harmonic motion with amplitude 2A\sin \left\\{ {\dfrac{{{\phi _2} - {\phi _1}}}{2}} \right\\}.
Given, maximum distance between the particles = $\left( {{X_0} + A} \right)$
That is, {X_0} + A - {X_0} = 2A\sin \left\\{ {\dfrac{{{\phi _2} - {\phi _1}}}{2}} \right\\}
\Rightarrow \dfrac{1}{2} = \sin \left\\{ {\dfrac{{{\phi _2} - {\phi _1}}}{2}} \right\\}
\Rightarrow \dfrac{{{\phi _2} - {\phi _1}}}{2} = {\sin ^{ - 1}}\left\\{ {\dfrac{1}{2}} \right\\}
$\Rightarrow \dfrac{{{\phi _2} - {\phi _1}}}{2} = \dfrac{\pi }{6}$
$\therefore {\phi _2} - {\phi _1} = \dfrac{\pi }{3}$
That is, the phase difference between the motion of the two particles, ${\phi _2} - {\phi _1} = \dfrac{\pi }{3}$

So the final answer is option (B), $\dfrac{\pi }{3}$.

Note: In physics, the simple harmonic motion of a particle is a special type of periodic motion in which the restoring force on the moving particle is directly proportional to the particles’ magnitude of displacement and acts towards the particle’s equilibrium position. This results in an oscillation which, if uninhibited by any dissipation of energy like friction, continues for indefinitely. A system that oscillates with simple harmonic motion is known as a simple harmonic oscillator.