Question

Two particles P and Q describe Simple Harmonic Motion (SHM) of same amplitude (a) and frequency (v) along the same straight line. The maximum distance between the two particles is $ \sqrt { 2 }a $ . The initial phase difference between the particles is:
$ A.Zero $
$ B.\dfrac{\pi }{2} $
$ C.\dfrac{\pi }{6} $
$ D.\dfrac{\pi }{3} $

Answer 1

Hint: Use the basic formula of Simple harmonic motion for a particle along an axis and use the same formula for the second particle. We don’t have to add phase angle to both the particles, since we only need to find the phase difference. Adding phase angle to either one is just enough as the phase difference will turn out to be that phase angle itself. Remember the trigonometric formulations for these kinds of questions as they play key roles in solving.

Complete step-by-step answer:
Let’s consider the 2 particles P and Q are executing SHM along the X axis. To further ease the problem we will consider only one of the particles to contain an additional phase angle while the other particle’s wave equation will not consider any phase angle.
SHM of P is $ {{x}_{1}}=a\sin \omega t $ and SHM of Q is $ {{x}_{2}}=a\sin (\omega t+\phi ) $ .
Here $ \omega =2\pi v $ .
The maximum distance between the 2 particles (d) is $ d=({ x }_{ 2 }-{ x }_{ 1 }) $ and $ d=\sqrt { 2 }a $ .
Therefore, \[\sqrt{2}a=[a\sin (\omega t+\phi )-asin(\omega t)]\]
 $ \sqrt{2}a=a[\sin (\omega t+\phi )-sin(\omega t)] $
Using the formula of $ \sin { C } -\sin { D } =2[\cos { (\dfrac { C+D }{ 2 } ) } \sin { (\dfrac { C-D }{ 2 } ) } ] $ ,
 $ \sqrt{2}=2[\cos (\dfrac{2\omega t+\phi }{2})\sin \dfrac{\phi }{2}] $
For maximum distance, we should have $ \cos (\dfrac{2\omega t+\phi }{2})=1 $ . This is necessary as “+1” is the maximum possible value of any cos angle.
Therefore, $ 1=\sqrt { 2 } [1\times \sin { \dfrac { \phi }{ 2 } } ] $
 $ \dfrac { 1 }{ \sqrt { 2 } } =\sin { \dfrac { \phi }{ 2 } } $
 $ \dfrac { 1 }{ \sqrt { 2 } } =\sin { \dfrac { (2n+1)\pi }{ 4 } } $ for n= any whole number.
We have been asked in the question to find the initial phase difference. Hence, we will take the smallest possible value of n. Therefore, n=0.
 $ \sin { \dfrac { \pi }{ 4 } =\sin { \dfrac { \phi }{ 2 } } } $
 $ \dfrac { \pi }{ 4 } =\dfrac { \phi }{ 2 } $
Therefore, $ \dfrac { \pi }{ 2 } =\phi $ .

Note: It’s important to remember here, that the SHM happens along the same direction.
Here, since we are trying to find the value of phase difference, we’ve considered the value of $ \cos (\dfrac{2\omega t+\phi }{2})=1 $ , as the numerator will always equate to $ \pi $ for maximum displacement between the particles condition.