Question

A particle is executing SHM of amplitude A, about the mean position x=0. Which of the following is a possible phase difference between the positions of the particle at $x = + \dfrac{A}{2}$ and $x = - \dfrac{A}{{\sqrt 2 }}$’.
A) ${75^\circ }$
B) ${165^\circ }$
C) ${135^\circ }$
D) ${195^\circ }$

Answer 1

In this question we have to find the phase difference between the positions of the particle at $x = + \dfrac{A}{2}$ and $x = - \dfrac{A}{2}$. For this we are going to use the equation of Simple Harmonic Motion and then put the values of the positions and find the value of phase difference between those positions.

Complete step by step solution:
Given,
Positions- $x = + \dfrac{A}{2}$and $x = - \dfrac{A}{2}$
The equation of Simple Harmonic Motion (SHM) is given below;
$x = A\cos (\omega t + \phi )$
Where, x is the position and
$\omega$ is angular frequency
$\phi$ is the phase difference
Let the phase difference of first position is ${\phi _1}$ and the phase difference of second position is ${\phi _2}$.
Now, putting the values of positions and phase difference in the above equations of simple harmonic motion;
When $x = \dfrac{A}{2}$ the equation of motion becomes,
$\dfrac{A}{2} = A\cos (\omega t + {\phi _1})$….. (1)
When $x = - \dfrac{A}{{\sqrt 2 }}$the equation of motion becomes,
$\Rightarrow - \dfrac{A}{{\sqrt 2 }} = A\cos (\omega t + {\phi _2})$….. (2)
From equation (1)
$\Rightarrow \dfrac{1}{2} = \cos (\omega t + {\phi _1})$
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = (\omega t + {\phi _1})$
$\Rightarrow 60^\circ = (\omega t + {\phi _1})$….. (3)
From equation (2)
$\Rightarrow - \dfrac{1}{{\sqrt 2 }} = \cos (\omega t + {\phi _2})$
$\Rightarrow {\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = (\omega t + {\phi _2})$
$\Rightarrow 45^\circ = (\omega t + {\phi _2})$….. (4)
Phase difference ${\phi _2} - {\phi _1} = 60^\circ - 45^\circ$
$\Rightarrow {\phi _2} - {\phi _1} = 15^\circ$
Now, if we consider the equation of simple harmonic motion in terms of sine then the equation of motion is given by,
$\Rightarrow x = A\sin (\omega t + \phi )$
Now, putting the values of positions and phase difference in the above equations of simple harmonic motion;
When $x = \dfrac{A}{2}$ the equation of motion becomes,
$\Rightarrow \dfrac{A}{2} = A\sin (\omega t + {\phi _1})$….. (5)
When $x = - \dfrac{A}{{\sqrt 2 }}$the equation of motion becomes,
$\Rightarrow - \dfrac{A}{{\sqrt 2 }} = A\sin (\omega t + {\phi _2})$….. (6)
From equation (5)
$\Rightarrow \dfrac{1}{2} = \sin (\omega t + {\phi _1})$
$\Rightarrow {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = (\omega t + {\phi _1})$
$\Rightarrow 30^\circ = (\omega t + {\phi _1})$….. (7)
From equation (6)
$\Rightarrow - \dfrac{1}{{\sqrt 2 }} = \sin (\omega t + {\phi _2})$
$\Rightarrow {\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = (\omega t + {\phi _2})$
$\Rightarrow 225^\circ = (\omega t + {\phi _2})$….. (4)
Phase difference ${\phi _2} - {\phi _1} = 225^\circ - 30^\circ$
$\Rightarrow {\phi _2} - {\phi _1} = 195^\circ$
Result- Hence, from above calculation we have seen that the phase difference between the positions $x = + \dfrac{A}{2}$ and $x = - \dfrac{A}{{\sqrt 2 }}$ is ${\phi _2} - {\phi _1} = 195^\circ$.

Note: In this question we have used the equation of motion for simple harmonic motion, for this we should have the knowledge of simple harmonic motion. Simple harmonic motion is a type of periodic motion where the restoring force on the moving object is directly proportional to the magnitude of the displacement of the object and the direction of restoring force is towards the mean position of the object.