Question

A particle is executing SHM of amplitude A about the mean position $x = 0$. Which of the following cannot be a possible phase difference between the positions of the particle at $x = + \dfrac{A}{2}$ and $x = - \dfrac{A}{{\sqrt 2 }}$
A. ${75^0}$
B. ${165^0}$
C. ${135^0}$
D. ${195^0}$

Answer 1

Hint: In case of simple harmonic motion there will be a mean position and two extreme positions. The particle executes the SHM with respect to the mean position and within the range of two extreme positions. Generally simple harmonic motions are denoted with the sinusoidal or cosecant functions.

Formula used:
$x = A\sin (\varphi )$

Complete answer:
In SHM acceleration of the particle executing the SHM is proportional to the displacement of the particle from the mean position. Generally if an object starts from the mean position then we write displacement as a sinusoidal function and if an object starts from the extreme position then we write the displacement of a body as a cosine function. We can assume in any way if it is not specified in the question. Let us assume that the given particle is executing the SHM about the mean position. Then the function of the SHM displacement will be $x = A\sin (\varphi )$ where $\varphi $ is the phase of the particle.
In the question positions given are $x = + \dfrac{A}{2}$ and $x = - \dfrac{A}{{\sqrt 2 }}$
If we equate $x = + \dfrac{A}{2}$ then we will get $\varphi $ as
$x = A\sin (\varphi )$
$\eqalign{
  & \Rightarrow + \dfrac{A}{2} = A\sin (\varphi ) \cr
  & \Rightarrow \sin (\varphi ) = + \dfrac{1}{2} \cr
  & \Rightarrow \varphi = {30^0},{150^0} \cr} $
If we equate $x = - \dfrac{A}{{\sqrt 2 }}$ then we will get $\varphi $ as
$x = A\sin (\varphi )$
$\eqalign{
  & \Rightarrow - \dfrac{A}{{\sqrt 2 }} = A\sin (\varphi ) \cr
  & \Rightarrow \sin (\varphi ) = - \dfrac{1}{{\sqrt 2 }} \cr
  & \Rightarrow \varphi = {225^0},{315^0} \cr} $
Now to get the possible values of phase difference we should subtract one value from the other
225-30=195
225-150=75
315-30=285
315-150=165
So except ${135^0}$ all the other options are possible.

Hence answer would be option C.

Note:
Even if we take the displacement as the cosine function then also we will get the same result. The other way to solve this problem will be draw a sinusoidal graph of amplitude one and find out the angles where sinusoidal function will become $\dfrac{1}{2}, - \dfrac{1}{{\sqrt 2 }}$ and find out the difference between those angles individually just as we did. We can draw phasor diagrams too to solve the problem.