Question

Two pendulums start oscillating in the same direction at the same time from the same mean position time period are respectively 2s and 1.5s. The phase difference between them, when the smaller pendulum is completed vibration, will be
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{2}$
C. $\dfrac{{2\pi }}{3}$
D. $\dfrac{{3\pi }}{2}$

Answer 1

Hint: In case of simple harmonic motion there will be a mean position and two extreme positions. The pendulum 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:
$\eqalign{
  & x = A\sin (\varphi ) \cr
  & \varphi = \omega t \cr} $

Complete answer:
In SHM acceleration of the pendulum executing the SHM is proportional to the displacement of the pendulum 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. It's given in the question that pendulum 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 pendulum.
Amplitude will be maximum at extreme position and minimum at the mean position. The phase difference between the successive extreme and mean positions will be 90 degrees.
Let us assume SHM equation for first pendulum is
$\eqalign{
  & {x_1} = {A_1}\sin ({\varphi _1}) \cr
  & {\varphi _1} = {\omega _1}t \cr} $
$ \Rightarrow {\omega _1} = \dfrac{{2\pi }}{{{T_1}}} = \dfrac{{2\pi }}{2} = \pi $
$ \Rightarrow {x_1} = {A_1}\sin (\pi t)$
Let us assume SHM equation for second pendulum is
$\eqalign{
  & {x_2} = {A_2}\sin ({\varphi _2}) \cr
  & {\varphi _2} = {\omega _2}t \cr} $
$ \Rightarrow {\omega _2} = \dfrac{{2\pi }}{{{T_2}}} = \dfrac{{2\pi }}{{1.5}} = \dfrac{4}{3}\pi $
$ \Rightarrow {x_2} = {A_2}\sin (\dfrac{4}{3}\pi t)$
Time taken for a smaller pendulum to complete one vibration will be 1.5 seconds.
Then phase difference between two pendulums will be
$\eqalign{
  & \Delta \varphi = {\omega _2}t - {\omega _1}t \cr
  & \Rightarrow \Delta \varphi = \dfrac{{4\pi }}{3}t - \pi t \cr
  & \Rightarrow \Delta \varphi = \dfrac{\pi }{3}t \cr
  & \Rightarrow \Delta \varphi = \dfrac{\pi }{3}\left( {\dfrac{3}{2}} \right) \cr
  & \Rightarrow \Delta \varphi = \dfrac{\pi }{2} \cr} $

Hence option B will be the answer.

Note: Another simple way to solve this will be, the time taken for one complete vibration of a smaller pendulum is 1.5 seconds. Time taken for a bigger pendulum is 2 seconds. That means for every 0.5 seconds it will complete one quarter of SHM. In 1.5 seconds it will complete 3 quarters of SHM. Which means this will be at an extreme position. The smaller pendulum will be at a mean position. Hence the phase difference will be 90 degrees.