Question

Two pendulums having time period T and $\dfrac{5T}{4}$. They start S.H.M. at the same position. What will be the phase difference between them after the bigger pendulum has completed one oscillation?
A. ${{45}^{\circ }}$
B. ${{90}^{\circ }}$
C. ${{60}^{\circ }}$
D. ${{30}^{\circ }}$

Answer 1

Hint: In this question, we are given two pendulums having SHM with different time periods and we have to find the phase difference after the bigger one completes one oscillation. Phase of SHM represents the state of a particle from the mean position at a certain instant. We need to understand the time period of oscillations which a pendulum completes and then we will be able to get our correct answer.

Complete step by step solution:
We analyze the oscillation of the pendulum by resolving the SHM into a circular motion of angular velocity. We have to find the phase difference between them after the bigger pendulum has completed one oscillation. For the pendulum with time period $\dfrac{5T}{4}$, we can split the time in the $\dfrac{5T}{4}$= $T+\dfrac{T}{4}$.

At time t = 0, the first pendulum have a time period = T
And the second have time period $\dfrac{5T}{4}$= $T+\dfrac{T}{4}$
This means that when the second pendulum completes 1 oscillation the first one completes $1+\dfrac{1}{4}$oscillation. First one completes extra $\dfrac{1}{4}$ oscillation.
Therefore, the phase difference between them is $\dfrac{2\pi }{4}$ = ${{90}^{\circ }}$

Thus, option B is the correct answer.

Note: Students must know about the phase to solve this question. We know displacement of a particle from the mean position is represented by sin and cos function. We know the general equation of SHM is $x=A\sin (wt+\theta )$ where $(wt+\theta )$is the phase at any instant t and the difference between two phases is the phase difference.