Question

Two simple pendulums of length $1m{\text{ }}and{\text{ }}4\,m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:
A. 2
B. 7
C. 5
D. 3

Answer 1

In order to answer this question, as we know that the oscillation or the time period is directly proportional to the angular frequency, so we need to first find the angular frequency for both the pendulums separately, and then we will find the relative angular frequency of two pendulum.

Complete step by step answer:
Length of two pendulums are $1m{\text{ }}and{\text{ }}4m$ respectively. As per the question, first we will find the angular frequency of the simple harmonic motion of the pendulum is given as:
$\omega = \sqrt {\dfrac{g}{L}}$
where, $\omega$ is the angular frequency,$g$ is the gravity downward, and $L$ is the length of the pendulum.
So, for the first pendulum, the angular frequency is:
${\omega _1} = \sqrt {\dfrac{g}{{{L_1}}}} \\\ \Rightarrow {\omega _1}= \sqrt {\dfrac{g}{1}} \\\ \Rightarrow {\omega _1} = {\omega _0}$

Now, angular frequency for the second pendulum is:
${\omega _2} = \sqrt {\dfrac{g}{{{L_2}}}} \\\ \Rightarrow {\omega _2} = \sqrt {\dfrac{g}{4}} \\\ \Rightarrow {\omega _2} = \dfrac{1}{2}{\omega _0}$
Now, the relative angular frequency of two pendulum:
${\omega _1} - {\omega _2} = \dfrac{1}{2}{\omega _0}$
So, in order to comeback in same phase again the time taken by it is given as:
$t = \dfrac{\theta }{\omega } \\\ \Rightarrow t = \dfrac{{2\pi }}{{\dfrac{1}{2}{\omega _0}}} \\\ \therefore t = 2(\dfrac{{2\pi }}{{{\omega _0}}})$
So, in this time shorter pendulum will complete its 2 oscillations.

Hence, the correct option is A.

Note: The velocity of a basic pendulum changes with time, but it does so at a regular interval determined by the angular frequency. The pendulum's acceleration and location both change over time, but with a periodicity determined by angular frequency.