Question

Two simple pendulums of length $1\text{ m}$ and $1.21\text{ m}$ are at their mean positions with their velocities in the same direction at some instant. After how many oscillations of the bigger pendulum they will be again in phase.
A) 9
B) 10
C) 11
D) 12

Answer 1

Hint: The time period of a simple pendulum depends upon the length of the string and the acceleration due to gravity. Finding the ratio between the time periods of the two simple pendulums gives an idea about the phase of the two pendulums.

Complete step by step answer:
The time period $\text{T}$of the simple pendulum is directly proportional to the square root of its length $\left( l \right)$and inversely proportional to the square root of acceleration due to gravity $\left( g \right)$. So we can write the equation of time period as,
$T=2\pi \sqrt{\dfrac{l}{g}}$
So the time period of the first pendulum can be expressed as,
${{T}_{1}}=2\pi \sqrt{\dfrac{{{l}_{1}}}{g}}$
The time period of the second pendulum can be expressed as,
${{T}_{2}}=2\pi \sqrt{\dfrac{{{l}_{2}}}{g}}$
Taking the ratio between ${{\text{T}}_{\text{2}}}\text{ and }{{\text{T}}_{1}}$ gives,
$\dfrac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\dfrac{{{l}_{2}}}{{{l}_{1}}}}$
$\Rightarrow \dfrac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\dfrac{1.21m}{1m}}$
$\therefore \dfrac{{{T}_{2}}}{{{T}_{1}}}=1.1=\dfrac{11}{10}$
$\Rightarrow 10{{T}_{2}}=11{{T}_{1}}$
Which means that by the time pendulum 2 with time period ${{\text{T}}_{\text{2}}}$takes 10 oscillations, pendulum 1 with time period ${{\text{T}}_{1}}$ would finish 11 oscillations. So, after 10 oscillations by the pendulum 2, the two pendulums will be in phase.
So the answer to the question is option (B) 10.

Note: The time period of a simple pendulum is independent of the mass of the bob. The equation for the time period that we used in the problem is only valid for small swings. (small-angle subtended by the string with the vertical)
If the swing taken is relatively low, then the motion of a simple pendulum can be approximated to a harmonic oscillator.