Question:
Two simple pendulums of lengths \(1\,\mathrm{m}\) and \(1.44\,\mathrm{m}\) are in phase when released from the same extreme position. The minimum time after which the two pendulums will again be in phase is
\[
(g=\pi^2\,\mathrm{m\,s^{-2}})
\]
Two simple pendulums of lengths \(1\,\mathrm{m}\) and \(1.44\,\mathrm{m}\) are in phase when released from the same extreme position. The minimum time after which the two pendulums will again be in phase is
\[
(g=\pi^2\,\mathrm{m\,s^{-2}})
\]
Show Hint
Two oscillators starting in phase will again be in phase after
\[
\boxed{
t=\operatorname{LCM}(T_1,T_2)
}
\]
provided the ratio of their periods is rational.
Updated On: Jul 15, 2026
- \(5~\mathrm{s}\)
- \(10~\mathrm{s}\)
- \(6~\mathrm{s}\)
- \(12~\mathrm{s}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Calculate the time periods.
The time period of a simple pendulum is
\[
T=2\pi\sqrt{\frac{L}{g}}.
\]
For \(L=1\,\mathrm{m}\),
\[
T_1
=2\pi\sqrt{\frac{1}{\pi^2}}
=2~\mathrm{s}.
\]
For \(L=1.44\,\mathrm{m}\),
\[
T_2
=2\pi\sqrt{\frac{1.44}{\pi^2}}
=2.4~\mathrm{s}.
\]
Step 2: Find the least common time. The pendulums will again be in phase after the least common multiple of their periods. Since \[ 2\times6=12, \qquad 2.4\times5=12, \] the minimum common time is \[ \boxed{12~\mathrm{s}}. \]
Step 3: Choose the correct option. Hence, \[ \boxed{(D)} \] is the correct answer.
Step 2: Find the least common time. The pendulums will again be in phase after the least common multiple of their periods. Since \[ 2\times6=12, \qquad 2.4\times5=12, \] the minimum common time is \[ \boxed{12~\mathrm{s}}. \]
Step 3: Choose the correct option. Hence, \[ \boxed{(D)} \] is the correct answer.
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