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Oscillations
Simple Pendulum
2 questions from 2 papers, each with its answer.
- questions
- 2
- papers
- 2
- years asked, 2017–2026
- 1 of 10
- of Oscillations
- 3.17%
Appears in
Concepts tested
- Frequency independent of mass
- in-phase condition of two pendulums
All Questions
2022 1 question
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Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :The longer pendulum (121 cm) has the larger time period. For them to be in phase again, the shorter one must complete one more oscillation: $n\,T_\ell = (n + 1)T_s$ $n \times 2\pi\sqrt{\frac{1.21}{g}} = (n + 1) \times 2\pi\sqrt{\frac{1}{g}}$ $1.1n = n + 1 \Rightarrow n = 10$ Number of oscillations of the shorter pendulum $= n + 1 = 11$
2000 AIPMT 1 question
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Two spherical bobs of masses $M_A$ and $M_B$ are hung vertically from two strings of length $l_A$ and $l_B$ respectively. They are executing SHM with frequency relation $f_A = 2f_B$, then :Frequency of a simple pendulum: $f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$, so $f \propto \dfrac{1}{\sqrt{l}}$; it does not depend on the mass of the bob.
$\dfrac{f_A}{f_B} = \sqrt{\dfrac{l_B}{l_A}}$
$2 = \sqrt{\dfrac{l_B}{l_A}} \Rightarrow 4 = \dfrac{l_B}{l_A}$
$l_A = \dfrac{l_B}{4}$