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JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2013
- Hard
- Numerical
Two simple pendulums of length \(1\, m\) and \(4\, m\) 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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) be the time period of the two
pendulums \(T_{1}=2 \pi \sqrt{\frac{1}{g}}\) and \(T_{2}=2 \pi \sqrt{\frac{4}{g}}\)
As \(\ell_{1}<\ell_{2}\) therefore \(\mathrm{T}_{1}<\mathrm{T}_{2}\)
Let at \(t=0\) they start swinging together. since their time periods are different, the swinging will not be in unison always. Only when number of completed oscillations
differ by an integer, the two pendulums will again begin to swing together
Let longer length pendulum complete \(n\) oscillation and shorter length pendulum complete \((\mathrm{n}+1)\) oscillation. For unison swinging
\((n+1) T_{1}=n T_{2}\)
\((n+1) \times 2 \pi \sqrt{\frac{1}{g}}=(n) \times 2 \pi \sqrt{\frac{4}{g}}\)
\(\Rightarrow n=1\)
\(\therefore n+1=1+1=2\)
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