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JEE Main 20139 AprilOnline
Exam
JEE Main
Year
2013
Date
9 April
Mode
Online
Subject
Physics
Chapter
Oscillations

JEE Main 2013 · 9 April · Online · Physics · Oscillations

Two simple pendulums of length 1 ~m and 4 ~m respectively are both...

Two simple pendulums of length 1 m1 \mathrm{~m} and 4 m4 \mathrm{~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:

Options

  1. A2
  2. B7
  3. C5
  4. D3

Correct answer

A. 2

Step-by-step solution

Let T1andT2T_1 and T_2 be the time period of the two pendulums T1=2π1 gand\mathrm{T}_1=2 \pi \sqrt{\frac{1}{\mathrm{~g}}} and\mathrm{T}_2=2 \pi \sqrt{\frac{4}{\mathrm{~g}}}AsAs\ell_1 \lt \ell_2thereforetherefore\mathrm{T}_1 \lt \mathrm{T}_2LetatLet at\mathrm{t}=0theystartswingingtogether.Sincetheirtimeperiodsaredifferent,theswingingwillnotbeinunisonalways.Onlywhennumberofcompletedoscillationsdifferbyaninteger,thetwopendulumswillagainbegintoswingtogetherLetlongerlengthpendulumcompletethey 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\mathrm{n}oscillationandshorterlengthpendulumcompleteoscillation and shorter length pendulum complete(n+1)oscillation.Forunisonswingingoscillation. For unison swinging(n+1)T1=n2(n+1)×2π1g=(n)×2π4g⇒n=1∴n+1=1+1=2\begin{aligned}& (n+1) T_1=n_2 \\ & (n+1) \times2 \pi \sqrt{\frac{1}{g}}=(n) \times2 \pi \sqrt{\frac{4}{g}}\\ & \Rightarrow n=1 \\ & \therefore n+1=1+1=2 \end{aligned}$
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