Let T1 and T2 be the time period of the two pendulums T1=2π g1 and T2=2π g4 As ℓ1<ℓ2 therefore T1<T2 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 (n+1) oscillation. For unison swinging (n+1)T1=n2(n+1)×2πg1=(n)×2πg4⇒n=1∴n+1=1+1=2
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…
JEE Main 2013 — Physics Waves & Oscillations
2013mcqmedium
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:
Official previous-year question
Held on 9 Apr 2013 · Verified 6 Jul 2026.
Options
- A
2
- B
7
- C
5
- D
3
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