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Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations. 

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detailed solution

Correct option is C

If t is the time taken by pendulums to come in same phase again first time after  t=0.and NS=Number of oscillations made by shorter length pendulum with time period TS.NL= Number of oscillations made by longer length pendulum with time period  TL.Then  t=NSTS=NLTL⇒NS2π5g=NL×2π20g              (∵T=2πlg)⇒NS=2NL  i.e. if  NL=1⇒NS=2


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