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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a
5
b
1
c
2
d
3
answer is C.
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Detailed Solution
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