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Question from Resonance DPP
JEE PROBLEMS
Answers - (D)
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✍ (D) 4
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EXPLANATION —
↪Actually welcome to the concept of Simple Harmonic Motion
Let shorter pendulam made n1 oscillation and bigger pendulam n2 oscillation
So total time elapsed T1 × n1 = T2 × n2
=> n1/n2 = (T2/T1)
T = 2π√(L/g)
Now T2/T1
= √(L2/L1) = √(16/1)
= > 4
Therefore, They will again be in phase for the first time when shorter pendulum cover 4 oscillation and bigger one onle 1 oscillation
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Answer :
Correct answer - option (D) 4
=> n = 4
Explanation :
Let the lengths of two pendulums be L1 & L2 with time period T1 & T2 respectively.
Here, L1 = 1m & L2 = 16 m
We have, Time period T = 2π√(L/g)
=> T1/T2 = √(L1/L2)
=> T1/T2 = √(1/16)
=> T1/T2 = 1/4
=> T1 = T2/4
So, when first pendulum will complete it's four oscillations the bigger one will complete it's one oscillation .
so they will be in phase for n = 4 .