Physics, asked by ZiaAzhar89, 11 months ago

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Answered by BrainlyWriter
4

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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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Answered by sagarnirapure914
35

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 .

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