Two oscillating simple pendulums with time periods T and 4T/3 are in phase at a given time. They will be again in phase after an elapse of time (a) 5T (b) 4T (c) 3T (d) 2T

Answer: (b). They are in phase again when the faster one has made exactly one more oscillation: (n+1)T = n·4T/3. n/3 = 1 ⇒ n = 3 Time = 3 × 4T/3 = 4T So the answer is (b) 4T.

Two oscillating simple pendulums with time periods T and 4T/3 are in… | MHT CET 2026 (20 Apr, Shift 1)

Q11MHT CET 2026 · 20 APR · SHIFT 1Medium

Simple Pendulum

Two oscillating simple pendulums with time periods $T$ and $\frac{4T}{3}$ are in phase at a given time. They will be again in phase after an elapse of time

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Answer: (b)

They are in phase again when the faster one has made exactly one more oscillation: $(n+1)T = n\cdot\frac{4T}{3}$ .
$\frac{n}{3} = 1 \Rightarrow n = 3$
Time $= 3 \times \frac{4T}{3} = 4T$
So the answer is (b) $4T$ .

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