First slide

Periodic Motion

Question

Two simple pendulums of length l m and 16 m start at t =0 from mean position. The minimum time after which they will be again in phase will be

Moderate

A

4T/3
B

4T
C

2T/3
D

T/3

Solution

The phase of simple pendulum is given by $θ = ωt$
$where ω is angular speed. Phase of first pendulum = ω_{1} t Phase of second pendulum = ω_{2} t Phase difference = 2 nπ ω_{1} t - ω_{2} t = 2 nπ or \frac{2 π}{T} t - \frac{2 π}{4 T} t = 2 nπ (∵ ω_{1} = 2 π / T and ω_{2} = 2 π / 4 T) Puttirg n = 1, we get t [\frac{1}{T} - \frac{1}{4 T}] = 1 or t [\frac{3}{4 T}] = 1 or t = \frac{4 T}{3}$

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