Asked by | 24th Jan, 2009, 04:05: PM
Use T = 2 pi (square root of L/g)
T1/T2 = (square root of L1)/square root of L2 = 11/10
So they will be in the same phase in 110 seconds which is the hcf of time period of the two pendulums.
Now you can find how many vibrations the shorter pendulum makes in 110s.
Answered by | 28th Jan, 2009, 02:40: PM
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