IIT JEE , AIEEE Forum - Physics , Mechanics

chetan_kp

let the time periods be T₁ and T₂

T = 2

( l/g)

T₁ / T_{2 =} 1/4

i.e. shorter pendulum moves 4 times faster than longer

now look at the diagram ...........

after the phase in the 3rd diagram the two pendulums come into the same phase after a short interval i.e. before the shorter pendulum completes the second oscillation

so no. of oscillations completed by the shorter should be between 1 and 2
in the given options it is clearly 4/3
this is just a shortcut i have found to save time(a lot of)
plz..... see if u can get me right or else nudge me

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