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Two simple pendulum `A` and `B` of lengths `1.69m` and `1.44m` start swinging at the time from a location where acceleration due to gravity is `10ms^(-1)`. Answer the following question. After how much time, the two pendulums will be in phase again ?A. `11.0s`B. `20.0s`C. `25.2s`D. `31.0s` |
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Answer» Correct Answer - D Let `t_(1)` be the time after which two pendulums are in same phase. It means, in time `t_(1)`, if `A`complets `n` vibrations, `B` will complte `(n+1)` vibrations. So, `t_(1)=nT_(1)=(n+1)T_(2)` or `(T_(1))/(T_(2))=(n+1)/(n)=(13)/(12)` o r `12n+12=13n` or `n=12` `t_(1)=12xx2pisqrt((1.69)/(10))=31s` |
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