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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. How many vibratinos pendulum `A` will compelte when it will be out of phase with `B` ?A. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - C Let `T_(1)` and `T_(2)` be the time period of vibrations of pendulums `A` and `B` respectively. Then, `T_(1)=2pisqrt((l_(1))/(y))` and `T_(2)=2pisqrt((l_(2))/(g))` or `(T_(1))/(T_(2))=sqrt((l_(1))/(l_(2)))=sqrt((1.69)/(1.44))=(13)/(12)` If the two pendulums go out of phase in time `t`, then in time `t`, if pendulum `A` completes `n` vibrations, the pendulum `B` will complete `(n+1//2)`vibrations. Therefore, `t=nT_(1)=(n+1//2)T_(2)` or `(T_(1))/(T_(2))=((n+1//2))/(n)=(13)/(12)` or `12n+6=13n` or `n=6` |
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