Two pendula of length $$121$$ cm and $$100$$ cm start vibrating. At some instant the two are in mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position

Question

Infinity Learn · Accepted Answer

Here, we $$havet1=2$$$$π$$$$121g$$ and $$t2=2$$$$π$$$$100g$$ Let shorterpendulum male n vibmtions. Then the longer will make one less tian n to come in phase again, i.e., $$nt2=(n$$−$$1)t1$$ or $$n2$$$$π$$$$100g=(n$$−$$1)2$$$$π$$$$121g$$ n×$$10=(n$$−$$1)11$$ Solving, we get n $$=$$ $$11$$

Two pendula of length 121 cm and 100 cm start vibrating. At some instant the two are in mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position

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