Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is $$5T4$$, where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.

Question

Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is $$5T4$$, where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.

Infinity Learn · Accepted Answer

Let $$T1$$ $$=$$ T and $$T2$$ $$=$$ KT, $$2$$$$π$$$$Tt-2$$$$π$$KTt $$=$$ $$2$$$$π$$$$t(K-1KT)$$ $$=$$ $$1$$; $$5T4(K-1KT)$$ $$=1$$ which gives K $$=5If$$ length of first pendulum $$=$$ l, $$T1$$ $$=$$ $$2$$$$π$$$$lgT2$$ $$=$$ $$5$$ ×$$2$$$$π$$lg $$=$$ $$2$$$$π$$$$25lgSo$$, ratio of lengths $$=$$ $$1$$ : $$25$$

Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is $\frac{5 T}{4}$ , where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.

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Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is 5T4, where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Two pendulum of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is $\frac{5 T}{4}$ , where T is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.