Two simple pendulums of length $$5$$ m and $$20$$ m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.

Question

Infinity Learn · Accepted Answer

If t is the time taken by pendulums to come in same phase again first time after $$t=$$  $$0$$. and Ns $$=$$ Number of oscillations made by shorter length pendulum with time period TS.   $$NL=$$ Number of oscillations made by longer length pendulum with time period TL.   Then t $$=$$ NSTS $$=$$ NLTL⇒$$NS2$$$$π$$$$5g$$ $$=$$ NL×$$2$$$$π$$$$20g$$ $$($$∵ T $$=$$ $$2$$$$π$$$$lg)$$⇒NS $$=$$ $$2NL$$   i.e., if NL $$=$$ $$1$$ ⇒NS $$=$$ $$2$$

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.

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