Two simple pendulums of lengths I m and $$16$$ m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is:

Question

Infinity Learn · Accepted Answer

Let shorter pendulum makes n vibrations then longer makes (n-1) vibrations to come in phase again then $$nT1=n-1T2$$                                                                                                                                          $$nn-1=T2T1=161=4$$⇒$$n=43$$

Two simple pendulums of lengths I m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is:

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