If $$T1$$ and $$T2$$ are the time periods of oscillation of a simple pendulum on the surface of earth (of$$ radius $$R) and at a depth d, the d is equal to

Question

Infinity Learn · Accepted Answer

We know that, the time period of a simple pendulum,                  $$T=2$$$$π$$lg⇒              T∝$$1gHere$$, in given condition,                   $$T1=Kg$$                                         …(i)and$$             $$T2=Kg1-dR$$                           …$$(ii)On$$ dividing Eq. $$(i) by Eq. (ii), we get                   $$T1T2=K/gK/g1-dR$$   or                 $$T1T2=1-dR$$   or   $$T12T22=1-dR$$⇒               $$d=1-T12T22R$$

If T₁ and T₂ are the time periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

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If T1 and T2 are the time periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

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If T₁ and T₂ are the time periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to