A pendulum that beats seconds on the surface of the earth were taken to a depth of $$(1/4)th$$ The radius of the earth is R. What will be its time period of oscillation?

Question

A pendulum that beats seconds on the surface of the earth were taken to a depth of $$(1/4)th$$  The radius of the earth is R. What will be its time period of oscillation?

Infinity Learn · Accepted Answer

from gravitation, acceleration due to gravity at a depth d from surface of earth is, $$g1=g1-dR$$, here given $$d=R4$$, substitute $$g1=g1$$−$$R/4R=3g/4$$  $$=$$ $$g2$$  (suppose) time period of simple pendulum is    $$T=2$$$$π$$lg  ⇒Tα$$1g$$, where length l of pendulum is constant ⇒$$Tg=constant$$ $$T1g1=T2g2$$  substitute $$g1=g$$ and $$g2=3g4$$ pendulum beats seconds means, $$T1=2$$ seconds $$2g=T23g4$$ $$2=T232$$ ∴$$T2=43=2.309$$ seconds

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th The radius of the earth is R. What will be its time period of oscillation?

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