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What is the percentage change in the time period, if the length of simple pendulum increases by `3%`. |
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Answer» Time period of simple pendulum of length l is `T=2pisqrt((l)/(g))` …(i) When length of simple pendulum increases by `3%` then `l_(1)=l+(3)/(100)l=((103)/(100))l` `(Deltal)/(l)=(l_(1)-1)/(l)=(3)/(100)` Taking log of (i) and differentiating it , we have `(DeltaT)/(T)=(1)/(2)(Deltal)/(l)` `:. % ` change in time period `=(DeltaT)/(T)xx100=(1)/(2)(Deltal)/(l)xx100=(1)/(2)xx3=1.5%` |
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