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The period of oscillation of a simple pendulum is given by T=2pi sqrt((l)/(g)). In finding the value of g, which quantity should be measured most accurately and why? |
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Answer» Solution :`T=2pi sqrt((L)/(g)) rArr T^(2)=(4pi^(2)l)/(g)` `or g=4pi^(2)(l)/(T^(2))` `therefore"Fractional ERROR in g"=(DELTAG)/(g)=(Deltal)/(l)+(2DeltaT)/(T)` Note that any error in T is doubled up. Therefore, time should be measured most accurately. |
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