Question

The period of oscillation of a simple pendulum is T = 2π√(L/g).
Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

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.・゜゜・kindly don't copy and paste answer from anywhere.
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Answer 1

as per the given formula,

$t = 2\pi \sqrt{ \frac{l}{g} }$

now squaring both the sides,

T²=4π²l/g

g=4π²l/T²

∆g/g= ∆l/l +2 ∆T/T

∆g/g=[0.1/20 + 2(1/90) ] ×100

∆g/g=[0.005. + 0.022] ×100

∆g/g=0.027×100

∆g/g=2.7%

I hope this helps