A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :
Solution
T = $2\pi \sqrt {{l \over g}}$
<br><br>$\Rightarrow$ $g = {{4{\pi ^2}l} \over {{T^2}}}$
<br><br>$\Rightarrow$ ${{\Delta g} \over g} = {{\Delta l} \over l} + 2{{\Delta T} \over T}$
<br><br>= ${{0.1} \over {25}} + 2{1 \over {50}}$ = ${{11} \over {250}}$
<br><br>$\therefore$ % accuracy = ${{11} \over {250}}$ $\times$ 100% = 4.40 %
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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