The time period of a simple pendulum is given by $T = 2\pi \sqrt {{l \over g}}$. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-
Solution
$T = 2\pi \sqrt {{l \over g}}$<br><br>${T^2} = 2\pi \left( {{l \over g}} \right)$<br><br>$g = 2\pi {l \over {{T^2}}}$<br><br>${{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}$<br><br>$${{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100}}$$<br><br>${{\Delta g} \over g} = 0.02 + 0.01 = 0.03$<br><br>$100 \times {{\Delta g} \over g} = 0.03 \times 100 = 3\%$<br><br>${{\Delta g} \over g} \times 100 = 3\%$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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