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Question

Physics Question on simple harmonic motion

The period of oscillation of a simple pendulum is T=2πLgT =2 \pi \sqrt{\frac{ L }{ g }}. Measured value of 'L' is 1.0m1.0 m from meter scale having a minimum division of 1mm1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01s0.01 s resolution. The percentage error in the determination of 'g' will be :

A

1.13%1.13 \%

B

1.03%1.03 \%

C

1.33%1.33 \%

D

1.30%1.30 \%

Answer

1.13%1.13 \%

Explanation

Solution

T=2πgT =2 \pi \sqrt{\frac{\ell}{ g }} g=4π2T2g =\frac{4 \pi^{2} \ell}{ T ^{2}} Δgg=Δ+2ΔTT\frac{\Delta g }{ g }=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T }{ T } Δgg=1×1031+2×0.011.95\frac{\Delta g }{ g }=\frac{1 \times 10^{-3}}{1}+2 \times \frac{0.01}{1.95} Δgg=0.0113\frac{\Delta g }{ g}=0.0113 or 1.13%1.13 \%