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Zorluk: ZorErrors in Measurement and Significant Figures

In a physics experiment to determine the acceleration due to gravity gg using a free-fall apparatus, the distance of fall is measured as h=(2.00±0.04) mh = (2.00 \pm 0.04)\text{ m} and the duration of fall is measured as t=(0.50±0.01) st = (0.50 \pm 0.01)\text{ s}. Given that g=2ht2g = \frac{2h}{t^2}, what is the percentage error in the calculated value of gg?

Cevap: 6 %

Cevap

The percentage error in the calculated value of gg is 6%.
For a physical quantity defined by g=2ht2g = \frac{2h}{t^2}, the maximum percentage error is determined by adding the percentage error in hh to twice the percentage error in tt. The percentage error in hh is 0.042.00×100%=2%\frac{0.04}{2.00} \times 100\% = 2\% and in tt is 0.010.50×100%=2%\frac{0.01}{0.50} \times 100\% = 2\%. Therefore, the total percentage error in gg is 2%+2(2%)=6%2\% + 2(2\%) = 6\%.

Adım Adım Çözüm

1
Calculate the percentage error in the distance measurement hh
2%
The fractional error in height is Δhh=0.042.00=0.02\frac{\Delta h}{h} = \frac{0.04}{2.00} = 0.02, which corresponds to 2%2\%.
2
Calculate the percentage error in the time measurement tt
2%
The fractional error in time is Δtt=0.010.50=0.02\frac{\Delta t}{t} = \frac{0.01}{0.50} = 0.02, which corresponds to 2%2\%.
3
Apply the power-law error propagation rule to find the maximum percentage error in gg
6%
For g=2ht2g = \frac{2h}{t^2}, the total relative error is Δgg=Δhh+2Δtt=2%+2(2%)=6%\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 2\% + 2(2\%) = 6\%, since the exponent of tt is 2.

Anahtar Kavram

Error propagation in physical formulas involving powers and quotients
Tahmini Süre:2m 0s
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Konular

Errors in Measurement and Significant Figures

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