Question #116985

The following observations we're made during an experiment to find the value of g using simple pendulum l=90.0 cm

Time t for 20 vibrations =36 sec

Find percentage error in measurement of g given that T=2π√l/g length is being measured to an accuracy of 0.1 cm and time 0.2 sec

Time t for 20 vibrations =36 sec

Find percentage error in measurement of g given that T=2π√l/g length is being measured to an accuracy of 0.1 cm and time 0.2 sec

Expert's answer

We know that

"T = 2\\pi\\sqrt{\\dfrac{l}{g}} \\Longrightarrow g = \\dfrac{4\\pi^2l}{T^2}."

T is equal to 36:20 = 1.8 seconds, so

"g = \\dfrac{4\\pi^2 \\cdot0.9}{1.8^2} \\approx 11.0\\,\\mathrm{m\/s^2}." The uncertainty of period will be "\\dfrac{0.2}{20} = 0.01\\,\\mathrm{s}."

Next, we should obtain the formula for calculating the percentage error (see part 5 in http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm)

"\\delta g = \\sqrt{\\left(\\dfrac{\\partial g}{\\partial T}\\delta T \\right)^2 + \\left(\\dfrac{\\partial g}{\\partial l}\\delta l \\right)^2} = \\sqrt{\\left(-\\dfrac{8\\pi^2l}{T^3}\\cdot {\\delta T} \\right)^2 + \\left(\\dfrac{4\\pi^2}{T^2}\\cdot{\\delta l} \\right)^2} = \\sqrt{\\left(-\\dfrac{8\\pi^2\\cdot0.9}{1.8^3}\\cdot{0.01} \\right)^2 + \\left(\\dfrac{4\\pi^2}{1.8^2}\\cdot{0.001} \\right)^2} \\approx 0.12\\,\\mathrm{m\/s^2}."

The percentage error will be "\\dfrac{0.12}{11.0} \\approx 1.1\\%."

The simplier way to calculate the percentage error is to use the formula

"\\dfrac{\\delta l}{l} + 2\\dfrac{\\delta T}{T} = \\dfrac{0.001}{0.9} + 2\\dfrac{0.01}{1.8} = 1.2\\%."

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