In an experiment with a simple pendulum, the following measurements are recorded
ID: 1406631 • Letter: I
Question
In an experiment with a simple pendulum, the following measurements are recorded: . L=110.5+-0.4cm T=2.1090+-0.0009sec
(a) Based on these measurements, what is the value of g and what is the value of g, the uncertainty in g? (b) What would be the value of g if we ignored the small uncertainty in T? [Remember to show all your work and to give final answers with the appropriate number of significant figures.] [Hint: in your calculations, should L be in cm or m?] L = ± 110.5 0.4 cm, T=2.1090±0.0009 sec
Explanation / Answer
L=110.5+-0.4 cm
T=2.1090+-0.0009 sec
T = 2 pi sqrt(L/g)
g = 4 pi^2 L / T^2
(a) value of g = 4 * pi^2 * 1.105 / 2.109^2 = 9.8077 m/s^2
value of g = g ( L / L + 2 T / T) = 9.8077 * ((0.4/110.5) + (2*0.0009/2.109)) = 0.0438 m/s^2
(b) g if uncertainity in T is ignored = g ( L / L ) = 9.8077 * (0.4/110.5) = 0.0355 m/s^2
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.