Physics, 11e WileyPLUS + Loose-leaf
11th Edition
ISBN: 9781119394112
Author: John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler
Publisher: Wiley (WileyPLUS Products)
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Chapter 1, Problem 11P
To determine
The dimension of
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We might assume that the period of a simple pendulum depends on the mass M, the length l of the string and g the acceleration due to gravity, i.e T is directly proportional to m l g. Consider the dimensions of the quantities involved to deduce the values of alpha, beta, gamma hence show that T is directly proportional to √l/g
The values of the period (T) and length (L) in the table were obtained experimentally.
T(s)
1,44±0,05
1,42±0,05
1,41±0,05
1,45±0,05
1,43±0,05
L (m)
0,51±0,01
0,50±0,01
0,49±0,01
0,52±0,01
0,51±0,01
Knowing that: T=2π√(L/g). Calculate:
The mean value of T (with 3 significant digits): T =
The mean value of L (2 significant digits): L =
The probable random error of T with the respective unit: Ea(T) =
The random error of L with the respective unit: Ea(L) =
The probable total error of T with the respective unit: ΔT =
The probable total error of L with the respective unit: ΔL =
The value of g with its error propagated: g =
The period p of a pendulum, or the time it takes for the pendulum to make one complete swing, varies directly as the square root of the length L of the pendulum. If the period of a pendulum is 1.1 s when the length is 2 ft, find the period when the length is 3 ft. Round to the nearest hundredth.
Chapter 1 Solutions
Physics, 11e WileyPLUS + Loose-leaf
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