Measuring the Acceleration Due to Gravity with a...
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Measuring the Acceleration Due to Gravity with a Pendulum Phil Rubin March 17, 2009 Abstract Because the period of a simple pendulum swinging with a small am- plitude is proportional to 1/ √ g, we use a pendulum to determine g, the acceleration due to gravity. The formula for the period of a small ampli- tude, simple pendulum is T =2π p ‘/g. We set up a simple pendulum by hanging a 50 g mass at the end of a string whose other end is tied to a thin bar. The length of the string, ‘, measured with a meter stick to 1 mm, is varied for eight trials between about 0.2 m and 1 m. The pendulum is released from an angle of (5 ± 2) ◦ , as measured with a compass. For each trial, a stopwatch measures the time interval for 50 oscillations of the pendulum to 0.01 s. The period, T , is then calculated by dividing the total time interval by 50. A regression of T 2 against ‘ gives the slope of the line relating these values, which is 4π 2 /g. We then calculate g and find g =9.63 ± 0.08 m/s 2 . We compare this value for g with the value measured in a free fall measurement, g =9.72 ± 0.02 m/s 2 , and find they differ by about 1 standard deviation. We conclude that the two values are consistent with one another. 1
Transcript of Measuring the Acceleration Due to Gravity with a...
Measuring the Acceleration Due to Gravity with
a Pendulum
Phil Rubin
March 17, 2009
Abstract
Because the period of a simple pendulum swinging with a small am-plitude is proportional to 1/
√g, we use a pendulum to determine g, the
acceleration due to gravity. The formula for the period of a small ampli-tude, simple pendulum is T = 2π
√`/g. We set up a simple pendulum by
hanging a 50 g mass at the end of a string whose other end is tied to a thinbar. The length of the string, `, measured with a meter stick to 1 mm,is varied for eight trials between about 0.2 m and 1 m. The pendulumis released from an angle of (5 ± 2)◦, as measured with a compass. Foreach trial, a stopwatch measures the time interval for 50 oscillations ofthe pendulum to 0.01 s. The period, T , is then calculated by dividing thetotal time interval by 50. A regression of T 2 against ` gives the slope ofthe line relating these values, which is 4π2/g. We then calculate g andfind g = 9.63 ± 0.08 m/s2. We compare this value for g with the valuemeasured in a free fall measurement, g = 9.72± 0.02 m/s2, and find theydiffer by about 1 standard deviation. We conclude that the two valuesare consistent with one another.
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