How Many Pendulum Periods Minimize Time Uncertainty in Gravity Calculations?

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SUMMARY

The discussion focuses on determining the number of pendulum periods required to ensure that the uncertainty in time measurements is smaller than the uncertainty in length when calculating gravitational acceleration (g). The relevant formula is T=2π(l/g)^(1/2), and the uncertainties in length (±0.2) and time (±0.1) are critical to the calculations. Participants suggest using calculus to derive the error in g, represented as δg, which incorporates the partial derivatives of g with respect to length and time. The goal is to find a relationship where the contribution of time uncertainty is minimized compared to length uncertainty.

PREREQUISITES
  • Understanding of pendulum motion and the formula T=2π(l/g)^(1/2)
  • Familiarity with error propagation and uncertainty analysis
  • Basic knowledge of calculus, specifically partial derivatives
  • Concept of precision in measurements and its impact on calculations
NEXT STEPS
  • Study error propagation techniques in physics experiments
  • Learn about the implications of measurement precision on experimental results
  • Explore advanced calculus applications in physics, particularly in uncertainty analysis
  • Investigate the relationship between pendulum length, time, and gravitational acceleration in detail
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Students in physics, educators teaching mechanics, and researchers involved in experimental physics who are interested in precision measurement and uncertainty analysis in gravitational calculations.

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Homework Statement



The length of a string attached to a pendulum is measured with a precision of (+or-)0.2. The time of the oscillation is measured to a precision of (+or-)0.1. How many periods must you measure so that the contribution of the uncertainty in time is smaller than the uncertainty in length, when calculating g?

Homework Equations



T=2pi(l/g)^(1/2)

The Attempt at a Solution



g=((2pi)^2(l+delta:l))/(T+delta:T)^2
I don't know where to go from here.
Delta l and T are the error in those measurements.
 
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nubey1 said:

Homework Statement



The length of a string attached to a pendulum is measured with a precision of (+or-)0.2. The time of the oscillation is measured to a precision of (+or-)0.1. How many periods must you measure so that the contribution of the uncertainty in time is smaller than the uncertainty in length, when calculating g?

Homework Equations



T=2pi(l/g)^(1/2)

The Attempt at a Solution



g=((2pi)^2(l+delta:l))/(T+delta:T)^2
I don't know where to go from here.
Delta l and T are the error in those measurements.

The next thing to do is assume that the error is really small relative to the actual value.
\delta l << l and \delta T << T. I think you might have an error in your equation for g:
g +\delta g= (2 \pi)^2 \frac{l+\delta l }{(T+\delta T)^2} where \delta g is the error in g.
If you are familiar with calculus then this comes out to:
\delta g = |\frac{\partial g}{\partial l}| \delta l + |\frac{\partial g}{\partial T}| \delta T
If you don't have the luxury of calculus we might need to know what relations for uncertainty you are given to clue you in
 

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