How Many Pendulum Periods Minimize Time Uncertainty in Gravity 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.
[tex]\delta l << l[/tex] and [tex]\delta T << T[/tex]. I think you might have an error in your equation for g:
[tex]g +\delta g= (2 \pi)^2 \frac{l+\delta l }{(T+\delta T)^2}[/tex] where [tex]\delta g[/tex] is the error in g.
If you are familiar with calculus then this comes out to:
[tex]\delta g = |\frac{\partial g}{\partial l}| \delta l + |\frac{\partial g}{\partial T}| \delta T[/tex]
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