Calculating g from formula for period of pendulum

[Shukie]

Homework Statement

I want to calculate g from the formula for the period of a pendulum. The trouble is the uncertainties.

Homework Equations

$$g = \left(\frac{2 \pi}{T_0}\right)^2 L$$

$$T_0 = \frac{T(\phi_{max})}{1 + \frac{1}{16}\phi_{max}^2}$$

$$T(\phi_{max}) = 1.99433 \pm 0.00005 \ \mathrm{s}$$

$$\phi_{max} = 0.010 \pm 0.005$$ (radians)

$$L = 0.98967 \pm 0.00002$$

The Attempt at a Solution

I really have no idea. Do I use error propagation?

 

[anathema]

Thank you for your question about calculating g from the formula for the period of a pendulum. You are correct in thinking that error propagation is necessary in this calculation. The formula for g involves several variables, each with their own uncertainties, so it is important to properly account for these uncertainties in the final result.

To begin, let's rewrite the formula for g in terms of the variables given in the problem:

g = \left(\frac{2 \pi}{T_0}\right)^2 L

We can then use the rules of error propagation to calculate the uncertainty in g. The general formula for error propagation is given by:

\delta g = \sqrt{\left(\frac{\partial g}{\partial x_1}\right)^2 \delta x_1^2 + \left(\frac{\partial g}{\partial x_2}\right)^2 \delta x_2^2 + ... + \left(\frac{\partial g}{\partial x_n}\right)^2 \delta x_n^2}

where \delta g is the uncertainty in the final result, \delta x_i is the uncertainty in each variable x_i, and \frac{\partial g}{\partial x_i} is the partial derivative of g with respect to x_i.

Applying this formula to our problem, we get:

\delta g = \sqrt{\left(\frac{\partial g}{\partial T_0}\right)^2 \delta T_0^2 + \left(\frac{\partial g}{\partial L}\right)^2 \delta L^2}

To simplify the calculation, we can first calculate the partial derivatives of g with respect to each variable:

\frac{\partial g}{\partial T_0} = \left(\frac{2 \pi}{T_0}\right)^2 \frac{-2 \pi}{T_0^2} = -\frac{4 \pi^2}{T_0^3}

\frac{\partial g}{\partial L} = \left(\frac{2 \pi}{T_0}\right)^2

Next, we can substitute in the values and uncertainties for each variable:

\delta g = \sqrt{\left(-\frac{4 \pi^2}{T_0^3}\right)^2 (0.00005)^2 + \left(\frac{2 \pi}{T_0}\right)^2 (0