To find the maximum error in the period \( T \) of a simple pendulum given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
where \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity, we need to consider how the errors in \( l \) and \( g \) affect the calculation of \( T \).
Steps to Find the Maximum Error:
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Differentiate the expression for \( T \):
To find the maximum error in \( T \), we will use partial derivatives to propagate the errors from \( l \) and \( g \) to \( T \).
The expression can be rewritten to emphasize the dependence on \( l \) and \( g \):
\[ T = 2\pi \left( l^{1/2} g^{-1/2} \right) \]
Taking the natural logarithm helps us in estimating relative errors:
\[ \ln T = \ln(2\pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g \]
Differentiating both sides gives:
\[ \frac{dT}{T} = \frac{1}{2} \frac{dl}{l} - \frac{1}{2} \frac{dg}{g} \]
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Finding the Maximum Error:
The maximum relative error in \( T \), denoted as \( \frac{\Delta T}{T} \), can be derived from the above expression:
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g} \]
where \( \Delta l \) and \( \Delta g \) are the absolute errors in the measurements of \( l \) and \( g \), respectively.
Rearranging this gives us the maximum absolute error in \( T \):
\[ \Delta T = T \left( \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g} \right) \]
Final Result
To summarize, the maximum error in the period \( T \) of a simple pendulum is given by:
\[ \Delta T = T \left( \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g} \right) \]
Thus, to find the actual maximum error, you simply need the values of the lengths and errors in the pendulum's length \( l \) and the error in \( g \). Substitute any specific values to calculate \( \Delta T \).
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