The period of oscillation of a simple pendulum is given by T=2π√(l/g). Deduce the formula for fractional error in g 

To deduce the formula for the fractional error in \( g \), we'll start with the given formula for the period of oscillation of a simple pendulum:

\[
T = 2\pi \sqrt{\frac{l}{g}}
\]

Here:
- \( T \) is the period of oscillation,
- \( l \) is the length of the pendulum, and
- \( g \) is the acceleration due to gravity.

### Step 1: Isolate \( g \) in the formula
First, we'll isolate \( g \) in the formula by squaring both sides to eliminate the square root:

\[
T^2 = 4\pi^2 \frac{l}{g}
\]

Next, solve for \( g \):

\[
g = \frac{4\pi^2 l}{T^2}
\]

### Step 2: Apply the concept of fractional error
The fractional error in a quantity is given by the relative error of the quantity. For a function \( f(x, y, \dots) \) with variables \( x \), \( y \), etc., the fractional error is:

\[
\frac{\Delta f}{f} = \sqrt{\left(\frac{\partial f}{\partial x} \frac{\Delta x}{x}\right)^2 + \left(\frac{\partial f}{\partial y} \frac{\Delta y}{y}\right)^2 + \dots}
\]

However, since we are dealing with a direct formula, we can differentiate the logarithm of both sides to obtain the fractional errors.

### Step 3: Differentiate to find fractional errors
Taking the natural logarithm of the equation \( g = \frac{4\pi^2 l}{T^2} \), we get:

\[
\ln g = \ln(4\pi^2) + \ln l - 2\ln T
\]

Now, differentiate with respect to \( g \), \( l \), and \( T \) to get the errors:

\[
\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
\]

### Step 4: Interpret the result
The formula above indicates that the fractional error in \( g \) depends on the fractional errors in \( l \) and \( T \). Specifically:

- The fractional error in \( l \) directly contributes to the fractional error in \( g \).
- The fractional error in \( T \) contributes twice as much to the fractional error in \( g \) because \( T \) appears squared in the original formula.

Thus, the formula for the fractional error in \( g \) is:

\[
\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
\]

Would you like more details or have any questions? Here are some related questions you might explore:

1. How can you minimize the fractional error in a pendulum experiment?
2. What is the physical significance of the fractional error?
3. How would the fractional error change if the length \( l \) is measured with higher precision?
4. How does the period \( T \) affect the accuracy of the calculated \( g \)?
5. How does air resistance influence the period of a simple pendulum?
6. What assumptions are made in deriving the period formula for a simple pendulum?
7. How would the formula for \( T \) change in a non-uniform gravitational field?
8. What are some practical applications of simple pendulum measurements?

**Tip:** In experiments, reducing the uncertainty in measurements (e.g., using precise timing methods) is key to minimizing fractional error and obtaining accurate results.

Learn how to deduce the formula for the fractional error in acceleration due to gravity (g) using the period formula of a simple pendulum. Understand the relationship between errors in pendulum length (l), period (T), and gravity (g). Suitable for undergraduate level physics and engineering students.

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