The period of oscillation of a simple pendulum is given by T=2π √(l/g). (i ) Deduce the formula of fractional error in "g. (Ii ) Which quantity in (i ) should be measured most accurately? Give reason for your answer

Let's break down the problem step by step:

### (i) Deduce the formula for fractional error in \( g \)

Given the formula for the period of oscillation of a simple pendulum:
\[
T = 2\pi \sqrt{\frac{l}{g}}
\]

We want to find the fractional error in \( g \). First, let's solve the equation for \( g \):

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

Square both sides to eliminate the square root:
\[
T^2 = (2\pi)^2 \frac{l}{g}
\]

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

Rearrange to solve for \( g \):
\[
g = \frac{4\pi^2 l}{T^2}
\]

Now, to find the fractional error in \( g \), we use the general formula for propagation of errors. The relative (or fractional) error in \( g \) can be found by considering the relative errors in \( T \) and \( l \).

Let's express the relative error in \( g \) as:
\[
\frac{\Delta g}{g} = \sqrt{\left(\frac{\partial g}{\partial T} \cdot \frac{\Delta T}{T}\right)^2 + \left(\frac{\partial g}{\partial l} \cdot \frac{\Delta l}{l}\right)^2}
\]

From the equation \( g = \frac{4\pi^2 l}{T^2} \):

- The partial derivative of \( g \) with respect to \( T \):
  \[
  \frac{\partial g}{\partial T} = \frac{\partial}{\partial T} \left(\frac{4\pi^2 l}{T^2}\right) = -\frac{8\pi^2 l}{T^3} = -\frac{2g}{T}
  \]

- The partial derivative of \( g \) with respect to \( l \):
  \[
  \frac{\partial g}{\partial l} = \frac{\partial}{\partial l} \left(\frac{4\pi^2 l}{T^2}\right) = \frac{4\pi^2}{T^2} = \frac{g}{l}
  \]

So, the relative error in \( g \) becomes:
\[
\frac{\Delta g}{g} = \sqrt{\left(\frac{-2g}{T} \cdot \frac{\Delta T}{g}\right)^2 + \left(\frac{g}{l} \cdot \frac{\Delta l}{g}\right)^2}
\]
\[
\frac{\Delta g}{g} = \sqrt{\left(-2 \frac{\Delta T}{T}\right)^2 + \left(\frac{\Delta l}{l}\right)^2}
\]
\[
\frac{\Delta g}{g} = \sqrt{4 \left(\frac{\Delta T}{T}\right)^2 + \left(\frac{\Delta l}{l}\right)^2}
\]

This is the formula for the fractional error in \( g \).

### (ii) Which quantity should be measured most accurately?

From the error formula derived above:
\[
\frac{\Delta g}{g} = \sqrt{4 \left(\frac{\Delta T}{T}\right)^2 + \left(\frac{\Delta l}{l}\right)^2}
\]

It is evident that the error term associated with \( T \) is multiplied by 4, while the error associated with \( l \) is not. This indicates that a small error in \( T \) has a larger impact on the error in \( g \) compared to the same fractional error in \( l \).

**Therefore, the period \( T \) should be measured most accurately.**

#### Reason:
The period \( T \) appears squared in the denominator in the formula for \( g \). Any error in measuring \( T \) is magnified (by a factor of 2) in the calculation of \( g \). Therefore, to minimize the overall error in \( g \), the period \( T \) must be measured with the highest precision.

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Would you like further details or have any questions?

### Follow-up Questions:
1. How would you calculate the absolute error in \( g \) if the errors in \( T \) and \( l \) are known?
2. What are common sources of error when measuring the period \( T \) of a pendulum?
3. How does the length \( l \) of the pendulum affect the period \( T \)?
4. Can you derive the formula for the period \( T \) from basic principles?
5. How would air resistance impact the period of a simple pendulum?

### Tip:
When conducting experiments involving pendulums, using a longer pendulum can make it easier to measure the period accurately, as the period will be longer and thus easier to time.

Understand how to deduce the formula for the fractional error in g for the period of oscillation of a simple pendulum. Learn about error propagation in physics experiments and why measuring the period accurately is crucial. Suitable for advanced high school and undergraduate physics students.

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