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.