Animated Solution for Physics - Physics and Measurement: The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$. Measured value of $L$ is $1.0\text{ m}$ from metre scale having a minimum division of $1\text{ mm}$ and time of one complete oscillation is $1.95\text{ s}$ measured from stopwatch of $0.01\text{ s}$ resolution. The percentage error in the determination of $g$ will be

Imagine you are standing in a grand physics laboratory. The simple pendulum swinging before you is not just a bob on a string; it is a timekeeping marvel that connects the length of a string directly to the gravitational pull of the Earth. But here is the catch: no measurement is perfect. Every instrument has its limits, and this is where the beautiful mathematics of error propagation comes into play.

Isolating $g$

The time period of a simple pendulum is governed by the classic equation:

Our goal is to find the percentage error in the determination of the acceleration due to gravity, $g$. To do this, we first need to make $g$ the subject of our formula. By squaring both sides, we strip away the square root:

Rearranging this to isolate $g$, we get our master equation:

Now, we apply the rules of error propagation. This is a favorite concept for JEE examiners! When physical quantities are multiplied or divided, their relative errors add up. Furthermore, if a quantity is raised to a power, that power becomes a multiplier for its relative error.

The term $4{\pi }^2$ is a pure mathematical constant. It has infinite precision, meaning its error contribution is exactly zero. Therefore, the relative error in $g$ is solely determined by the measurements of $L$ and $T$:

Notice how the $T^2$ in the denominator translates to a factor of $2$ in the error equation. Errors always compound; they never cancel each other out.

Precision Matters

Let's carefully extract the values from the problem. The measured length $L$ is $1.0\text{ m}$. The absolute error $\Delta L$ is the least count of the metre scale, which is $1\text{ mm}$ or $10^{-3}\text{ m}$.

The measured time $T$ for one oscillation is $1.95\text{ s}$. The absolute error $\Delta T$ is the resolution of the stopwatch, which is $0.01\text{ s}$.

Substituting these into our error equation:

Let's break down the calculation. The length error term is simply $0.001$. The time error term requires a bit more care:

Adding these together gives us the total relative error:

To find the percentage error, we simply multiply the relative error by $100$:

Looking at our options, we round this off to two decimal places, giving us a final answer of $1.13\%$.

Chasing Perfection

Did you get the feel of it? Think about how we could improve this experiment. If we measured the time for $100$ oscillations instead of just $1$, the total time $T$ would be much larger. However, the least count $\Delta T$ of the stopwatch would remain the same $0.01\text{ s}$. This would make the fraction $\frac{\Delta T}{T}$ significantly smaller, drastically reducing the overall error in $g$. In experimental physics, more observations always lead to higher accuracy!