Animated Solution for Physics - Physics and Measurement: A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0\text{ cm}$ and a stop watch with $1\text{ s}$ resolution measures the time taken for $40$ oscillations to be $50\text{ s}$. The accuracy in $g$ is

Imagine you are standing in a physics laboratory, holding a simple pendulum. Your mission is to determine the acceleration due to gravity, $g$, at your exact location. You have a ruler and a stopwatch. But here is the catch: no measuring instrument is perfect. Every measurement you take carries a tiny seed of uncertainty. This problem is a beautiful exercise in understanding how those tiny uncertainties propagate through a mathematical formula to affect your final result.

The problem states that the length of the pendulum is $l=25.0\text{ cm}$. Why didn't they just write $25\text{ cm}$? That `.0` is not just for show; it is a profound statement about the precision of the ruler. It tells us that the ruler can measure down to a tenth of a centimeter. Therefore, the absolute error or least count in the length measurement is $\Delta l=0.1\text{ cm}$.

Similarly, you use a stopwatch with a resolution of $1\text{ s}$ to measure the time for $40$ oscillations, which comes out to be $50\text{ s}$. This means the total time measured is $t=50\text{ s}$, and the absolute error in this time measurement is $\Delta t=1\text{ s}$.

To find $g$, we rely on the classic formula for the time period of a simple pendulum:

We need to isolate $g$. By squaring both sides and rearranging the terms, we get our master equation:

Now, we apply the rules of error propagation. When quantities are multiplied or divided, their fractional errors add up. Furthermore, if a quantity is raised to a power, that power becomes a multiplier for its fractional error. Applying this to our master equation, we get the expression for the maximum fractional error in $g$:

Here is where many students fall into a trap. What is $\frac{\Delta T}{T}$? We know that the time period $T$ is the total time $t$ divided by the number of oscillations $n$ ($T=\frac{t}{n}$).

If we look at the fractional error, we see something magical happen:

The number of oscillations $n$ is an exact integer; it has no uncertainty. Therefore, it completely cancels out! The fractional error in the time period is exactly equal to the fractional error in the total time measured. This means we don't need to divide our $1\text{ s}$ error by $40$. We can directly use the total time values.

Now, we substitute our known values into the error equation to find the percentage error (accuracy):

Let's break down the arithmetic carefully:

Adding the terms inside the parenthesis gives $0.044$. Multiplying by $100\%$ yields our final answer:

This tells us that our calculated value of $g$ could be off by up to $4.40\%$ due to the limitations of our measuring instruments. To improve this accuracy in a real lab, we would need a more precise ruler or we would need to measure the time for a much larger number of oscillations to make the fraction $\frac{\Delta t}{t}$ smaller!