Animated Solution for Physics - Physics and Measurement: A student uses a simple pendulum of exactly $1\text{ m}$ length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of $1\text{ s}$ for this and records $40\text{ s}$ for $20$ oscillations. For this observation, which of the following statement(s) is/are true?

The measurement of physical quantities is the bedrock of experimental physics. But no measurement is perfect; every instrument has its limitations. In this problem, we embark on a journey to determine the acceleration due to gravity, $g$, using a simple pendulum, and more importantly, we will meticulously track the errors in our measurement.

Imagine you are in a physics laboratory. You have a simple pendulum suspended from a rigid support. The length of this pendulum is given as exactly $1\text{ m}$. The word "exactly" is a crucial hint here—it means we can assume the error in the length measurement, $\Delta l$, is zero.

To find $g$, we need to measure the time period of the pendulum. We use a stopwatch with a least count of $1\text{ s}$. The least count is the smallest value the instrument can measure, and it represents the maximum possible absolute error in any single reading. Therefore, the error in our total time measurement, $\Delta t$, is $1\text{ s}$.

We record the time for $20$ oscillations and find it to be $40\text{ s}$. Let's break this down.

First, we need the time period $T$ for a single oscillation. Since $20$ oscillations take $40\text{ s}$, the time for one oscillation is simply the total time divided by the number of oscillations:

Now, what about the error in this time period, $\Delta T$? The total time $t$ is the sum of $n$ individual time periods, so $t=nT$. If we differentiate this (or apply error propagation for a constant multiplier), we get:

Rearranging this to solve for $\Delta T$, we find:

This is a beautiful result! By measuring the time for multiple oscillations, we effectively divide the error of our stopwatch by the number of oscillations. Substituting our values:

So, the error in measuring the time period $T$ is $0.05\text{ s}$. This matches one of our options perfectly.

Now, let's connect our measurements to the acceleration due to gravity, $g$. The time period of a simple pendulum is given by the well-known formula:

To isolate $g$, we square both sides of the equation:

Rearranging for $g$, we get our master equation:

We want to find the percentage error in $g$. According to the rules of error propagation, when 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.

Applying this to our master equation, the relative error in $g$ is:

Notice the factor of $2$ in front of the time period error. This comes from the $T^2$ term in the denominator. The constants $4$ and ${\pi }^2$ have no error, so they disappear from the error equation.

We established earlier that the length is exactly $1\text{ m}$, meaning $\Delta l=0$. This simplifies our error equation significantly:

To find the percentage error, we multiply both sides by $100$:

Now, we substitute the values we calculated for $\Delta T$ and $T$:

The $2$ in the numerator and denominator cancel out gracefully:

The percentage error in the determination of $g$ is $5\%$.

Through careful analysis of the given data and systematic application of error propagation rules, we have determined that the error in the time period $\Delta T$ is $0.05\text{ s}$, and the percentage error in $g$ is $5\%$. This problem beautifully illustrates how taking multiple readings (like $20$ oscillations instead of just $1$) can significantly reduce the error in the final calculated result.