Three students S1, S2 and S3 perform an experiment for determining the acceleration due to gravity (

Q: Why is the relative error in time period $T$ the same as the relative error in total time $t$?

Since $t = nT$, the error in total time is $\Delta t = n\Delta T$. When we find the relative error, $\frac{\Delta T}{T} = \frac{\Delta t / n}{t / n} = \frac{\Delta t}{t}$. The factor of $n$ cancels out perfectly!

Q: Why did we ignore $4\pi^2$ when calculating the error?

Constants like $4$ and $\pi^2$ are exact mathematical values. They do not have any uncertainty or measurement error associated with them, so their relative error is strictly zero.

Q: The least count of length is given as 0.1 m in some books, but we used 0.1 cm. Why?

Looking at the table, the lengths are 64.0 cm and 20.0 cm. The ".0" indicates precision up to the first decimal place in centimeters, which means the least count is 0.1 cm. A least count of 0.1 m (10 cm) would not allow for a reading like 64.0 cm.

Q: Why does taking more oscillations reduce the error?

The absolute error (least count) of the stopwatch remains constant at 0.1 s. By taking more oscillations, the total time $t$ increases. Since relative error is $\frac{\Delta t}{t}$, a larger denominator $t$ results in a smaller overall relative error.

Q: Could we have just guessed the answer without calculating?

Yes! Student 1 and 2 have the same length, but Student 1 took more oscillations, so $E_1 < E_2$. Student 3 has a much shorter length and fewer oscillations, leading to larger relative errors in both length and time. Thus, Student 1 is the clear winner.

Question

Three students $S_1$, $S_2$ and $S_3$ perform an experiment for determining the acceleration due to gravity ($g$) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

\begin{array}{|c|c|c|c|c|}
\hline
	ext{Student No.} & 	ext{Length of pendulum (cm)} & 	ext{No. of oscillations } (n) & 	ext{Total time for } n 	ext{ oscillations} & 	ext{Time period (s)} \
\hline
1. & 64.0 & 8 & 128.0 & 16.0 \
2. & 64.0 & 4 & 64.0 & 16.0 \
3. & 20.0 & 4 & 36.0 & 9.0 \
\hline
\end{array}

(Least count of length $= 0.1	ext{ cm}$, least count for time $= 0.1	ext{ s}$)
If $E_1, E_2$ and $E_3$ are the percentage errors in $g$ for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student number ……… .

Sigma Percentile · Accepted Answer

Imagine you are standing in a quiet physics laboratory. The only sound is the rhythmic *tick-tock* of a simple pendulum swinging back and forth. You hold a stopwatch in one hand and a meter scale in the other. Your mission? To determine the fundamental constant of our universe: the acceleration due to gravity, $g$.

But here is the catch—every measurement you make is flawed. The meter scale can only measure down to $0.1 	ext{ cm}$, and your stopwatch can only measure down to $0.1 	ext{ s}$. These are your least counts, the unavoidable uncertainties in your experiment. The question is, how do you minimize the impact of these errors?

Let's dive into the mathematics of errors and see which student performed the smartest experiment.

### The Master Equation

We start with the classic formula for the time period of a simple pendulum:

$$T = 2\pi \sqrt{\frac{l}{g}}$$

To find $g$, we square both sides and rearrange the terms:

$$g = \frac{4\pi^2 l}{T^2}$$

Now, we need to find the percentage error in $g$. Using the rules of error analysis, we take the relative error of each variable. Since $4$ and $\pi^2$ are exact constants, they contribute zero error. The power of $2$ on $T$ comes down as a multiplier:

$$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$$

### The Catch: Time Period vs Total Time

Here is where a massive conceptual trap lies. Many students assume that $\Delta T$ is simply the least count of the stopwatch ($0.1 	ext{ s}$). **This is incorrect!**

The stopwatch does not measure the time for a single oscillation; it measures the *total time* $t$ for $n$ oscillations. Therefore, the least count of $0.1 	ext{ s}$ is actually $\Delta t$, the error in the total time.

Since total time $t = nT$, the error in total time is $\Delta t = n\Delta T$.

What happens when we look at the relative error in the time period?

$$\frac{\Delta T}{T} = \frac{\Delta t / n}{t / n} = \frac{\Delta t}{t}$$

This is a beautiful result! The relative error in the time period is exactly equal to the relative error in the total time. This means that if you want to reduce your error, you simply need to increase your total time $t$. How do you do that? By measuring for more oscillations!

### Calculating the Errors

Now, let's evaluate the performance of our three students.

**Student 1:**
They used a length of $64.0 	ext{ cm}$ and measured $8$ oscillations for a total time of $128.0 	ext{ s}$.

$$E_1 = \frac{0.1}{64} + 2\left(\frac{0.1}{128}ight) = \frac{0.1}{64} + \frac{0.1}{64} = \frac{0.2}{64}$$

**Student 2:**
They used the same length of $64.0 	ext{ cm}$ but only measured $4$ oscillations, resulting in a total time of $64.0 	ext{ s}$.

$$E_2 = \frac{0.1}{64} + 2\left(\frac{0.1}{64}ight) = \frac{0.3}{64}$$

Notice how Student 2's error is larger simply because they were impatient and recorded fewer oscillations!

**Student 3:**
They used a much shorter pendulum of $20.0 	ext{ cm}$ and measured $4$ oscillations for a total time of $36.0 	ext{ s}$.

$$E_3 = \frac{0.1}{20} + 2\left(\frac{0.1}{36}ight) = \frac{1}{200} + \frac{1}{180}$$

This value is significantly larger than the first two.

### The Final Verdict

Comparing the three values, it is crystal clear that **Student 1** has the minimum error. By choosing a reasonably long pendulum and having the patience to record a large number of oscillations, they maximized their total time $t$, thereby drastically shrinking the relative error $\frac{\Delta t}{t}$.

Physics rewards patience and precision. The correct answer is **1**.

Enter Numerical Value:

Visualized Solution

\text{The Simple Pendulum Experiment}

\text{Formula for } g

\text{Relative Error Equation}

\text{The Catch: Total Time vs Time Period}

\text{Error in Time Period}

\text{Error for Student 1}

\text{Error for Student 2}

\text{Error for Student 3}

\text{Conclusion}

The Sigma Insight: Errors in Measurement

The Master Equation

The Catch

Calculating the Errors

The Final Verdict

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