Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using

Q: Why do we use the total time $t$ instead of the time period $T$ for error calculation?

The stopwatch measures the total time $t$ directly, so the least count applies to $t$. The time period $T$ is a calculated value ($T = t/n$), so its error depends on the error in $t$.

Q: Why is the error in time multiplied by 2?

In the formula for $g$, the time $t$ appears as $t^2$ in the denominator. According to the rules of error propagation, the power of a quantity becomes the coefficient of its fractional error.

Q: Why don't we consider the error in the number of oscillations $n$?

The number of oscillations $n$ is an exact counted integer, not a measured continuous quantity. Therefore, it has absolute certainty and zero error.

Q: Why does Student I have the minimum error even though the length is the same as Student II?

Student I recorded time for more oscillations (8 instead of 4), which doubled the total time $t$. A larger denominator in the fractional error $\frac{\Delta t}{t}$ reduces the overall percentage error.

Q: Can the percentage error ever be zero?

No, as long as there is a non-zero least count in the measuring instruments, there will always be some measurement error. It can be minimized, but never truly zero.

Question

Students I, II and III perform an experiment for measuring the acceleration due to gravity ($g$) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length = $0.1	ext{ cm}$, Least count for time = $0.1	ext{ s}$

$$\begin{array}{|c|c|c|c|c|} \hline 	ext{Student} & 	ext{Length of the pendulum (cm)} & 	ext{Number of oscillations } (n) & 	ext{Total time for } (n) 	ext{ oscillations (s)} & 	ext{Time period (s)} \ \hline 	ext{I} & 64.0 & 8 & 128.0 & 16.0 \ 	ext{II} & 64.0 & 4 & 64.0 & 16.0 \ 	ext{III} & 20.0 & 4 & 36.0 & 9.0 \ \hline \end{array}$$

If $E_{	ext{I}}$, $E_{	ext{II}}$ and $E_{	ext{III}}$ are the percentage errors in $g$, i.e. $\left(\frac{\Delta g}{g} 	imes 100ight)$ for students I, II and III, respectively

Options:

(A) $E_{	ext{I}} = 0$

(B) $E_{	ext{I}}$ is minimum

(C) $E_{	ext{I}} = E_{	ext{II}}$

(D) $E_{	ext{II}}$ is maximum

Sigma Percentile · Accepted Answer

Imagine you are in a physics lab, tasked with finding the acceleration due to gravity, $g$, using a simple pendulum. You have a string, a bob, a meter scale, and a stopwatch. The formula that connects these physical quantities is the well-known time period equation:

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

However, measuring the time for a single oscillation is highly prone to human reaction error. To counter this, we measure the total time $t$ for $n$ oscillations. Thus, $T = \frac{t}{n}$. Substituting this back and rearranging for $g$, we get our master equation:

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

### The Error Propagation Law

In the real world, no measurement is perfect. The meter scale and the stopwatch have limitations, known as their least counts. Here, $\Delta l = 0.1 	ext{ cm}$ and $\Delta t = 0.1 	ext{ s}$.

To find the percentage error in $g$, we apply the rules of error propagation. Constants like $4\pi^2$ and exact counted numbers like $n$ have zero error. For measured quantities, the fractional errors add up, and any exponent becomes a multiplier:

$$\frac{\Delta g}{g} 	imes 100 = \left( \frac{\Delta l}{l} + 2\frac{\Delta t}{t} ight) 	imes 100$$

Notice that the error in time is multiplied by 2. This means any mistake in measuring time will hurt us twice as much!

### Crunching the Numbers

Let's evaluate the experimental technique of our three students.

**Student I:**
They used a length of $64.0 	ext{ cm}$ and recorded 8 oscillations, taking $128.0 	ext{ s}$.

$$E_{	ext{I}} = \left( \frac{0.1}{64.0} + 2 	imes \frac{0.1}{128.0} ight) 	imes 100$$

$$E_{	ext{I}} = \left( \frac{0.1}{64} + \frac{0.1}{64} ight) 	imes 100 = \frac{0.2}{64} 	imes 100 = 0.3125\%$$

**Student II:**
They used the same length but got lazy and only recorded 4 oscillations, taking $64.0 	ext{ s}$.

$$E_{	ext{II}} = \left( \frac{0.1}{64.0} + 2 	imes \frac{0.1}{64.0} ight) 	imes 100$$

$$E_{	ext{II}} = \left( \frac{0.1}{64} + \frac{0.2}{64} ight) 	imes 100 = \frac{0.3}{64} 	imes 100 = 0.46875\%$$

See the difference? By halving the total time, the fractional error in time doubled, increasing the overall error.

**Student III:**
They used a short pendulum of $20.0 	ext{ cm}$ and recorded 4 oscillations in $36.0 	ext{ s}$.

$$E_{	ext{III}} = \left( \frac{0.1}{20.0} + 2 	imes \frac{0.1}{36.0} ight) 	imes 100$$

$$E_{	ext{III}} = \left( 0.005 + 0.0055 ight) 	imes 100 = 1.055\%$$

This is a disaster! Both the length and total time were small, making their respective fractional errors massive.

### The Grand Conclusion

Comparing the results, $E_{	ext{I}} < E_{	ext{II}} < E_{	ext{III}}$. Student I has the minimum error.

This problem beautifully illustrates a core principle of experimental physics: to minimize relative error, maximize the magnitude of your measured quantities. Use a longer pendulum and record time for a larger number of oscillations!

Select Answer:

Visualized Solution

$g = \frac{4 π ^{2} n ^{2} l}{t ^{2}}$

$\frac{Δ g}{g} \times 100$

$E_{I}$

$E_{I} = 0.3125%$

$E_{II}$

$E_{II} = 0.46875%$

$E_{III}$

$E_{III} = 1.055%$

$E_{I} < E_{II} < E_{III}$

The Sigma Insight: Errors in Measurement

The Error Propagation Law

Crunching the Numbers

The Grand Conclusion

Similar Questions

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

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

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is $90 s$ , $91 s$ , $92 s$ and $95 s$ . If the minimum division in the measuring clock is $1 s$ , then the reported mean time should be

The period of oscillation of a simple pendulum is $T = 2 π \frac{L}{g}$ . Measured value of $L$ is $20.0 cm$ known to $1 mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90 s$ using a wrist watch of $1 s$ resolution. The accuracy in the determination of $g$ is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g=t24π2n2l​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } gΔg​×100

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EI​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EI​=0.3125%

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EII​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EII​=0.46875%

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EIII​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EIII​=1.055%

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EI​<EII​<EIII​