The period of oscillation of a simple pendulum is T = 2\pi \sqrt{L/g}. Measured value of L is 20.0 c

Q: Why do we add the errors instead of subtracting them?

In error analysis, we always want to find the maximum possible error. Therefore, relative errors are always added, even if the quantities are divided in the original formula.

Q: Why is the error in $T$ multiplied by 2?

According to the rules of error propagation, if a physical quantity is raised to a power, that power becomes a multiplier for its relative error. Since $T$ is squared in the formula for $g$, its relative error is multiplied by 2.

Q: Why do we ignore $4\pi^2$ in the error calculation?

Constants like $4$, $\pi$, or any pure number have exact values. They do not have any uncertainty or measurement error, so their relative error is zero.

Q: Why did we convert 1 mm to 0.1 cm?

To calculate a relative error like $\frac{\Delta L}{L}$, both the absolute error $\Delta L$ and the measured value $L$ must be in the exact same units so that the units cancel out, leaving a dimensionless ratio.

Question

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

Options:

(A) $2\%$

(B) $3\%$

(C) $1\%$

(D) $5\%$

Sigma Percentile · Accepted Answer

Imagine you are in a physics lab, tasked with measuring the acceleration due to gravity, $g$. You have a simple pendulum, a ruler, and a stopwatch. It sounds simple, right?

But here is the catch: every measurement you make has some inherent uncertainty. Your ruler can only measure down to a millimeter, and your stopwatch only ticks every second. How do these tiny imperfections affect your final calculation of $g$?

This is the beautiful and rigorous world of error analysis! Let's embark on this journey to decode the precision of gravity.

### The Master Equation

To find $g$, we first need a relationship between the things we can measure (length and time) and the thing we want to find ($g$). The time period of a simple pendulum is given by the famous equation:

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

Since we are interested in the accuracy of $g$, we need to rearrange this equation to make $g$ the subject. By squaring both sides and rearranging the terms, we get:

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

This is our master equation. It tells us exactly how $g$ depends on the length $L$ and the time period $T$.

### The Rules of Error Propagation

Now, how do we find the error in $g$? We must use 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 these rules to our master equation, we get the error equation:

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

Notice that the constant $4\pi^2$ does not appear in this equation. **Constants have zero uncertainty**, so they do not contribute to the relative error.

### Decoding the Length Error

Let's look at the length of the pendulum. We are given the measured length $L = 20.0 	ext{ cm}$. The accuracy of the measurement is $1 	ext{ mm}$, which is our absolute error, $\Delta L$.

But wait! We cannot mix millimeters and centimeters in our calculation. We must convert $1 	ext{ mm}$ to $0.1 	ext{ cm}$ to keep the units consistent.

Now, we can calculate the relative error in length:

$$\frac{\Delta L}{L} = \frac{0.1}{20.0} = \frac{1}{200}$$

### The Catch with the Time Error

Next, let's find the relative error in the time period $T$. We measured the time for $100$ oscillations to be $90 	ext{ s}$. The resolution of the stopwatch is $1 	ext{ s}$, which is our absolute error in total time, $\Delta t$.

The time period $T$ is the total time $t$ divided by the number of oscillations $n$. So, $T = \frac{t}{n}$ and $\Delta T = \frac{\Delta t}{n}$.

When we calculate the relative error $\frac{\Delta T}{T}$, the $n$ cancels out completely! **The relative error in the time period $T$ is exactly the same as the relative error in the total time $t$.**

$$\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{90}$$

### The Final Calculation

We have all our pieces now. Let's substitute these relative errors back into our error equation to find the percentage error in $g$. We multiply the whole equation by $100$ to get percentages.

$$\frac{\Delta g}{g} 	imes 100\% = \left( \frac{1}{200} + 2 	imes \frac{1}{90} ight) 	imes 100\%$$

Let's break down the calculation. The length contributes $\frac{1}{200} 	imes 100\% = 0.5\%$ to the total error. The time contributes $2 	imes \frac{1}{90} 	imes 100\% \approx 2.22\%$ to the total error.

Adding them up, we get a total percentage error of $0.5\% + 2.22\% = 2.72\%$.

Looking at our options, we round this off to the nearest integer. **The accuracy in the determination of $g$ is approximately $3\%$.**

Select Answer:

Visualized Solution

$g = 4 π^{2} \frac{L}{T ^{2}}$

$\frac{Δ g}{g} = \frac{Δ L}{L} + 2 \frac{Δ T}{T}$

$\frac{Δ L}{L} = \frac{0.1}{20.0}$

$\frac{Δ T}{T} = \frac{1}{90}$

$\frac{Δ g}{g} \times 100%$

\approx 3\%

\text{Improving Accuracy}

The Sigma Insight: Errors in Measurement

The Master Equation

The Rules of Error Propagation

Decoding the Length Error

The Catch with the Time Error

The Final Calculation

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