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The problem states:

"If the length of a pendulum is \( 6 \pm 0.01 \, \text{m} \) and the time period is \( 12 \pm 0.10 \, \text{sec} \), then what is the percentage of error in the measurement of acceleration due to gravity?"

### Step 1: Formula for the time period of a pendulum
The time period \( T \) of a simple pendulum is related to its length \( l \) and the acceleration due to gravity \( g \) by the formula:
\[
T = 2\pi \sqrt{\frac{l}{g}}
\]

Rearranging the formula to find \( g \):
\[
g = \frac{4\pi^2 l}{T^2}
\]

### Step 2: Error propagation in gravity
We need to find the percentage error in the calculation of gravity \( g \). To do this, we apply the rule for the propagation of relative errors. The relative error in \( g \) is given by the sum of the relative errors in \( l \) and \( T^2 \):
\[
\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}
\]
Where:
- \( \frac{\Delta l}{l} \) is the relative error in the length of the pendulum.
- \( \frac{\Delta T}{T} \) is the relative error in the time period.

### Step 3: Calculate the relative errors

Given:
- \( l = 6 \, \text{m} \), \( \Delta l = 0.01 \, \text{m} \)
- \( T = 12 \, \text{sec} \), \( \Delta T = 0.10 \, \text{sec} \)

The relative error in length \( l \) is:
\[
\frac{\Delta l}{l} = \frac{0.01}{6} \approx 0.00167
\]

The relative error in the time period \( T \) is:
\[
\frac{\Delta T}{T} = \frac{0.10}{12} \approx 0.00833
\]

Since \( T \) is squared, the relative error in \( T^2 \) is:
\[
2\frac{\Delta T}{T} = 2 \times 0.00833 = 0.01667
\]

### Step 4: Calculate the total percentage error

The total relative error in \( g \) is the sum of the relative errors in \( l \) and \( T^2 \):
\[
\frac{\Delta g}{g} = 0.00167 + 0.01667 = 0.01834
\]

Converting this to a percentage error:
\[
\text{Percentage error} = 0.01834 \times 100 \approx 1.834\%
\]

### Final Answer:
The percentage error in the measurement of acceleration due to gravity is approximately **1.83%**.

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Would you like any further explanation, or do you have any questions about the steps involved?

Here are 5 related questions for deeper understanding:
1. How would the percentage error change if the length of the pendulum were shorter?
2. What is the significance of the time period being squared in the error propagation formula?
3. Could this method be used to calculate the error in gravity for a different kind of pendulum?
4. How would air resistance affect the accuracy of these measurements in a real experiment?
5. What other factors besides measurement errors can cause variations in gravitational acceleration?

**Tip:** In error propagation, always remember that squaring a term doubles the relative error contribution from that term.

If the length of a pendulum is 6 ± 0.01 m and time period is 12 ± 0.10 sec, then what is the percentage of error in the measurement of acceleration due to gravity?

Find the percentage error in gravitational acceleration when the length of a pendulum is 6 ± 0.01 m and the time period is 12 ± 0.10 sec. This problem uses error propagation formulas and concepts of pendulum motion to derive the final answer. Suitable for Grades 10-12.

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