Math Problem Statement

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?

Solution

The problem states:

"If the length of a pendulum is 6±0.01m6 \pm 0.01 \, \text{m} and the time period is 12±0.10sec12 \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 TT of a simple pendulum is related to its length ll and the acceleration due to gravity gg by the formula: T=2πlgT = 2\pi \sqrt{\frac{l}{g}}

Rearranging the formula to find gg: g=4π2lT2g = \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 gg. To do this, we apply the rule for the propagation of relative errors. The relative error in gg is given by the sum of the relative errors in ll and T2T^2: Δgg=Δll+2ΔTT\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} Where:

  • Δll\frac{\Delta l}{l} is the relative error in the length of the pendulum.
  • ΔTT\frac{\Delta T}{T} is the relative error in the time period.

Step 3: Calculate the relative errors

Given:

  • l=6ml = 6 \, \text{m}, Δl=0.01m\Delta l = 0.01 \, \text{m}
  • T=12secT = 12 \, \text{sec}, ΔT=0.10sec\Delta T = 0.10 \, \text{sec}

The relative error in length ll is: Δll=0.0160.00167\frac{\Delta l}{l} = \frac{0.01}{6} \approx 0.00167

The relative error in the time period TT is: ΔTT=0.10120.00833\frac{\Delta T}{T} = \frac{0.10}{12} \approx 0.00833

Since TT is squared, the relative error in T2T^2 is: 2ΔTT=2×0.00833=0.016672\frac{\Delta T}{T} = 2 \times 0.00833 = 0.01667

Step 4: Calculate the total percentage error

The total relative error in gg is the sum of the relative errors in ll and T2T^2: Δgg=0.00167+0.01667=0.01834\frac{\Delta g}{g} = 0.00167 + 0.01667 = 0.01834

Converting this to a percentage error: Percentage error=0.01834×1001.834%\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%.


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.

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Math Problem Analysis

Mathematical Concepts

Error Propagation
Pendulum Motion
Gravitational Acceleration

Formulas

T = 2π√(l/g)
g = (4π²l) / T²
Percentage error formula: Δg/g = Δl/l + 2ΔT/T

Theorems

Error Propagation Theorem

Suitable Grade Level

Grades 10-12