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Grade 11Mechanics

The period of oscillation of a simple pendulum is given by:
T=2×pie×under root of L/g
where L-Length of the pendulum
100vibrations are taken and the total time measured is 200s. The least count of stop watch is 0.1 s. the length of pendulum is 1m. the least count of the scale is 1mm. what is the maximum error in the measurement of g?

Profile image of Saurav Chahar
11 Years agoGrade 11
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer0 Years ago

To determine the maximum error in the measurement of the acceleration due to gravity (g) using the given data about the simple pendulum, we need to analyze the formula for the period of oscillation and how the measurements contribute to the overall uncertainty.

Understanding the Formula

The period of a simple pendulum is expressed as:

T = 2π√(L/g)

Where:

  • T = period of oscillation (time for one complete vibration)
  • L = length of the pendulum
  • g = acceleration due to gravity

Given Data

From your question, we have:

  • Number of vibrations = 100
  • Total time measured = 200 s
  • Length of the pendulum (L) = 1 m
  • Least count of the stopwatch = 0.1 s
  • Least count of the scale = 1 mm = 0.001 m

Calculating the Period (T)

First, we need to calculate the period of one oscillation:

T = Total time / Number of vibrations

Substituting the values:

T = 200 s / 100 = 2 s

Finding g

Next, we can rearrange the formula for T to solve for g:

g = (4π²L) / T²

Substituting L = 1 m and T = 2 s:

g = (4π² × 1) / (2²) = (4π²) / 4 = π² ≈ 9.87 m/s²

Calculating Errors

Now, we need to consider the errors in our measurements. The maximum error in g can be derived from the errors in L and T.

Maximum Error in Length (ΔL)

The least count of the scale is 1 mm, which means the maximum error in measuring L is:

ΔL = 0.001 m

Maximum Error in Time (ΔT)

The least count of the stopwatch is 0.1 s, so the maximum error in measuring T is:

ΔT = 0.1 s

Applying Error Propagation

To find the maximum error in g, we can use the formula for error propagation. The relative error in g can be expressed as:

Δg/g = 2(ΔT/T) + (ΔL/L)

Substituting the values:

  • ΔT = 0.1 s
  • T = 2 s
  • ΔL = 0.001 m
  • L = 1 m

Calculating the relative errors:

Δg/g = 2(0.1/2) + (0.001/1) = 0.1 + 0.001 = 0.101

Calculating Maximum Error in g

Now, we can find the maximum error in g:

Δg = g × Δg/g = 9.87 × 0.101 ≈ 0.999 m/s²

Final Result

The maximum error in the measurement of g is approximately:

Δg ≈ 1 m/s²

This means that the value of g could realistically range from about 8.87 m/s² to 10.87 m/s², depending on the uncertainties in your measurements. Understanding these errors is crucial in experimental physics, as it helps us gauge the reliability of our results.