Source: Assignment 1, Experimental Physics and Engineering Lab, Texas A&M University
Tags: error propagation, pendulum period, acceleration uncertainty, Atwood machine, agreement of measurements, ENGR 216, propagation of error, comparing measurements
These problems apply the general error propagation formula to real lab scenarios: predicting a pendulum's period from a length measurement, extracting acceleration from velocity data, and deriving uncertainty for an Atwood machine. A recurring theme is checking whether two values "agree" by seeing if their uncertainty ranges overlap.
Consistency of measurements (agreement)
Two measurements are consistent if their uncertainty ranges overlap. More precisely, if the difference between them is smaller than the combined uncertainty (added in quadrature), they agree within experimental error.
Pendulum period formula
T = 2π√(L/g), where L is the pendulum length and g = 9.81 m/s² (or 9.80 m/s², depending on convention).
Atwood machine
A system of two masses connected by a string over a pulley. The acceleration is a = g(m₁ − m₂)/(m₁ + m₂). Classic setup for studying Newton's second law.
Partial derivative method
Taking the derivative of a function with respect to each measured variable in turn, treating others as constants. The core technique for deriving uncertainty formulas from scratch.
Setup:
T = 2π√(L/g)
L = 1.40 ± 0.01 m
g = 9.81 m/s² (exact for this problem)
Part (a) – Predicted T and its uncertainty:
Best estimate:
T = 2π√(1.40/9.81) = 2π√(0.14271) = 2π × 0.37783 = 2.374 s
For the uncertainty, take the partial derivative. Since T = 2π(L/g)^(1/2):
∂T/∂L = 2π · (1/2) · (1/g)^(1/2) · L^(−1/2) = T/(2L)
Alternatively, use the power rule: T depends on L^(1/2), so the fractional uncertainty is:
δT/T = ½ · (δL/L) = ½ × (0.01/1.40) = 0.003571
δT = 0.003571 × 2.374 = 0.00848 s ≈ 0.01 s
Answer: T = 2.37 ± 0.01 s
Part (b) – Does T = 2.39 ± 0.01 s agree?
The predicted value is 2.37 ± 0.01 s. The measured value is 2.39 ± 0.01 s.
The difference is: 2.39 − 2.37 = 0.02 s
The combined uncertainty (discrepancy test): √(0.01² + 0.01²) = √(0.0002) = 0.014 s
Since the difference (0.02) is larger than the combined uncertainty (0.014), the ranges do not quite overlap. The predicted range is 2.36 to 2.38, and the measured range is 2.38 to 2.40, so they barely touch at 2.38 but do not properly overlap.
Answer: The measured value is marginally inconsistent with the prediction. The discrepancy is small, so you might call it borderline, but strictly speaking the values do not agree within their stated uncertainties.
Setup:
v₁ = 0.21 ± 0.05 m/s
v₂ = 0.85 ± 0.05 m/s
t = 8.0 ± 0.1 s
a = (v₂ − v₁)/t
Part (a) – Calculate a and δa:
Best estimate:
a = (0.85 − 0.21)/8.0 = 0.64/8.0 = 0.080 m/s²
For the uncertainty, treat this in two stages.
First, let Δv = v₂ − v₁. For a difference:
δ(Δv) = √(δv₂² + δv₁²) = √(0.05² + 0.05²) = √(0.005) = 0.0707 m/s
Now a = Δv/t, which is a quotient. Use fractional uncertainties:
δa/a = √[ (δ(Δv)/Δv)² + (δt/t)² ]
δa/a = √[ (0.0707/0.64)² + (0.1/8.0)² ]
δa/a = √[ (0.1105)² + (0.0125)² ]
δa/a = √[ 0.01221 + 0.000156 ]
δa/a = √(0.01237) = 0.1112
δa = 0.1112 × 0.080 = 0.0089 m/s² ≈ 0.009 m/s²
Answer: a = 0.080 ± 0.009 m/s²
Notice how the velocity uncertainties dominate: the fractional uncertainty from Δv is about 11%, while the time contributes only about 1.3%. This is typical when you subtract two similar-magnitude quantities, as the difference is small relative to the individual uncertainties.
Part (b) – Does the air resistance model agree?
Measured: 0.080 ± 0.009 m/s² Predicted: 0.13 ± 0.01 m/s²
Difference = 0.13 − 0.080 = 0.050 m/s²
Combined uncertainty = √(0.009² + 0.01²) = √(0.000081 + 0.0001) = √(0.000181) = 0.013 m/s²
The difference (0.050) is roughly 3.7 times the combined uncertainty, which is far outside the overlap range.
Answer: No, the measurement does not agree with the prediction. The discrepancy is much larger than the combined uncertainty.
Setup:
a = g(m₁ − m₂)/(m₁ + m₂)
m₁ = 100 ± 1 g
m₂ = 50 ± 1 g
g = 9.81 m/s² (treated as exact)
Step 1 – Compute the partial derivatives:
Let S = m₁ + m₂ (the sum) and D = m₁ − m₂ (the difference), so a = gD/S.
∂a/∂m₁ = g · [ S · 1 − D · 1 ] / S² = g(S − D)/S²
Substituting back: S − D = (m₁ + m₂) − (m₁ − m₂) = 2m₂
So: ∂a/∂m₁ = 2gm₂ / (m₁ + m₂)²
∂a/∂m₂ = g · [ S · (−1) − D · 1 ] / S² = −g(S + D)/S²
Substituting: S + D = (m₁ + m₂) + (m₁ − m₂) = 2m₁
So: ∂a/∂m₂ = −2gm₁ / (m₁ + m₂)²
Step 2 – Write the uncertainty formula:
δa = √[ (∂a/∂m₁ · δm₁)² + (∂a/∂m₂ · δm₂)² ]
δa = √[ (2gm₂/(m₁+m₂)² · δm₁)² + (2gm₁/(m₁+m₂)² · δm₂)² ]
Factor out the common terms:
δa = [2g/(m₁+m₂)²] · √[ (m₂ · δm₁)² + (m₁ · δm₂)² ]
Step 3 – Plug in numbers:
m₁ + m₂ = 150 g
(m₁ + m₂)² = 22500 g²
2g/(m₁+m₂)² = 2 × 9.81 / 22500 = 19.62 / 22500 = 0.000872 m/s² per g² (but we need consistent units)
Working in grams throughout, then converting: since g = 9.81 m/s² and masses are in grams, keep the ratio dimensionless.
Best estimate: a = 9.81 × (100 − 50)/(100 + 50) = 9.81 × 50/150 = 9.81 × 1/3 = 3.27 m/s²
For the uncertainty:
The terms inside the square root:
m₂ · δm₁ = 50 × 1 = 50
m₁ · δm₂ = 100 × 1 = 100
√(50² + 100²) = √(2500 + 10000) = √12500 = 111.8
2g/(m₁+m₂)² = 2 × 9.81 / (150)² = 19.62 / 22500 = 8.72 × 10⁻⁴ (in units of m/s² per gram)
δa = 8.72 × 10⁻⁴ × 111.8 = 0.0975 m/s²
Answer: δa ≈ 0.10 m/s², so a = 3.27 ± 0.10 m/s²
The general formula is:
δa = [2g / (m₁ + m₂)²] · √[ (m₂ · δm₁)² + (m₁ · δm₂)² ]
Pendulum period: T = 2π√(L/g), with δT/T = ½ · (δL/L)
Acceleration from velocities: a = (v₂ − v₁)/t
Atwood machine acceleration: a = g(m₁ − m₂)/(m₁ + m₂)
Atwood uncertainty: δa = [2g/(m₁+m₂)²] · √[(m₂·δm₁)² + (m₁·δm₂)²]
Agreement test: Two values agree if |value₁ − value₂| < √(δ₁² + δ₂²)
⚠️ The "do these values agree?" question appears constantly in lab courses. The method: compute the difference, compute the combined uncertainty in quadrature, and check whether the difference fits within it.
⚠️ When propagating through a square root (like the pendulum formula), the fractional uncertainty is halved. Through a square, it doubles. The exponent always multiplies the fractional uncertainty.
⚠️ In Problem 4, the subtraction v₂ − v₁ is the dominant source of error because the difference is small compared to the uncertainties. This "catastrophic cancellation" effect is worth mentioning in lab reports.
⚠️ For the Atwood machine derivation, the quotient rule does not apply directly because m₁ appears in both the numerator and denominator. You must use the full partial derivative approach. This is a common exam trap: if a variable appears more than once in the formula, you cannot treat them as separate variables.
Q: A pendulum has length L = 0.80 ± 0.02 m. What is the predicted period and its uncertainty?
A: T = 2π√(0.80/9.81) = 2π × 0.2855 = 1.794 s. δT/T = ½ × (0.02/0.80) = 0.0125. δT = 0.0125 × 1.794 = 0.022 s. Answer: T = 1.79 ± 0.02 s.
Q: You predict a value of 5.0 ± 0.3 and measure 5.8 ± 0.4. Do they agree?
A: Difference = 0.8. Combined uncertainty = √(0.3² + 0.4²) = √(0.09 + 0.16) = 0.50. Since 0.8 > 0.50, the values do not agree within their uncertainties.
Q: In the Atwood machine formula, why can you not simply use the fractional uncertainty product/quotient rule?
A: Because m₁ and m₂ each appear in both the numerator (m₁ − m₂) and the denominator (m₁ + m₂). The product/quotient rule assumes each variable appears in only one place. When a variable appears multiple times, you must use the general partial derivative method.
Q: Two velocities are measured as 1.20 ± 0.05 m/s and 1.30 ± 0.05 m/s. What is the uncertainty in their difference?
A: δ(Δv) = √(0.05² + 0.05²) = 0.071 m/s. The difference is only 0.10 m/s, so the fractional uncertainty is 71%, showing how subtraction of similar values amplifies relative error.
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