Tags: propagation of error examples, card thickness uncertainty, angular momentum error, pendulum period uncertainty, acceleration uncertainty, Atwood machine, partial derivatives, error propagation worked problems
These worked examples cover the main patterns you will encounter in error propagation: dividing by an exact constant, products of powers (with the fractional uncertainty rule), square root functions, subtraction followed by division, and the full partial-derivative method for more complex expressions. Each example maps to one of the core rules from the theory notes.
Setup: Calipers measure a deck of 52 cards as 0.590 ± 0.005 in. Find the thickness of one card.
Part (a) – thickness and uncertainty of one card
The thickness of one card is:
t = 0.590 / 52 = 0.01135 in
Since 52 is an exact number (no uncertainty), the constant-multiple rule applies:
δt = 0.005 / 52 ≈ 0.0001 in (rounded to 1 significant figure in the uncertainty)
Result: t = 0.0114 ± 0.0001 in
The key idea is that dividing by an exact integer simply divides the uncertainty by that same integer.
Part (b) – how many decks to reach a target uncertainty
Target uncertainty for one card: 0.00002 in. The instrument still has δ = 0.005 in regardless of how many cards you stack.
If you measure N decks (52N cards total):
δ_card = 0.005 / (52N)
Set this equal to the target:
0.005 / (52N) = 0.00002
52N = 0.005 / 0.00002 = 250
N = 250 / 52 ≈ 4.8
You need 5 decks (round up, since you cannot measure a fractional deck).
Rule used: q = cx → δq = |c| · δx, where c = 1/52N.
Setup: L = ½MR²ω with measurements:
M = 1.10 ± 0.01 kg
R = 0.250 ± 0.005 m
ω = 21.5 ± 0.4 rad/s
Calculating L:
L = 0.5 × 1.10 × (0.250)² × 21.5 L = 0.5 × 1.10 × 0.0625 × 21.5 L = 0.742 kg·m²/s
Propagating the uncertainty:
The formula L = ½MR²ω is a constant (½) times a product of powers: M¹ · R² · ω¹. Use the fractional uncertainty rule:
δL/L = √[(1 · δM/M)² + (2 · δR/R)² + (1 · δω/ω)²]
Compute each fractional contribution:
δM/M = 0.01/1.10 = 0.00909
2 · δR/R = 2 × 0.005/0.250 = 0.0400
δω/ω = 0.4/21.5 = 0.0186
Combine:
δL/L = √(0.00909² + 0.0400² + 0.0186²)
δL/L = √(0.0000826 + 0.001600 + 0.000346)
δL/L = √0.002029 = 0.0450
So δL = 0.0450 × 0.742 = 0.033 kg·m²/s
Result: L = 0.74 ± 0.03 kg·m²/s
Notice that the R term dominates because the exponent 2 doubles its fractional uncertainty, making it the largest contributor even though δR/R by itself is only 2%.
Rule used: Product of powers → fractional uncertainties (with exponents) add in quadrature.
Setup: T = 2π√(L/g), with L = 1.40 ± 0.01 m. Use g = 9.81 m/s² (exact for this purpose).
Part (a) – predicted T and its uncertainty
Calculate T:
T = 2π√(1.40/9.81) = 2π√(0.14272) = 2π × 0.37779 = 2.374 s
For the uncertainty, rewrite T = 2π · g^(−1/2) · L^(1/2). The exponent on L is ½:
δT/T = ½ · δL/L = 0.5 × 0.01/1.40 = 0.00357
So δT = 0.00357 × 2.374 = 0.0085 s
Rounding to one significant figure in the uncertainty:
Result: T = 2.37 ± 0.01 s
The square root halves the fractional uncertainty. This is a useful pattern to remember: if a variable appears under a square root, its fractional contribution to uncertainty is cut in half.
Part (b) – consistency check
Measured value: T = 2.39 ± 0.01 s. Predicted value: T = 2.37 ± 0.01 s.
Discrepancy: |2.39 − 2.37| = 0.02 s
Sum of uncertainties: 0.01 + 0.01 = 0.02 s
The discrepancy equals the sum of uncertainties, so the values are marginally consistent (just barely within range). Most instructors would accept this as consistent.
Rule used: Product of powers with exponent ½ (square root).
Setup: a = (v₂ − v₁)/t, with:
v₁ = 0.21 ± 0.05 m/s
v₂ = 0.85 ± 0.05 m/s
t = 8.0 ± 0.1 s
Part (a) – acceleration and its uncertainty
Calculate a:
a = (0.85 − 0.21) / 8.0 = 0.64 / 8.0 = 0.080 m/s²
This requires two steps because a = (v₂ − v₁)/t involves subtraction followed by division.
Step 1, uncertainty in the numerator (v₂ − v₁):
δ(v₂ − v₁) = √(δv₂² + δv₁²) = √(0.05² + 0.05²) = √0.005 = 0.0707 m/s
Step 2, use the general formula for a = Δv / t. This is a quotient, so use fractional uncertainties:
δa/a = √[(δΔv/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 = 0.111
So δa = 0.111 × 0.080 = 0.0089 m/s²
Result: a = 0.080 ± 0.009 m/s²
The velocity uncertainties dominate here. The time uncertainty contributes almost nothing because δt/t is so small relative to δΔv/Δv.
Part (b) – does it agree with the air resistance model?
Model prediction: 0.13 ± 0.01 m/s².
Discrepancy: |0.13 − 0.080| = 0.050 m/s²
Sum of uncertainties: 0.009 + 0.01 = 0.019 m/s²
The discrepancy (0.050) is much larger than the combined uncertainty (0.019). These values do not agree. The measurement is inconsistent with the air resistance model's prediction.
Rule used: Subtraction rule (absolute uncertainties in quadrature) then division rule (fractional uncertainties in quadrature).
Setup: An Atwood machine has acceleration a = g(m₁ − m₂)/(m₁ + m₂), with:
m₁ = 100 ± 1 g
m₂ = 50 ± 1 g
This formula has m₁ and m₂ appearing in both the numerator and denominator, so the simple product-of-powers rule does not apply directly. Use partial derivatives.
Deriving the uncertainty formula:
Compute ∂a/∂m₁:
Using the quotient rule on f(m₁) = (m₁ − m₂)/(m₁ + m₂):
∂a/∂m₁ = g · [(m₁ + m₂) − (m₁ − m₂)] / (m₁ + m₂)²
∂a/∂m₁ = g · 2m₂ / (m₁ + m₂)²
Compute ∂a/∂m₂:
∂a/∂m₂ = g · [−(m₁ + m₂) − (m₁ − m₂)] / (m₁ + m₂)²
∂a/∂m₂ = g · (−2m₁) / (m₁ + m₂)²
The uncertainty formula is:
δa = √[(∂a/∂m₁ · δm₁)² + (∂a/∂m₂ · δm₂)²]
δa = √[(g · 2m₂/(m₁+m₂)² · δm₁)² + (g · 2m₁/(m₁+m₂)² · δm₂)²]
δa = (2g/(m₁+m₂)²) · √[(m₂ · δm₁)² + (m₁ · δm₂)²]
Plugging in numbers:
m₁ + m₂ = 150 g, so (m₁ + m₂)² = 22500 g²
g = 980 cm/s² (using CGS since masses are in grams)
First, the central value:
a = 980 × (100 − 50)/(100 + 50) = 980 × 50/150 = 326.7 cm/s² ≈ 3.27 m/s²
Now δa:
δa = (2 × 980 / 22500) × √[(50 × 1)² + (100 × 1)²]
δa = 0.08711 × √[2500 + 10000]
δa = 0.08711 × √12500
δa = 0.08711 × 111.8
δa = 9.74 cm/s² ≈ 0.097 m/s²
Result: a = 3.27 ± 0.10 m/s²
Note that m₁ contributes more to the uncertainty than m₂ does (the term with m₁ · δm₂ is larger than the term with m₂ · δm₁) because m₁ is the larger mass, even though both have the same absolute uncertainty.
Rule used: General partial derivative formula, required because the same variables appear in both numerator and denominator.
⚠️ Know when to use the product-of-powers shortcut versus the full partial derivative method. If a variable appears in both the numerator and denominator (like in the Atwood machine), you must use partial derivatives.
⚠️ When a formula involves subtraction followed by division (like acceleration = Δv/t), break it into two steps: find the uncertainty in the difference first, then propagate through the division.
⚠️ Square roots halve fractional uncertainty; squares double it. Commit this pattern to memory for quick estimates.
⚠️ For consistency checks, always compare the discrepancy to the combined uncertainties of both values, not just one.
⚠️ Round uncertainties to one significant figure, then round the central value to match the decimal place of the uncertainty.
Q: A quantity q = A · B² / C has fractional uncertainties δA/A = 1%, δB/B = 2%, and δC/C = 3%. What is the fractional uncertainty in q?
A: δq/q = √[(1%)² + (2 × 2%)² + (3%)²] = √[1 + 16 + 9] = √26 ≈ 5.1%. The B² term dominates because the exponent doubles its contribution.
Q: Why can you not use the simple fractional uncertainty rule for the Atwood machine formula a = g(m₁ − m₂)/(m₁ + m₂)?
A: Because m₁ and m₂ each appear in both the numerator and denominator. The product-of-powers rule assumes each variable appears only once (as a single power). When a variable is entangled in both parts of a fraction, you must use partial derivatives.
Q: If you measure a quantity x = 10.0 ± 0.5 and compute q = x², what is δq?
A: Fractional uncertainty in q = 2 × (0.5/10.0) = 0.10, so δq = 0.10 × 100 = 10. Result: q = 100 ± 10.
Q: Two measurements give 7.2 ± 0.3 and 8.1 ± 0.4. Are they consistent?
A: Discrepancy = |8.1 − 7.2| = 0.9. Sum of uncertainties = 0.3 + 0.4 = 0.7. Since 0.9 > 0.7, they are not consistent.
Q: You want to measure the thickness of a single piece of paper to within ±0.001 mm. Your ruler has a precision of ±0.5 mm. How many sheets should you stack?
A: N = 0.5 / 0.001 = 500 sheets.
propagation of error examples, error propagation worked problems, card thickness uncertainty, angular momentum uncertainty, pendulum period error, Atwood machine uncertainty, partial derivative error propagation, fractional uncertainty product rule, subtraction uncertainty, consistency check physics, physics lab worked examples, quadrature addition examples, PHYS 206 error analysis