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Measurements and Uncertainties - Random and Systematic Errors

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Random Errors are caused by unpredictable fluctuations in environmental conditions or difficulties in reading instruments. They affect the precision of a measurement and cause data to scatter around a mean value.

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Systematic Errors are caused by flawed experimental design or incorrectly calibrated instruments (e.g., zero offset error). They affect the accuracy of a measurement, shifting all readings in the same direction from the true value.

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Precision refers to how close a series of measurements are to one another. High precision is indicated by a small range/standard deviation and relates to small random errors.

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Accuracy refers to how close a measured value is to the accepted or 'true' value. High accuracy relates to small systematic errors.

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To reduce Random Errors, one should take multiple repeat readings and calculate the mean value xˉ\bar{x}.

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To eliminate Systematic Errors, instruments must be calibrated (zeroed) or the experimental technique must be adjusted. They cannot be reduced by repeating measurements.

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On a linear graph, a systematic error is often indicated if the line of best fit does not pass through the origin (0,0)(0,0) when theory suggests it should.

📐Formulae

Fractional Uncertainty=Δxx\text{Fractional Uncertainty} = \frac{\Delta x}{x}

Percentage Uncertainty=Δxx×100%\text{Percentage Uncertainty} = \frac{\Delta x}{x} \times 100\%

Addition/Subtraction: y=a±b  ⟹  Δy=Δa+Δb\text{Addition/Subtraction: } y = a \pm b \implies \Delta y = \Delta a + \Delta b

Multiplication/Division: y=abc  ⟹  Δyy=Δaa+Δbb+Δcc\text{Multiplication/Division: } y = \frac{ab}{c} \implies \frac{\Delta y}{y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}

Power Rule: y=an  ⟹  Δyy=∣n∣Δaa\text{Power Rule: } y = a^n \implies \frac{\Delta y}{y} = |n| \frac{\Delta a}{a}

Uncertainty from Repeat Readings: Δx=xmax−xmin2\text{Uncertainty from Repeat Readings: } \Delta x = \frac{x_{max} - x_{min}}{2}

💡Examples

Problem 1:

A student measures the mass of an object as m=(200±2) gm = (200 \pm 2)\text{ g} and its volume as V=(100±5) cm3V = (100 \pm 5)\text{ cm}^3. Calculate the density ρ\rho and its absolute uncertainty Δρ\Delta \rho.

Solution:

  1. Calculate density: ρ=mV=200100=2.0 g cm−3\rho = \frac{m}{V} = \frac{200}{100} = 2.0\text{ g cm}^{-3}.
  2. Calculate fractional uncertainties: Δmm=2200=0.01\frac{\Delta m}{m} = \frac{2}{200} = 0.01 and ΔVV=5100=0.05\frac{\Delta V}{V} = \frac{5}{100} = 0.05.
  3. Sum fractional uncertainties: Δρρ=0.01+0.05=0.06\frac{\Delta \rho}{\rho} = 0.01 + 0.05 = 0.06.
  4. Calculate absolute uncertainty: Δρ=0.06×2.0=0.12 g cm−3\Delta \rho = 0.06 \times 2.0 = 0.12\text{ g cm}^{-3}. Result: ρ=(2.0±0.1) g cm−3\rho = (2.0 \pm 0.1)\text{ g cm}^{-3} (rounded to appropriate significant figures).

Explanation:

Since density is calculated via division, we sum the fractional uncertainties of mass and volume to find the fractional uncertainty of the density.

Problem 2:

During a simple pendulum experiment, a student records the time for 20 oscillations as t=(40.0±0.2) st = (40.0 \pm 0.2)\text{ s}. If the formula for the period is T=t20T = \frac{t}{20}, find the absolute uncertainty in the period TT.

Solution:

T=40.020=2.00 sT = \frac{40.0}{20} = 2.00\text{ s}. Since 2020 is a constant (no uncertainty), the fractional uncertainty in TT is the same as in tt: ΔT=0.220=0.01 s\Delta T = \frac{0.2}{20} = 0.01\text{ s}. Result: T=(2.00±0.01) sT = (2.00 \pm 0.01)\text{ s}.

Explanation:

When a value is divided by a constant, the absolute uncertainty is also divided by that same constant.