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Experimental Programme - Tools and Techniques

Grade 12IBPhysics

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

🔑Concepts

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Random and Systematic Errors: Random errors are caused by unpredictable fluctuations and can be reduced by repeated trials. They affect the precision of measurements. Systematic errors are due to faulty equipment or experimental design (e.g., zero offset) and affect the accuracy of the result.

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Accuracy and Precision: Accuracy is how close a measured value is to the true value. Precision is how close repeated measurements are to each other, indicated by the number of significant figures or the spread of data.

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Absolute, Fractional, and Percentage Uncertainties: Absolute uncertainty Δx\Delta x is the margin of error in a measurement. Fractional uncertainty is Δxx\frac{\Delta x}{x}, and percentage uncertainty is Δxx×100%\frac{\Delta x}{x} \times 100\%.

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Propagation of Uncertainties: When adding or subtracting quantities, add the absolute uncertainties. When multiplying or dividing, add the percentage/fractional uncertainties. When a value is raised to a power nn, multiply the percentage uncertainty by ∣n∣|n|.

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Graphical Analysis: The line of best fit should pass through all error bars. The uncertainty in the gradient mm is estimated using Δm=mmax−mmin2\Delta m = \frac{m_{max} - m_{min}}{2}, where mmaxm_{max} and mminm_{min} are the gradients of the steepest and shallowest possible lines of fit.

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Linearization: To analyze non-linear relationships such as y=axny = ax^n, physicists use logarithms: log⁡(y)=nlog⁡(x)+log⁡(a)\log(y) = n \log(x) + \log(a). This allows the determination of nn (gradient) and aa (vertical intercept) via a linear plot.

📐Formulae

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

Δy=Δa+Δb(for y=a±b)\Delta y = \Delta a + \Delta b \quad \text{(for } y = a \pm b\text{)}

Δyy=Δaa+Δbb(for y=ab or y=ab)\frac{\Delta y}{y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} \quad \text{(for } y = ab \text{ or } y = \frac{a}{b}\text{)}

Δyy=∣n∣Δxx(for y=xn)\frac{\Delta y}{y} = |n| \frac{\Delta x}{x} \quad \text{(for } y = x^n\text{)}

Δm=mmax−mmin2\Delta m = \frac{m_{max} - m_{min}}{2}

Δc=cmax−cmin2\Delta c = \frac{c_{max} - c_{min}}{2}

💡Examples

Problem 1:

An object travels a distance s=(20.0±0.5) ms = (20.0 \pm 0.5) \text{ m} in a time t=(4.0±0.2) st = (4.0 \pm 0.2) \text{ s}. Calculate the average speed vv and its absolute uncertainty Δv\Delta v.

Solution:

  1. Calculate speed: v=st=20.04.0=5.0 m s−1v = \frac{s}{t} = \frac{20.0}{4.0} = 5.0 \text{ m s}^{-1}.
  2. Calculate fractional uncertainty: Δvv=Δss+Δtt=0.520.0+0.24.0=0.025+0.05=0.075\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t} = \frac{0.5}{20.0} + \frac{0.2}{4.0} = 0.025 + 0.05 = 0.075.
  3. Calculate absolute uncertainty: Δv=v×0.075=5.0×0.075=0.375 m s−1\Delta v = v \times 0.075 = 5.0 \times 0.075 = 0.375 \text{ m s}^{-1}.
  4. Round to appropriate significant figures: v=(5.0±0.4) m s−1v = (5.0 \pm 0.4) \text{ m s}^{-1}.

Explanation:

Since speed is a quotient, we add the fractional uncertainties of distance and time. The final absolute uncertainty is typically rounded to one significant figure, and the value is rounded to the same decimal place.

Problem 2:

The radius of a sphere is measured as r=(10.0±0.2) cmr = (10.0 \pm 0.2) \text{ cm}. Find the percentage uncertainty in the volume VV of the sphere.

Solution:

  1. Volume formula: V=43πr3V = \frac{4}{3} \pi r^3.
  2. The constant 43π\frac{4}{3} \pi has no uncertainty.
  3. Use the power rule for uncertainties: ΔVV=3×Δrr\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}.
  4. Calculate percentage uncertainty: 3×(0.210.0×100%)=3×2%=6%3 \times (\frac{0.2}{10.0} \times 100\%) = 3 \times 2\% = 6\%.

Explanation:

According to the propagation rules, when a variable is raised to the power of 3, its percentage uncertainty is tripled.

Problem 3:

A student measures the period TT of a pendulum for different lengths ll to find gg using T=2πlgT = 2\pi \sqrt{\frac{l}{g}}. How should the data be plotted to obtain a straight line, and what would the gradient represent?

Solution:

  1. Square both sides: T2=4π2glT^2 = \frac{4\pi^2}{g} l.
  2. This matches the linear form y=mx+cy = mx + c, where y=T2y = T^2 and x=lx = l.
  3. Plot T2T^2 on the yy-axis and ll on the xx-axis.
  4. The gradient m=4π2gm = \frac{4\pi^2}{g}. Therefore, g=4π2mg = \frac{4\pi^2}{m}.

Explanation:

Linearization allows for the use of a line of best fit to average out random errors and provides a clear method to calculate physical constants from the gradient.

Tools and Techniques - Revision Notes & Key Formulas | IB Grade 12 Physics