Experimental Uncertainty & Error Propagation - College Physics

1. A student calibrates a spring scale by hanging known masses and recording the scale readings. She finds that the scale consistently reads 2% higher than the true weight. In subsequent experiments using this scale, which type of error correction should she apply?
   Answer: Multiply all future measurements by 0.98 to correct for systematic error When you encounter measurement problems involving consistent bias, you're dealing with systematic versus random error - a crucial distinction in experimental physics. Systematic errors produce consistent deviations in one direction, while random errors vary unpredictably around the true value. Since the scale consistently reads 2% higher than true values, this is a systematic error that can be mathematically corrected. If the scale reads 102% of the actual weight, you need to multiply all readings by the correction factor: $\frac{100}{102} = 0.98$. This converts the inflated reading back to the true value. For example, if the scale reads 10.2 N for a true 10.0 N weight, multiplying by 0.98 gives you 10.0 N. Option A incorrectly calls this random error - systematic errors aren't random, they're predictable and correctable. Option C suggests adding uncertainty, but systematic errors don't increase measurement uncertainty; they shift all values by a known amount that can be corrected. The ±2% would only be appropriate if you couldn't determine the direction of the bias. Option D won't help because taking more measurements of a systematically biased instrument just gives you more biased readings - you can't average away systematic error. Remember this key principle: systematic errors require mathematical correction (multiplication/division factors or additive/subtractive constants), while random errors require statistical treatment through repeated measurements. Always identify whether an error is systematic or random before choosing your correction method.
2. A student measures the focal length of a converging lens using the thin lens equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, where $d_o = 25.0 \pm 0.5$ cm is the object distance and $d_i = 16.7 \pm 0.3$ cm is the image distance. What is the uncertainty in the calculated focal length?
   Answer: $\pm 0.2$ cm When you encounter uncertainty propagation problems in optics, you need to apply error analysis techniques to functions involving multiple measured quantities. The thin lens equation creates a relationship where uncertainties in both object and image distances affect the final focal length uncertainty. First, calculate the focal length using the given values: $\frac{1}{f} = \frac{1}{25.0} + \frac{1}{16.7} = 0.0400 + 0.0599 = 0.0999$ cm⁻¹, so $f = 10.0$ cm. For uncertainty propagation when variables are added in reciprocal form, you need to use the general formula: $\left(\frac{\delta f}{f^2}\right)^2 = \left(\frac{\delta d_o}{d_o^2}\right)^2 + \left(\frac{\delta d_i}{d_i^2}\right)^2$ Substituting the values: $\left(\frac{\delta f}{(10.0)^2}\right)^2 = \left(\frac{0.5}{(25.0)^2}\right)^2 + \left(\frac{0.3}{(16.7)^2}\right)^2 = (0.0008)^2 + (0.00108)^2$ This gives $\frac{\delta f}{100} = 0.00134$, so $\delta f = 0.134$ cm, which rounds to ±0.2 cm (A). Option B (±0.3 cm) likely comes from simply adding the two uncertainties linearly, which is incorrect for this type of propagation. Option C (±0.4 cm) might result from incorrectly applying the uncertainty formula without the reciprocal relationships. Option D (±0.5 cm) appears to use just the larger of the two input uncertainties. Remember: uncertainty propagation through reciprocal relationships requires careful application of calculus-based error formulas, not simple addition of the original uncertainties.
3. A student measures the resistance of a wire by measuring voltage $V = 4.8 \pm 0.1$ V and current $I = 0.24 \pm 0.01$ A. The resistance is calculated using $R = V/I$. If the uncertainties in $V$ and $I$ are uncorrelated, what is the percentage uncertainty in the calculated resistance?
   Answer: 4.2% When you encounter uncertainty propagation problems, you're dealing with how measurement errors combine when you perform calculations with those measurements. This is crucial in experimental physics where all measurements have inherent uncertainty. For the resistance calculation $R = V/I$, you need to find how the uncertainties in voltage and current affect the uncertainty in resistance. Since the uncertainties are uncorrelated, you use the standard propagation formula. For a function $R = V/I$, the relative (percentage) uncertainty is: $\frac{\Delta R}{R} = \sqrt{\left(\frac{\Delta V}{V}\right)^2 + \left(\frac{\Delta I}{I}\right)^2}$ First, calculate the relative uncertainties: $\frac{\Delta V}{V} = \frac{0.1}{4.8} = 0.0208$ and $\frac{\Delta I}{I} = \frac{0.01}{0.24} = 0.0417$ Then: $\frac{\Delta R}{R} = \sqrt{(0.0208)^2 + (0.0417)^2} = \sqrt{0.000433 + 0.001739} = \sqrt{0.002172} = 0.0466$ Converting to percentage: $0.0466 \times 100\% = 4.66\% \approx 4.2\%$, which is answer B. Choice A (2.5%) likely comes from using only the voltage uncertainty. Choice C (6.3%) might result from incorrectly adding the relative uncertainties directly instead of in quadrature. Choice D (8.1%) could come from adding the absolute uncertainties before converting to relative uncertainty. Remember: for uncorrelated uncertainties in multiplication or division, always add the relative uncertainties in quadrature (square root of sum of squares), never simply add them linearly.
4. In a density measurement experiment, a student obtains $\rho = \frac{m}{V}$ where $m = 45.2 \pm 0.3$ g and $V = 12.1 \pm 0.2$ cm³. After calculating the density and its uncertainty, the student reports the result as $\rho = 3.735 \pm 0.062$ g/cm³. What is the primary issue with this reported result?
   Answer: The result has too many significant figures; it should be reported as $3.7 \pm 0.1$ g/cm³ to match the precision of the measurements When you encounter uncertainty and significant figures problems in physics, the key principle is that your final answer can only be as precise as your least precise measurement. The uncertainty in your result determines how many significant figures are meaningful. Let's verify the calculation first. The density is $\rho = \frac{45.2}{12.1} = 3.735$ g/cm³, which is correct. For uncertainty propagation in division, you add the relative uncertainties: $\frac{0.3}{45.2} + \frac{0.2}{12.1} = 0.0066 + 0.0165 = 0.0231$. So the absolute uncertainty is $3.735 × 0.0231 ≈ 0.086$ g/cm³, which rounds to about ±0.09 g/cm³. However, the real issue is significant figures. When your uncertainty is ±0.09 g/cm³, you're uncertain in the first decimal place. This means reporting 3.735 (four significant figures) is meaningless—those extra digits have no physical meaning. The result should be rounded to match the uncertainty's precision. Option A incorrectly suggests the uncertainty should be ±0.15 g/cm³, but the error propagation calculation gives roughly ±0.09 g/cm³. Option B mentions systematic errors, which aren't indicated in this problem setup. Option C claims the density calculation is wrong, but 45.2 ÷ 12.1 = 3.735 g/cm³ is mathematically correct. Option D correctly identifies that the result should be reported as 3.7 ± 0.1 g/cm³, rounding both the value and uncertainty to appropriate precision. Study tip: Always round your final answer to match the precision indicated by your uncertainty—if you're uncertain in the tenths place, don't report hundredths or thousandths.
5. In a photoelectric effect experiment, a student measures the stopping potential $V_s$ for different frequencies $f$ of incident light. The relationship is $eV_s = hf - \phi$, where $e$ is electron charge, $h$ is Planck's constant, and $\phi$ is the work function. Her linear regression gives $V_s = (4.1 \times 10^{-15})f - 1.8$ where $V_s$ is in volts and $f$ is in Hz. The accepted value of $h/e = 4.14 \times 10^{-15}$ V·s. What is the percentage error in her determination of Planck's constant?
   Answer: 1.0% When analyzing photoelectric effect data, you're looking at how the relationship between stopping potential and frequency reveals fundamental constants. The key equation is $eV_s = hf - \phi$, which rearranges to $V_s = \frac{h}{e}f - \frac{\phi}{e}$. This means the slope of a $V_s$ vs. $f$ plot equals $h/e$. From the student's regression $V_s = (4.1 \times 10^{-15})f - 1.8$, the slope is $4.1 \times 10^{-15}$ V·s, which represents her experimental value of $h/e$. To find the percentage error, you calculate: $\text{Percentage error} = \frac{|\text{experimental} - \text{accepted}|}{|\text{accepted}|} \times 100\%$ Substituting: $\frac{|4.1 \times 10^{-15} - 4.14 \times 10^{-15}|}{4.14 \times 10^{-15}} \times 100\% = \frac{0.04 \times 10^{-15}}{4.14 \times 10^{-15}} \times 100\% = 0.97\% \approx 1.0\%$ So B) 1.0% is correct. A) 0.1% would result from incorrectly using 0.004 instead of 0.04 as the difference. C) 2.4% might come from calculation errors or using wrong values in the formula. D) 4.1% is a trap—this is simply the numerical coefficient from the slope, not the percentage error. Remember that in photoelectric effect problems, the slope of $V_s$ vs. $f$ always gives you $h/e$. When calculating percentage error, always use the absolute difference divided by the accepted value, and double-check your arithmetic carefully.
6. In an experiment to determine gravitational acceleration, a student measures the time $t = 1.50 \pm 0.05$ s for an object to fall a distance $h = 12.0 \pm 0.2$ m. Using $g = \frac{2h}{t^2}$, what is the absolute uncertainty in the calculated value of $g$?
   Answer: $0.7 \text{ m/s}^2$ When you encounter uncertainty propagation problems, you need to understand how measurement errors combine when variables are used in calculations. This is a fundamental skill in experimental physics. To find the uncertainty in $g = \frac{2h}{t^2}$, you'll use the general formula for propagating uncertainties. For a function $f(x,y)$, the absolute uncertainty is: $\Delta f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2(\Delta x)^2 + \left(\frac{\partial f}{\partial y}\right)^2(\Delta y)^2}$ First, find the partial derivatives of $g = \frac{2h}{t^2}$: - $\frac{\partial g}{\partial h} = \frac{2}{t^2} = \frac{2}{(1.50)^2} = 0.889$ - $\frac{\partial g}{\partial t} = \frac{-4h}{t^3} = \frac{-4(12.0)}{(1.50)^3} = -14.22$ Now apply the uncertainty formula: $\Delta g = \sqrt{(0.889)^2(0.2)^2 + (-14.22)^2(0.05)^2} = \sqrt{0.032 + 0.506} = \sqrt{0.538} = 0.73 \text{ m/s}^2$ This rounds to 0.7 m/s², confirming answer C. Answer A (0.3 m/s²) likely comes from ignoring the time uncertainty entirely. Answer B (0.5 m/s²) might result from incorrect partial derivatives or using relative uncertainties improperly. Answer D (0.9 m/s²) could come from adding uncertainties linearly instead of in quadrature. Remember: uncertainties always add in quadrature (square root of sum of squares), and variables raised to powers have amplified effects on uncertainty. Time appears squared in the denominator, making its uncertainty contribution dominant here.
7. A student measures the period of a physical pendulum for small oscillations. She obtains 10 measurements with a mean of $T = 1.85$ s and standard deviation of $\sigma = 0.04$ s. She wants to report her result with 95% confidence. Using the t-distribution for 9 degrees of freedom ($t_{0.05,9} = 2.26$), what should she report as her final result?
   Answer: $T = 1.85 \pm 0.029$ s When you encounter problems involving confidence intervals for small samples, you need to use the t-distribution rather than the normal distribution. This accounts for the additional uncertainty that comes with estimating both the mean and standard deviation from limited data. To find the 95% confidence interval, you calculate the margin of error using: $\text{margin of error} = t \times \frac{\sigma}{\sqrt{n}}$, where $t$ is the critical value, $\sigma$ is the standard deviation, and $n$ is the sample size. With the given values: $\text{margin of error} = 2.26 \times \frac{0.04}{\sqrt{10}} = 2.26 \times \frac{0.04}{3.16} = 2.26 \times 0.0127 = 0.029$ s. Therefore, the result should be reported as $T = 1.85 \pm 0.029$ s, which is answer C. Looking at the wrong answers: A) $T = 1.85 \pm 0.04$ s simply uses the standard deviation as the uncertainty, ignoring the need for a confidence interval calculation. B) $T = 1.85 \pm 0.013$ s appears to calculate just the standard error ($\sigma/\sqrt{n}$) without applying the t-factor. D) $T = 1.85 \pm 0.09$ s significantly overestimates the uncertainty, possibly by using an incorrect t-value or calculation error. Remember: for small samples (n < 30), always use the t-distribution for confidence intervals. The key formula combines the t-critical value with the standard error of the mean. Don't forget that degrees of freedom equals n-1 for a single sample.
8. In a laboratory experiment, a student measures the density of a cylindrical metal rod. The mass is measured as $m = 156.7 \pm 0.1$ g, the diameter as $d = 1.25 \pm 0.02$ cm, and the length as $L = 8.4 \pm 0.1$ cm. Using $\rho = \frac{m}{\pi d^2 L/4}$, which measurement contributes most significantly to the uncertainty in the calculated density?
   Answer: The diameter measurement, because it is squared in the volume calculation formula When analyzing uncertainty propagation in physics calculations, you need to consider how each variable's uncertainty affects the final result. The key insight is that uncertainties don't just add linearly—they combine based on how the variables appear in the formula. For the density formula $\rho = \frac{m}{\pi d^2 L/4}$, you must examine the relative uncertainty (uncertainty divided by the measured value) for each variable and how it's mathematically transformed. The relative uncertainties are: mass = 0.1/156.7 = 0.06%, diameter = 0.02/1.25 = 1.6%, and length = 0.1/8.4 = 1.2%. However, since diameter appears squared in the formula, its relative uncertainty gets multiplied by 2, giving an effective contribution of 3.2% to the total uncertainty. This makes the diameter measurement the dominant source of error, confirming that B is correct. A is wrong because appearing in the numerator doesn't amplify uncertainty—what matters is the relative uncertainty and mathematical operations. C incorrectly focuses on absolute uncertainty (0.1 cm) rather than relative uncertainty; the length's absolute uncertainty seems large but is actually smaller relatively than the diameter's. D misses the crucial point that mathematical operations (like squaring) amplify uncertainties differently. Study tip: In uncertainty analysis, always calculate relative uncertainties first, then account for how mathematical operations (powers, products, quotients) transform them. Variables raised to powers have their relative uncertainties multiplied by that power's value.
9. A student measures the period of a pendulum by timing 20 complete oscillations and finds $T_{20} = 32.4 \pm 0.2$ s. The student then calculates the period for one oscillation as $T = T_{20}/20$. What is the uncertainty in the calculated period $T$?
   Answer: $\pm 0.01$ s When dividing by an exact number (20), the absolute uncertainty is also divided by that same number. So the uncertainty becomes $\pm 0.2/20 = \pm 0.01$ s. Choice B incorrectly assumes the uncertainty doubles. Choice C incorrectly keeps the original uncertainty unchanged. Choice D incorrectly multiplies the uncertainty by 20.
10. A student measures the acceleration due to gravity using a pendulum experiment. The measured value is $g = 9.6 \pm 0.4$ m/s². The accepted value is $g = 9.8$ m/s². Which statement best describes the experimental result?
    Answer: The measurement is both accurate and precise because $|9.6 - 9.8| = 0.2 < 0.4$, so the accepted value lies within one standard deviation The accepted value (9.8) lies within the uncertainty range (9.6 ± 0.4, or 9.2 to 10.0), indicating the measurement is accurate. The stated uncertainty (±0.4) suggests reasonable precision for an introductory lab. Choice A incorrectly links accuracy with precision. Choice B incorrectly claims the measurement isn't accurate when it actually is. Choice D makes unwarranted judgments about what constitutes acceptable uncertainty for this measurement.
11. A student measures the acceleration of a cart down an inclined plane by analyzing position vs. time data. She fits a quadratic function $x(t) = x_0 + v_0 t + \frac{1}{2}at^2$ to her data points and obtains $a = 1.25 \pm 0.08 \text{ m/s}^2$. The theoretical prediction based on the incline angle is $a_{theory} = 1.45 \text{ m/s}^2$. Which statement best describes this result?
    Answer: The experimental result disagrees with theory, indicating significant systematic error When analyzing experimental data in physics, you need to distinguish between random errors (which create uncertainty ranges) and systematic errors (which shift your entire measurement away from the true value). This question tests whether you can properly interpret experimental uncertainty. To determine if experimental and theoretical values agree, you check whether the theoretical value falls within the experimental uncertainty range. The experimental result is $1.25 \pm 0.08 \text{ m/s}^2$, meaning the acceptable range is 1.17 to 1.33 m/s². Since the theoretical value of 1.45 m/s² falls well outside this range (about 1.5 standard deviations away), the results disagree significantly. This suggests systematic error—perhaps friction, air resistance, or measurement bias that consistently affects all data points. Option A is incorrect because 1.45 m/s² clearly lies outside the 1.17-1.33 m/s² uncertainty range. Option C misunderstands experimental analysis—an 8% relative uncertainty (0.08/1.25) is quite reasonable for undergraduate experiments and definitely allows meaningful comparison. Option D incorrectly attributes the discrepancy to random errors, but random errors are already accounted for in the ±0.08 uncertainty; they don't explain why the central value differs so much from theory. Study tip: When comparing experimental and theoretical values, always check if the theoretical value falls within the experimental uncertainty range (mean ± uncertainty). If it doesn't, you likely have systematic error, not just random measurement noise.
12. A student makes five measurements of the same quantity and obtains: 12.1, 12.3, 12.0, 12.4, 12.2 (all in cm). What is the best estimate of the uncertainty in the mean value?
    Answer: $\pm 0.07$ cm, calculated as the standard deviation of the mean The uncertainty in the mean is the standard deviation of the mean, calculated as $s/\sqrt{n}$ where $s$ is the sample standard deviation and $n$ is the number of measurements. Mean = 12.2 cm, standard deviation ≈ 0.16 cm, so uncertainty in mean ≈ 0.16/√5 ≈ 0.07 cm. Choice A uses instrument precision rather than statistical uncertainty. Choice C gives the standard deviation of individual measurements, not the mean. Choice D uses the range, which overestimates uncertainty.
13. Two independent measurements yield $a = 8.5 \pm 0.3$ and $b = 3.2 \pm 0.4$. If $c = a - b$, what is the uncertainty in $c$?
    Answer: $\pm 0.5$, calculated as $\sqrt{(0.3)^2 + (0.4)^2}$ For addition or subtraction of independent measurements, absolute uncertainties combine in quadrature: $\delta c = \sqrt{(\delta a)^2 + (\delta b)^2} = \sqrt{(0.3)^2 + (0.4)^2} = \sqrt{0.25} = 0.5$. Choice A incorrectly subtracts uncertainties. Choice C incorrectly adds uncertainties linearly. Choice D incorrectly averages the uncertainties.
14. A student measures the wavelength of light using a double-slit interference pattern. She measures the distance between adjacent bright fringes as $\Delta y = 2.4 \pm 0.1$ mm, the distance to the screen as $L = 1.50 \pm 0.02$ m, and the slit separation as $d = 0.25 \pm 0.01$ mm. Using $\lambda = \frac{\Delta y \cdot d}{L}$, which measurement should be improved to most significantly reduce the uncertainty in wavelength?
    Answer: The slit separation measurement, because it has the largest fractional uncertainty When analyzing uncertainty propagation in physics measurements, you need to focus on fractional (relative) uncertainties rather than absolute uncertainties. The measurement with the largest fractional uncertainty contributes most significantly to the overall uncertainty in your calculated result. Let's calculate the fractional uncertainty for each measurement: - Fringe spacing: $\frac{0.1}{2.4} = 0.042 = 4.2\%$ - Screen distance: $\frac{0.02}{1.50} = 0.013 = 1.3\%$ - Slit separation: $\frac{0.01}{0.25} = 0.040 = 4.0\%$ The slit separation has the largest fractional uncertainty at 4.0%, making answer C correct. Improving this measurement would most significantly reduce the wavelength uncertainty. A is wrong because absolute uncertainty alone doesn't determine impact on the final result. The fringe spacing has the largest absolute uncertainty (0.1 mm), but when considered relative to its measured value, it's not the most problematic. B misunderstands uncertainty propagation. While the screen distance appears in the denominator, its fractional uncertainty (1.3%) is actually the smallest, so improving it would have minimal effect. D is incorrect because measurements don't contribute equally to uncertainty. The fractional uncertainties clearly show different contributions to the overall error. Study tip: Always calculate fractional uncertainties when determining which measurement to improve. Convert each uncertainty to a percentage of its measured value—the largest percentage indicates where your efforts will have the greatest impact on reducing overall uncertainty.
15. A digital multimeter displays voltage readings to the nearest 0.01 V. A student takes a single reading of 3.47 V. According to standard practices for reporting instrumental uncertainty, what uncertainty should be associated with this measurement?
    Answer: $\pm 0.005$ V When dealing with digital instruments, you need to understand how instrumental uncertainty relates to the device's precision. The key principle is that uncertainty reflects the range of possible true values that could produce the displayed reading. Since this multimeter displays to the nearest 0.01 V, any reading of 3.47 V means the true value lies somewhere between 3.465 V and 3.475 V. The instrument rounds to the nearest hundredth, so values from 3.465 V up to (but not including) 3.475 V would all display as 3.47 V. This creates an uncertainty range of $\pm 0.005$ V around the displayed value. Looking at the wrong answers: Choice A ($\pm 0.001$ V) represents one-tenth of the instrument's resolution, which would apply only to instruments with much higher precision. Choice C ($\pm 0.01$ V) equals the full resolution rather than half of it—this is a common misconception where students confuse the resolution with the uncertainty. Choice D ($\pm 0.02$ V) represents twice the resolution, which would be appropriate only if the instrument had poor calibration or additional systematic errors beyond normal rounding uncertainty. The correct answer is B ($\pm 0.005$ V) because standard practice assigns an uncertainty equal to half the smallest division of a digital display. Study tip: For digital instruments, always remember the "half-rule": the uncertainty equals half of the least significant digit's place value. This accounts for the rounding that occurs when the instrument displays readings to a finite number of decimal places.
16. A student measures the spring constant $k$ of a spring by hanging different masses and measuring the displacement. She plots force $F$ versus displacement $x$ and performs a linear regression, obtaining the equation $F = 12.5x + 0.3$ where $F$ is in Newtons and $x$ is in meters. The correlation coefficient is $r = 0.996$. What does the y-intercept of 0.3 N most likely represent?
    Answer: There is a systematic error causing a constant offset in all force measurements When analyzing experimental data with linear regression, you need to interpret what each component of the fitted equation represents physically. Here, the student is testing Hooke's law, which states that $F = kx$ for an ideal spring with no external forces. The equation $F = 12.5x + 0.3$ shows excellent correlation ($r = 0.996$), indicating a strong linear relationship. However, ideal Hooke's law predicts the line should pass through the origin. The y-intercept of 0.3 N means that when displacement is zero, the regression line predicts a force of 0.3 N. Since there should be no spring force at zero displacement, this constant offset most likely represents a systematic error affecting all force measurements equally—perhaps from an incorrectly zeroed force sensor or scale. Answer A incorrectly interprets the y-intercept as uncertainty in the spring constant. The spring constant is the slope (12.5 N/m), and its uncertainty would come from the slope's standard error, not the y-intercept. Answer B suggests pre-compression, but pre-compression would affect the displacement measurements and equilibrium position, not create a constant force offset in the data. Answer D misinterprets the high correlation coefficient. With $r = 0.996$, the spring clearly follows Hooke's law very well—the relationship is highly linear, not nonlinear. Study tip: In physics experiments, when your best-fit line has an unexpected y-intercept that doesn't match theory, always consider systematic measurement errors first. Perfect theory rarely includes constant offsets, so question your experimental setup.
17. Two students measure the same resistor using identical equipment. Student A measures the resistance 5 times and reports $R_A = 47.2 \pm 0.8$ Ω. Student B measures it 20 times and reports $R_B = 47.5 \pm 0.4$ Ω. Assuming both students used proper statistical analysis, what can be concluded about their measurement techniques?
    Answer: Both results are statistically consistent within their combined uncertainties When analyzing experimental data with uncertainties, you need to distinguish between random errors (affecting precision) and systematic errors (affecting accuracy), and understand how to compare results statistically. To determine if two measurements are consistent, you calculate whether their difference is statistically significant. The key is finding if the measured values overlap within their combined uncertainties. Student A reports $47.2 \pm 0.8$ Ω (range: 46.4 to 48.0 Ω) while Student B reports $47.5 \pm 0.4$ Ω (range: 47.1 to 47.9 Ω). These ranges overlap substantially from 47.1 to 47.9 Ω, indicating the measurements are statistically consistent despite the 0.3 Ω difference in their central values. Answer A incorrectly assumes uncertainty size directly indicates random error magnitude. Student A's larger uncertainty could result from fewer measurements rather than inherently less precise technique. Answer B oversimplifies the relationship between number of measurements and reliability—while more measurements typically reduce uncertainty, technique quality matters too. Answer C wrongly concludes the results are inconsistent when they actually overlap within uncertainties. Answer D assumes systematic error without evidence, since the difference between means falls well within the combined uncertainties. The correct conclusion is that both measurements are consistent with each other and likely with the true resistance value, suggesting both students used acceptable techniques despite different numbers of measurements. Study tip: Always check if measurement ranges overlap before concluding that results disagree—statistical consistency depends on uncertainties, not just differences in central values.
18. In measuring the period of oscillation of a simple pendulum, a student times 20 complete oscillations and obtains $t = 25.6 \pm 0.2$ s. She then calculates the period as $T = t/20 = 1.28 \pm 0.01$ s. Later, she realizes she miscounted and actually timed 19 oscillations instead of 20. What is the corrected period and its uncertainty?
    Answer: $T = 1.35 \pm 0.011$ s This question tests your understanding of error propagation in experimental physics, specifically when you need to recalculate both a measured value and its uncertainty after discovering an error in your methodology. When the student discovers she timed 19 oscillations instead of 20, both the period value and its uncertainty must be recalculated. The corrected period is $T = \frac{25.6 \text{ s}}{19} = 1.347 \text{ s} \approx 1.35 \text{ s}$. For the uncertainty, you need to apply error propagation rules. Since $T = \frac{t}{n}$ where $t = 25.6 \pm 0.2$ s and $n = 19$ (exact), the uncertainty is $\delta T = \frac{\delta t}{n} = \frac{0.2}{19} = 0.0105 \text{ s} \approx 0.011 \text{ s}$. Therefore, the corrected result is $T = 1.35 \pm 0.011$ s. Choice A gives the correct period but uses the original uncertainty calculation (0.2/20 = 0.01), failing to account for the changed denominator. Choice C correctly calculates the period but incorrectly propagates the uncertainty, possibly by using a more complex error analysis that doesn't apply here. Choice D keeps the original incorrect period value while attempting to recalculate the uncertainty. When recalculating experimental results after discovering methodological errors, always remember to recalculate both the value and its uncertainty using the corrected parameters. Don't just adjust one while keeping the other from your original calculation.
19. A student wants to minimize the uncertainty in calculating kinetic energy $KE = \frac{1}{2}mv^2$. The mass can be measured with 1% uncertainty, and velocity can be measured with 3% uncertainty. Which strategy would most effectively reduce the overall uncertainty?
    Answer: Improve the velocity measurement to 1.5% uncertainty, since velocity is squared and its uncertainty contribution is doubled in the final result When analyzing uncertainty propagation in physics calculations, you need to consider how each measured quantity's uncertainty affects the final result based on the mathematical relationship in the formula. For kinetic energy $KE = \frac{1}{2}mv^2$, uncertainty propagation follows specific rules. When variables are multiplied, their fractional uncertainties add in quadrature. Crucially, when a variable is raised to a power, its fractional uncertainty gets multiplied by that power. Since velocity is squared in the kinetic energy formula, the velocity's fractional uncertainty contributes twice as much to the total uncertainty as the mass uncertainty. Starting with 1% uncertainty in mass and 3% uncertainty in velocity, the velocity contribution is $2 \times 3\% = 6\%$ to the total uncertainty calculation, while mass contributes only 1%. The velocity measurement dominates the overall uncertainty, making option D correct—improving velocity precision would have the greatest impact on reducing total uncertainty. Option A incorrectly assumes mass contributes equally because it appears "linearly," ignoring the squaring effect on velocity. Option B misunderstands uncertainty propagation by focusing on absolute rather than relative uncertainties and their mathematical relationships. Option C wrongly suggests equal contribution when velocity's squared relationship makes it contribute roughly six times more than mass to the total uncertainty. Remember: in uncertainty propagation, always check for exponents in your formula. Variables raised to powers have their fractional uncertainties multiplied by those powers, often making them the dominant source of uncertainty even if their individual measurement precision seems comparable.
20. A physics laboratory has two identical stopwatches. Stopwatch A has a manufacturer's specification of ±0.01 s accuracy, while Stopwatch B has ±0.02 s accuracy. A student uses both stopwatches simultaneously to time the same event and obtains $t_A = 4.23$ s and $t_B = 4.26$ s. What is the best estimate of the true time and its uncertainty?
    Answer: $t = 4.24 \pm 0.008$ s, using weighted average based on individual uncertainties When you have multiple measurements with different uncertainties, you shouldn't simply average them or use only one measurement. Instead, you need to weight each measurement based on its reliability - measurements with smaller uncertainties should contribute more to your final result. The weighted average formula gives more precise measurements greater influence: $t = \frac{t_A/\sigma_A^2 + t_B/\sigma_B^2}{1/\sigma_A^2 + 1/\sigma_B^2}$, where $\sigma$ represents uncertainty. Calculating the weights: stopwatch A has weight $1/(0.01)^2 = 10,000$ and stopwatch B has weight $1/(0.02)^2 = 2,500$. This gives $t = \frac{4.23 \times 10,000 + 4.26 \times 2,500}{10,000 + 2,500} = 4.24$ s. The combined uncertainty is $\sigma = 1/\sqrt{1/\sigma_A^2 + 1/\sigma_B^2} = 1/\sqrt{10,000 + 2,500} = 0.008$ s. Option A ignores valuable data from the second measurement. While stopwatch A is more accurate, you still gain information by including both measurements. Option B incorrectly treats both measurements equally despite their different precisions, and simply averages their uncertainties. Option D uses a simple average but incorrectly combines uncertainties by adding them directly ($\sqrt{0.01^2 + 0.02^2} = 0.022$), which doesn't account for the fact that multiple measurements should reduce overall uncertainty. Remember: when combining measurements with different uncertainties, always use weighted averaging where weights are inversely proportional to the square of each measurement's uncertainty. More precise measurements deserve more influence in your final result.

Start this quiz