Best-fit lines, gradient and intercept
Adjust lines for illustrative pendulum, spring-length and cell data. Compare limiting gradients under fixed-x, vertical-allowance assumptions and interpret coefficients with units.
Learning goals
- Use SI quantities, units, prefixes and dimensional analysis.
Pendulum: T² against l. Small-angle pendulum: T² = (4π²/g)l; l is pivot-to-centre-of-mass length. A positive gradient is required. Fixed x values; separate vertical allowances; free intercept. These bounds do not include x uncertainty or shared calibration errors and are not statistical confidence limits. Your solid line: gradient 2.82 s^2 m^{−1}, intercept 0.998 s^2. Actual illustrative observations (x, y ± allowance): 0.200 m, 0.8197 s^2 ± 0.0541 s^2; 0.350 m, 1.390 s^2 ± 0.0723 s^2; 0.500 m, 2.001 s^2 ± 0.0904 s^2; 0.650 m, 2.655 s^2 ± 0.108 s^2; 0.800 m, 3.227 s^2 ± 0.127 s^2; 0.950 m, 3.823 s^2 ± 0.145 s^2.
- Gradient of your line
- 2.82 s2 m−1
- y-intercept of your line
- 0.998 s2
- Gradient of the limiting-line trial
- 4.44 s2 m−1
- Gradient allowance covering both limits
- —
- Allowed gradient range
- 3.740 s2 m−1 to 4.270 s2 m−1
- Limiting-line trial
- Misses at least one vertical allowance
- Model, units and fit condition
- Small-angle pendulum: T² = (4π²/g)l; l is pivot-to-centre-of-mass length. A positive gradient is required. The solid line misses at least one vertical allowance; its gradient allowance is unavailable.
- Illustrative data: x, y ± vertical allowance
- 0.200 m: 0.8197 s2 ± 0.0541 s2; 0.350 m: 1.390 s2 ± 0.0723 s2; 0.500 m: 2.001 s2 ± 0.0904 s2; 0.650 m: 2.655 s2 ± 0.108 s2; 0.800 m: 3.227 s2 ± 0.127 s2; 0.950 m: 3.823 s2 ± 0.145 s2
- From your line
- g = 14.0 m s⁻²
Try this
0 of 4 doneDrag the ends of the line until it is a good best fit. (not done yet)
Judge the overall trend and residuals. Here a compatible best-fit line also crosses every vertical allowance; it need not pass through any point or the origin. The dotted reference is unweighted least squares, not an uncertainty-weighted fit.
Bring the trial line close to either limiting gradient while crossing every vertical allowance. (not done yet)
For fixed x values and a free intercept, the allowed range is calculated using every vertical allowance. The task accepts a line within 5% of that slope span from either limit. A symmetric allowance around a compatible best fit must cover the farther limit; these are not confidence limits.
Use the pendulum data to find g within 2% of 9.81 m s⁻². (not done yet)
T² = (4π² / g) l, so g = 4π² / gradient. A large gradient triangle reduces the error in reading it.
Use the cell data to find the e.m.f. from the y-intercept, within 0.03 V. (not done yet)
V = E − I r: the intercept on the V-axis (I = 0) is the e.m.f. and the gradient is −r.
You’ve tried every task.