IB · A · data analysis
Space, time and motion · Question 7
Space, time and motion · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 12
- Topics
- 1
- Answer
- Complete
A student determines the acceleration of free fall using a simple pendulum. She measures the period T for several lengths L and processes the data as shown. Theory predicts T = 2π√(L/g).
| L / m | 0.400 | 0.600 | 0.800 | 1.000 | 1.200 |
|---|---|---|---|---|---|
| T / s | 1.269 | 1.554 | 1.794 | 2.006 | 2.198 |
| T² / s² | 1.610 | 2.415 | 3.219 | 4.024 | 4.829 |
- (a)
Show (that) Show that a graph of T² against L should be a straight line through the origin, and state an expression for its gradient.
3 marks - (b)
Determine Determine the gradient of the line, and hence determine a value for g.
3 marks - (c)
Outline The uncertainty in each value of L is ±0.005 m and the uncertainty in each value of T is ±0.002 s. Outline why the uncertainty in T² is not ±0.002 s², and determine the uncertainty in T² for the shortest pendulum.
3 marks - (d)
Evaluate The student's line of best fit has a small positive intercept on the T² axis rather than passing through the origin. Evaluate two possible reasons for this.
3 marks
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(a)
Show (that) Show that a graph of T² against L should be a straight line through the origin, and state an expression for its gradient.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
squares the relationship: T² = 4π²L/g
- 2
this is of the form y = mx with no constant term, so the line passes through the origin
- 3
Comparing T² = (4π²/g)L with y = mx gives gradient = 4π²/g.
(b)
Determine Determine the gradient of the line, and hence determine a value for g.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
gradient = (4.829 − 1.610) / (1.200 − 0.400) = 3.219 / 0.800
- 2
gradient = 4.02 s² m⁻¹
- 3
g = 4π²/4.02 = 9.8 m s⁻²
(c)
Outline The uncertainty in each value of L is ±0.005 m and the uncertainty in each value of T is ±0.002 s. Outline why the uncertainty in T² is not ±0.002 s², and determine the uncertainty in T² for the shortest pendulum.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
squaring a quantity doubles its fractional (percentage) uncertainty, so the absolute uncertainty is not carried through unchanged
- 2
fractional uncertainty in T = 0.002/1.269 = 0.16%, so fractional uncertainty in T² = 0.32%
- 3
uncertainty in T² = 0.0032 × 1.610 = ±0.005 s²
(d)
Evaluate The student's line of best fit has a small positive intercept on the T² axis rather than passing through the origin. Evaluate two possible reasons for this.
Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.
- 1
a systematic error in the measurement of L — for example measuring to the top of the bob rather than to its centre, so every length is recorded too short
- 2
a systematic error in timing — for example consistently starting the stopwatch late, though this would need to affect longer pendulums proportionately less to produce an intercept rather than a change of gradient
- 3
evaluates which is more likely, noting that a constant offset in L produces exactly a constant intercept and is therefore the better explanation
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