| Date | May 2012 | Marks available | 5 | Reference code | 12M.2.SL.TZ2.1 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine and Draw | Question number | 1 | Adapted from | N/A |
Question
Data analysis question.
Metal girders are often used in buildings that have been constructed to withstand earthquakes. To aid the design of these buildings, experiments are undertaken to measure how the natural frequency ƒ of horizontal oscillations of metal girders varies with their dimensions. In an experiment, ƒ was measured for vertically supported girders of the same cross-sectional area but with different heights h.
The graph shows the plotted data for this experiment. Uncertainties in the data are not shown.
Draw a best-fit line for the data.
It is hypothesized that the frequency ƒ is inversely proportional to the height h.
By choosing two well separated points on the best-fit line that you have drawn in (a), show that this hypothesis is incorrect.
Another suggestion is that the relationship between ƒ and h is of the form shown below, where k is a constant.
\[f = \frac{k}{{{h^2}}}\]
The graph shows a plot of ƒ against h−2.
The uncertainties in h−2 are too small to be shown.
(i) Draw a best-fit line for the data that supports the relationship \(f = \frac{k}{{{h^2}}}\).
(ii) Determine, using the graph, the constant k.
State one reason why the results of the experiment could not be used to predict the natural frequency of oscillation for girders of height 50 m.
Markscheme
Do not allow curve to “curl round” at low or high h.
Single “non-hairy” line only is acceptable.
choice of points separated by (Δh ≥ 7.5) e.g. [6.0, 37] [15, 6.0];
recognize ƒh = constant for an inverse relation;
calculates ƒh correctly for both points;
state that two calculated numbers are not equal (therefore not inverse);
Award [3 max] if data points are not on line.
Award [3 max] if data points are too close together (Δh ≥ 7.5).
Award [2 max] if both of above.
(i) a straight-line that goes through all the error bars; and drawn through the origin; (allow ±½ square)
(ii) read-off of suitable point(s) on line separated by at least half of drawn line;
(allow implicit use of origin)
calculation of gradient to give 1.5(±0.2)×103;
s-1m2 or Hzm2;
the relation might not hold/extrapolate for larger values of h / outside range of experiment / values would be close to origin and with large (percentage experimental) error / girders of this height could buckle under their own weight / OWTTE;
Examiners report
Candidates were required to draw a smooth curve through a series of points. Few could do this adequately and it was rare to see a good construction. Lines were usually point-to-point, kinked in some way, or “hairy” (meaning that the candidate had a number of separate attempts to draw the line). This is evidently not a skill that all candidates possess.
There is still a widespread misconception that when the question asks for the candidate to “draw a best-fit line” this implies that the line is straight. This error is seen in work representing all the IB languages and is a simple point that traps candidates year after year.
The question required a test of inverse proportionality. Examiners were expecting candidates to show that fh was not a constant for two well separated points. Only about 75% the cohort could manage this. Many tried to show that f/h was constant and gained little credit other than for choosing two well separated data points.
(i) This part required a straight line going through all the error bars. Here candidates made good attempts. A common error was to fail to draw the line through the origin.
(ii) It should have been a simple matter to determine the gradient of this graph with its intercept at the origin. Many candidates missed the 10-3 in the axis scaling and went on to omit the unit from their answer. A widespread failure to add units to a gradient calculation has been a feature of several recent paper 2 examinations.
Candidates understood the dangers of extrapolation but could not express them well.
