AP Physics 1 · Free-response skill

Test a claim with a timing experiment

The move. Plan a fair test of a claim about an oscillator, measure a period precisely, and decide what the data show by comparing them with their own spread and with the model.

How to do it

  1. Start by naming the variable you will change on purpose and the quantity you will measure, then list anything else that could change the result and what in the setup keeps each of those fixed.
  2. For anything you time by hand, think about where the timing error comes from and how the way you take the measurement can shrink that error’s share of the result.
  3. To judge a claim, set the size of the change in the variable beside the size of the change in the result, using numbers from the data, and say whether the result follows the variable.
  4. Finish by comparing the data with the model: compute the prediction for your setup and say how close the measurements come to it.

Practice prompt Try it

A student hears the claim that a pendulum with a heavier bob swings more slowly. To test it, she uses four solid spheres that all have a diameter of 3.0 cm but are made of different materials. She hangs each sphere in turn from the same light string, so the distance from the pivot to the center of each sphere is 0.85 m. She pulls each sphere aside to a small angle, releases it, and times 20 complete swings with a stopwatch. Use g = 9.8 m/s².

Mass of each sphere and the time for 20 complete swings

SphereMass (kg)Time for 20 swings (s)
Wood0.01037.2
Aluminum0.03836.9
Steel0.11137.1
Brass0.12037.0

Use the description and the data to answer each part.

  1. Which quantity is the student changing on purpose, and which quantity is she measuring to see whether it responds?
  2. The student timed 20 swings rather than 1. Explain how this makes her value for each period more precise.
  3. From the wood sphere to the brass sphere, by what factor does the mass increase?
  4. What is the largest difference between any two of the times for 20 swings in the table?
  5. Considering your answers to parts (c) and (d), do the data support the claim that a heavier bob swings more slowly? Answer yes or no.
  6. Use the simple pendulum model to calculate the period, the time for one complete swing, that it predicts for a length of 0.85 m. Give your answer to three significant figures.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) She is changing the mass of the bob by using spheres of different materials. She is measuring the period, by timing 20 complete swings.

  • Point 1 Names the bob’s mass as the quantity changed on purpose.
  • Point 2 Names the period (through the 20-swing time) as the quantity measured.

(b) Her reaction time when she starts and stops the stopwatch adds roughly the same error whether she times 1 swing or 20. Against a 37 s measurement that error is a much smaller fraction than against a 1.85 s one, so the period she gets by dividing by 20 is about 20 times less affected.

  • Point 3 Says the start-stop error is a much smaller fraction of the longer measured time.

(c) The mass goes from 0.010 kg to 0.120 kg, and 0.120 ÷ 0.010 = 12, so the brass sphere has 12 times the mass of the wood sphere.

  • Point 4 Divides the two masses from the table.

(d) The longest time is 37.2 s (wood) and the shortest is 36.9 s (aluminum), so the largest difference is 37.2 − 36.9 = 0.3 s.

  • Point 5 Subtracts the shortest time from the longest.

(e) No. The mass grows by a factor of 12, but the times for 20 swings differ by only 0.3 s at most, so the swing time does not follow the mass.

  • Point 6 Answers no, and gives the reason from parts (c) and (d).

(f) T = 2π√(L/g) = 2π√(0.85/9.8) = 1.85 s. This is close to the measured periods, which run from 36.9/20 = 1.845 s to 37.2/20 = 1.86 s.

  • Point 7 Uses the small-angle pendulum equation with L = 0.85 m and g = 9.8 m/s², and gives three significant figures.