Time the swing — answer key

1. SQ6 · Practice

Model answer. From the marker moving right until it next passes the marker moving right. One oscillation goes from the marker moving right, out to the right, back through the marker, out to the left and back to the marker moving right again. The marker is at the lowest point because the bob moves fastest there and passes it in a flash, so the moment is easy to judge; at the ends of the swing the bob slows down and turns, so it is hard to tell exactly when it stops.

A full answer includes

• I chose: from the marker moving right until it next passes the marker moving right.
• I said the bob moves fastest at the lowest point, so the moment it passes is easiest to judge; at the ends it slows and stops.

Also accept

• From the marker moving right back to the marker moving right; the bob is fastest at the lowest point, so the moment is clearest.

Watch for

• “From the marker to the next time it passes the marker, whichever way it moves.” — That is only half an oscillation: the bob passes the marker twice in each complete swing, once each way.
• “From one end of the swing to the other end.” — That is half an oscillation too. A complete one comes back to the start, moving the same way.

2. SQ7 · Practice

Model answer. Period = 28.40 s ÷ 20 = 1.42 s.

Checked automatically

• Period of the pendulum: 1.42 s (3 significant figures)
  • Common wrong answer 28.4 : That is the time for 20 oscillations. Divide by 20 for one.
  • Common wrong answer 568 : You multiplied by 20. One oscillation takes less time than twenty: divide.
  • Common wrong answer 0.704 : You divided 20 by 28.40. Divide the time by the number of oscillations.

A full answer includes

• I divided the time by the number of oscillations: 28.40 s ÷ 20 = 1.42 s.

Also accept

• 1.42 s

Watch for

• “The period is 28.40 s.” — 28.40 s is the time for 20 oscillations. The period is the time for one: divide by 20.
• “Multiply by 20: 568 s.” — One oscillation takes less time than twenty, so divide by 20.

3. SQ12 · Practice

Model answer. Twenty swings of a pendulum. They take about 25 s, so an error of about 0.2 s from her reactions is less than 1 % of the reading. A ball falling 0.5 m takes about 0.3 s, a single bounce of a ball on a table a fraction of a second and a camera flash far less: each is about as short as a reaction time or shorter, so a hand-timed reading would be mostly error. Those need light gates or an electronic timer.

A full answer includes

• I chose twenty swings of a pendulum: they take about 25 s, so a 0.2 s reaction error is less than 1 % of it.
• I explained that the other events last about as long as a reaction time or less, so a hand-timed reading would be mostly reaction error.

Also accept

• Twenty swings of a pendulum; they last much longer than a reaction time.

Watch for

• “The stop-clock shows hundredths, so it can time anything to 0.01 s.” — The display shows 0.01 s, but the person pressing it is about 0.2 s slow, and by a different amount each time. That limits what a hand-held clock can time.

4. SQ13 · Practice

Model answer. length / m · time for 20 oscillations / s · period / s. The length then goes in the table as 0.250, the time as 20.1 and the period as 1.01 (the same precision for every row).

A full answer includes

• I wrote three headings with quantity / unit: length / m, time for 20 oscillations / s, period / s.

Also accept

• length / m; time for 20 oscillations / s (or t / s); period / s (or T / s).

Watch for

• “Put the unit next to each number in the table: “20.1 s”.” — The unit belongs in the heading, after a slash. The numbers in the column are then plain numbers.
• “Head the column “length (25 cm)”.” — A heading names a quantity and its unit, not a value. Write “length / m” and put 0.250 in the column.

5. SQ8 · Practice

Model answer. Time 20 oscillations and divide, time at a marker at the lowest point, and repeat and take the mean. Timing 20 oscillations shares the reaction error over 20 swings; a marker at the lowest point marks a moment that is easy to judge because the bob moves fastest there; repeating and averaging reduces random errors (the press-to-press spread). Counting “one” as the clock starts times only 19 swings, and starting at the end of the swing makes the moment harder to judge.

A full answer includes

• I chose: time 20 oscillations and divide by 20; start and stop as the bob passes a marker at its lowest point; repeat the timing and take the mean.
• I did not choose “Say ‘one’ as you start the clock” (that gives only 19 swings) or “Start and stop the clock at the end of the swing” (there the moment is harder to judge).

Also accept

• Twenty oscillations ÷ 20, a marker at the lowest point, repeat and average.

Watch for

• “Start counting at one when you start the clock.” — Then you stop after 19 complete oscillations. The pass when you start is zero.
• “Start at the end of the swing, where the bob is still for a moment.” — The bob slows down gradually at the end, so it is hard to tell the exact moment it stops. At the lowest point it passes quickly and clearly.

6. SQ9 · Practice

Model answer. (a) Mean = (25.32 + 25.48 + 25.40) s ÷ 3 = 76.20 s ÷ 3 = 25.40 s. (b) Period = 25.40 s ÷ 20 = 1.27 s.

Checked automatically

• (a) Mean time for 20 oscillations: 25.40 s (4 significant figures)
  • Common wrong answer 76.2 : That is the total of the three times. Divide it by 3 for the mean.
• (b) Period: 1.27 s (3 significant figures)
  • Common wrong answer 25.4 : That is the mean time for 20 oscillations. Divide by 20 for one.
  • Common wrong answer 3.81 : You divided the total of the three runs by 20. They cover 60 oscillations: use the mean, then divide by 20.

A full answer includes

• (a) I found the mean: (25.32 + 25.48 + 25.40) s ÷ 3 = 25.40 s.
• (b) I divided by 20: 25.40 s ÷ 20 = 1.27 s.

Also accept

• (a) 25.40 s; (b) 1.27 s

Watch for

• “The period is the mean time, 25.40 s.” — 25.40 s is still the time for 20 oscillations. Divide by 20 for the period.
• “Divide the total 76.20 s by 20.” — 76.20 s covers 60 oscillations (three runs of 20). Take the mean first, or divide the total by 60.

7. SQ10 · Practice

Model answer. The pendulum swings with the same period each time, so the spread comes from the person timing: their reaction time, about 0.2 s, at the start and at the stop, is a little different every time. That is a large part of a swing of about 1.3 s. Timing 20 swings still has one reaction error at each end, but it is shared out over 20 swings when you divide, so the period is about 20 times more precise.

A full answer includes

• I explained that the readings differ because of the timer’s reaction time at the start and the stop, which is about 0.2 s and changes from press to press.
• I explained that over 20 swings the same size of error is shared out, so the error in one period is about 20 times smaller.

Also accept

• Reaction time varies (about 0.2 s), a big fraction of one swing; dividing a 20-swing time shares the error out.

Watch for

• “The pendulum swings faster or slower each time.” — For small swings the period stays the same. The spread comes from the timing, not from the pendulum.
• “Timing 20 swings removes the reaction time.” — The reaction error is still there at the start and at the stop, but dividing by 20 makes it 20 times smaller per swing.

8. SQ14 · Practice

Model answer. (a) Length along the bottom (for example 0 to 0.5 m, 0.1 m per large square) and period up the side (0 to 1.6 s, 0.2 s per large square); four small crosses. (b) The points lie on a curve that rises less and less steeply, so draw one smooth curve through them. (c) From 0.300 m up to the curve and across: about 1.10 s.

A full answer includes

• (a) I put length / m on the horizontal axis and period / s on the vertical axis, with equal steps and the points spread over more than half the grid, plotted as small crosses.
• (b) I drew one smooth curve through or close to all the points, not a straight line and not dot-to-dot.
• (c) I read off about 1.1 s for 0.300 m (anything from 1.07 to 1.13 s).

Also accept

• (c) 1.07–1.13 s from a smooth curve.

Watch for

• “Join the points with straight lines, dot to dot.” — Draw one smooth best-fit curve. Dot-to-dot lines suggest the pattern changes suddenly at each reading.
• “Rule a straight line through the origin.” — These points bend: the period grows less steeply as the length grows. Draw the curve the points follow.

9. ST2 · Apply

Options as numbered on the pupil sheet: 1 — A clock or watch with a second hand; 2 — A hand-held stop-clock; 3 — Light gates and an electronic timer.

Model answer. Kettle boiling (minutes): a clock with a second hand. 100 m race (about 15 s): a stop-clock. Steel ball falling 0.50 m: light gates and an electronic timer, because the fall takes only about 0.3 s, about the same as a reaction time; a hand-held reading would be mostly reaction error, while a light gate starts and stops with no human delay.

A full answer includes

• I matched: kettle → clock or watch with a second hand; 100 m race → stop-clock; steel ball falling 0.50 m → light gates and an electronic timer (each timer used once).
• I explained that the ball falls in about 0.3 s, about the same as a reaction time, so a hand-held reading would be mostly reaction error.

Also accept

• kettle–clock, race–stop-clock, ball–light gates; the fall is as short as a reaction time.

Watch for

• “A stop-clock is best for everything because it reads to 0.01 s.” — The display reads 0.01 s, but a person can’t press it that precisely: the reaction time is about 0.2 s.
• “Light gates are best for the kettle because they are most accurate.” — For minutes-long times a fraction of a second doesn’t matter, and nothing passes a light gate. A clock is the sensible choice.

10. ST3 · Apply

Model answer. (a) Counting the starting drip as zero, she timed 50 gaps: 31.0 s ÷ 50 = 0.620 s. (b) For example: time for 50 drips / s (three repeats) and time between drips / s.

A full answer includes

• (a) I divided by the 50 gaps: 31.0 s ÷ 50 = 0.620 s.
• (b) I wrote headings with quantity / unit, such as “time for 50 drips / s” and “time between drips / s”.

Also accept

• (a) 0.620 s (0.62 s); (b) quantity / unit headings in seconds.

Watch for

• “Divide by 51, because 51 drips fell.” — Fifty-one drips make 50 gaps, and she counted the first drip as zero. The time between drips is 31.0 s ÷ 50.
• “Divide by 49.” — She counted from zero to 50, so she timed 50 gaps, not 49.