Oscillations.

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 m

Show circle and energy

Off

Accepted within

± 2%

Computed from the Scene at this step’s setting

How far the mass stretches the spring at rest, mg/k (0 for a pendulum)/nodes/spring/outputs/staticStretch

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.0167 m ± 5%

  Misconception: Uses the mass in kilograms as the force in Hooke's law: x = m/k instead of mg/k.

  Feedback from the catalogue

  In Hooke’s law, $F = kx$⁠, the force is in newtons, and a mass is in kilograms. What stretches the spring is the weight of the mass, $mg$⁠: a 0.5 kg mass pulls with about 4.9 N, not $0.5$⁠. So $x = \dfrac{mg}{k}$⁠, not $\dfrac{m}{k}$⁠.

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.05 m

Show circle and energy

Off

Choices

• Correct It will stay the same

• Wrong It will get longer, because the path is longer

  Misconception: Thinks a bigger swing takes longer, because it has farther to go.

  Feedback from the catalogue

  A bigger swing has farther to go, but it also moves faster: the restoring force grows in proportion to the distance from the middle, so the mass speeds up more. The two effects cancel exactly, and the period of a mass on a spring, $T = 2\pi\sqrt{m/k}$⁠, does not depend on the amplitude. For a pendulum it is almost true: at small angles the period hardly changes, and only at wide angles does it grow a little, by about $14\%$⁠ at $80^\circ$⁠, not several times over.

• Wrong It will get shorter, because the spring pulls harder

  Feedback

  The spring does pull harder, but the mass also has a longer way to go. The two effects cancel exactly.

Once the prediction is locked in, the Scene changes

Pull down

0.2 m

The Scene then shows: 0.8112 s · The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/spring/outputs/period

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Spring stiffness

20 N/mfixed

Mass

0.8 kgfixed

Pull down

0.12 m

Show circle and energy

Off

Other settings

Timer/nodes/spring/props/timer = false

Accepted within

± 2%

Computed from the Scene at this step’s setting

The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/spring/outputs/period

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.251 s ± 5%

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• 31.4 s ± 5%

  Misconception: Turns the ratio under the root upside down: 2π√(k/m) or 2π√(g/L).

  Feedback from the catalogue

  Check the direction with common sense: a heavier mass bounces more slowly, so $m$⁠ goes on top; a stiffer spring bounces faster, so $k$⁠ goes underneath. For a pendulum a long string slows it ($L$⁠ on top) and strong gravity speeds it up ($g$⁠ underneath). You can check with units too: $\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} = \mathrm{s}$$\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} =$$\mathrm{s}$⁠, while the other way up gives $\mathrm{1/s}$⁠, a unit of frequency, not of time.

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.2 kg

Pull down

0.12 m

Show circle and energy

Off

Choices

Shown in a random order.

• Correct It doubles

• Wrong It grows four times

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• Wrong It does not change

  Feedback

  That is true of a pendulum, not of a spring. Here the restoring force comes from the spring and does not grow with the mass, so a heavier mass speeds up more slowly.

• Wrong It halves

  Feedback

  A heavier mass is harder to speed up, so the bounce slows down, not up.

Worked solution

The Scene at this step

Spring stiffness

30 N/m

Mass

0.5 kg

Pull down

0.12 m

Show circle and energy

Off

Read from the Scene at

The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/spring/outputs/period

One setting that reaches it

Spring stiffness

20 N/m

Mass

0.5 kg

Pull down

0.05 m

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 m

Show circle and energy

Off

Choices

• Correct As it passes the equilibrium position, in the middle

• Wrong At the bottom, where the spring pulls hardest

  Misconception: Thinks the mass moves fastest at the bottom of its bounce, where the spring pulls hardest.

  Feedback from the catalogue

  The force sets how fast the velocity changes, not how big it is. At the bottom the spring pulls hardest, but that is where the mass has just stopped and is starting back. All the way to the middle the spring keeps speeding it up, so it is fastest as it passes the equilibrium position, where the net force on it is zero.

• Wrong At the top, after the spring has pushed it up

  Feedback

  At the top the mass stops for an instant and starts down again, so its speed there is zero too.

• Wrong At the same speed all the way

  Feedback

  If the speed were constant, the mass could not stop and turn round at the ends. It speeds up on the way to the middle and slows down after it.

Once the prediction is locked in, the Scene changes

Show circle and energy

On

The Scene then shows: 0.9295 m/s · Speed through the middle of the free motion: ω₀·pull, or √(2gL(1 − cos θ₀))/nodes/spring/outputs/maxSpeed

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 m

Show circle and energy

On

Accepted within

± 2%

Computed from the Scene at this step’s setting

Swings per second, 1 / period/nodes/spring/outputs/frequency

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 7.75 Hz ± 3%

  Misconception: Mixes up the frequency f with the angular frequency ω = 2πf.

  Feedback from the catalogue

  There are two ways to say how fast an oscillation repeats. The frequency $f$⁠ counts whole oscillations per second, in hertz. The angular frequency $\omega$⁠ counts the angle the point on the circle turns through per second, in radians per second. One oscillation is a full turn, $2\pi$⁠ radians, so $\omega = 2\pi f$⁠. The formulas $\omega = \sqrt{k/m}$⁠ and $v_{\max} = A\omega$⁠ use $\omega$⁠, not $f$⁠.

• 0.811 Hz ± 3%

  Misconception: Gives the period as the frequency, or confuses the two.

  Feedback from the catalogue

  The period $T$⁠ is how many seconds one oscillation takes. The frequency $f$⁠ is how many oscillations there are in one second, in hertz ($\mathrm{Hz}$⁠). Each is the inverse of the other: $f = \dfrac{1}{T}$⁠. An oscillation that takes 0.5 s repeats twice a second: 2 Hz.

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 mfixed

Show circle and energy

On

Accepted within

± 2%

Computed from the Scene at this step’s setting

Speed through the middle of the free motion: ω₀·pull, or √(2gL(1 − cos θ₀))/nodes/spring/outputs/maxSpeed

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.148 m/s ± 5%

  Misconception: Mixes up the frequency f with the angular frequency ω = 2πf.

  Feedback from the catalogue

  There are two ways to say how fast an oscillation repeats. The frequency $f$⁠ counts whole oscillations per second, in hertz. The angular frequency $\omega$⁠ counts the angle the point on the circle turns through per second, in radians per second. One oscillation is a full turn, $2\pi$⁠ radians, so $\omega = 2\pi f$⁠. The formulas $\omega = \sqrt{k/m}$⁠ and $v_{\max} = A\omega$⁠ use $\omega$⁠, not $f$⁠.

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 m

Show circle and energy

On

Choices

Shown in a random order.

• Correct At the turning points, top and bottom

• Wrong In the middle, where it moves fastest

  Misconception: Thinks the acceleration is largest where the speed is largest, in the middle.

  Feedback from the catalogue

  The acceleration follows the force, and the spring’s force, measured from the equilibrium position, is $F = -kx$⁠: zero in the middle and largest at the ends. In the middle the mass is fastest, but there its velocity is momentarily not changing, so the acceleration is zero. At the turning points it stops, and that is where its velocity changes fastest.

• Wrong In the middle; at the turning points the acceleration is zero, because the mass stops there

  Misconception: Thinks the acceleration is zero at the turning points, because the mass stops there.

  Feedback from the catalogue

  Zero speed is not zero acceleration. At a turning point the mass stops for a single instant, and there the spring pulls on it hardest, with $kA$⁠. If the acceleration were zero, the mass would stay at the end and never come back. It is like a ball at the top of a vertical throw: it stops for an instant, but $g$⁠ keeps acting.

• Wrong It is the same everywhere

  Feedback

  The spring force changes with the stretch, and so does the acceleration: $a = -\omega^2 x$⁠.

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.12 mfixed

Show circle and energy

On

Accepted within

± 2%

Computed from the Scene at this step’s setting

Energy of the free motion: ½k·pull² about equilibrium, or mgL(1 − cos θ₀)/nodes/spring/outputs/totalEnergy

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.432 J ± 3%

  Misconception: Leaves out the half: takes the spring's energy to be kA².

  Feedback from the catalogue

  The energy stored in a spring is $\tfrac12 kx^2$⁠. The half comes from the force growing from zero to $kA$⁠ while you stretch it, so the average force is only $\tfrac12 kA$⁠, and the work is the average force times the distance: $\tfrac12 kA \cdot A = \tfrac12 kA^2$$\tfrac12 kA \cdot A =$$\tfrac12 kA^2$⁠.

• 1.8 J ± 3%

  Misconception: Thinks the energy grows in proportion to the pull, not to its square.

  Feedback from the catalogue

  The energy is $\tfrac12 kA^2$⁠, so it depends on the square of the amplitude: twice the pull gives four times the energy. The reason: with twice the pull, the average force is doubled and so is the distance, and the work is their product. The top speed only doubles, because $\tfrac12 m v^2$⁠ grows four times when the speed doubles.

Hint

Worked solution

The Scene at this step

Spring stiffness

30 N/mfixed

Mass

0.5 kgfixed

Pull down

0.06 m

Show circle and energy

On

Choices

Shown in a random order. More than one choice is correct: every one must be selected.

• Correct The top speed

• Correct The acceleration at the turning points

• Wrong The total energy

  Misconception: Thinks the energy grows in proportion to the pull, not to its square.

  Feedback from the catalogue

  The energy is $\tfrac12 kA^2$⁠, so it depends on the square of the amplitude: twice the pull gives four times the energy. The reason: with twice the pull, the average force is doubled and so is the distance, and the work is their product. The top speed only doubles, because $\tfrac12 m v^2$⁠ grows four times when the speed doubles.

• Wrong The period

  Misconception: Thinks a bigger swing takes longer, because it has farther to go.

  Feedback from the catalogue

  A bigger swing has farther to go, but it also moves faster: the restoring force grows in proportion to the distance from the middle, so the mass speeds up more. The two effects cancel exactly, and the period of a mass on a spring, $T = 2\pi\sqrt{m/k}$⁠, does not depend on the amplitude. For a pendulum it is almost true: at small angles the period hardly changes, and only at wide angles does it grow a little, by about $14\%$⁠ at $80^\circ$⁠, not several times over.

Worked solution

The Scene at this step

String length

0.25 m

Gravity

9.81 m/s²fixed

Release angle

10 °fixed

Show energy and trace

Off

Choices

• Correct It will double

• Wrong It will grow four times

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• Wrong It will not change

  Feedback

  A longer string makes a longer arc, and gravity pulls along it just as hard, so the bob needs more time.

• Wrong It will halve

  Feedback

  A long string makes a longer arc with the same restoring force, so the swing slows down, not up.

Once the prediction is locked in, the Scene changes

String length

1 m

The Scene then shows: 2.01 s · The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

String length

1.5 mfixed

Gravity

9.81 m/s²fixed

Release angle

10 °fixed

Show energy and trace

Off

Other settings

Timer/nodes/pendulum/props/timer = false

Accepted within

± 2%

Computed from the Scene at this step’s setting

The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.961 s ± 5%

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• 16.07 s ± 5%

  Misconception: Turns the ratio under the root upside down: 2π√(k/m) or 2π√(g/L).

  Feedback from the catalogue

  Check the direction with common sense: a heavier mass bounces more slowly, so $m$⁠ goes on top; a stiffer spring bounces faster, so $k$⁠ goes underneath. For a pendulum a long string slows it ($L$⁠ on top) and strong gravity speeds it up ($g$⁠ underneath). You can check with units too: $\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} = \mathrm{s}$$\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} =$$\mathrm{s}$⁠, while the other way up gives $\mathrm{1/s}$⁠, a unit of frequency, not of time.

Hint

Worked solution

The Scene at this step

String length

1 mfixed

Gravity

9.81 m/s²fixed

Release angle

10 °fixed

Show energy and trace

Off

Choices

• Correct It will not change

• Wrong It will get longer, because a heavy bob is harder to move

  Misconception: Thinks a heavier pendulum bob swings more slowly.

  Feedback from the catalogue

  Gravity does pull harder on a heavier bob, but it is also harder to speed up, in exactly the same proportion, as in free fall. So the mass cancels, and a pendulum’s period, $T = 2\pi\sqrt{L/g}$⁠, does not depend on it. A spring is different: there the restoring force comes from the spring, not from gravity, so a heavier mass does slow the bounce.

• Wrong It will get shorter, because gravity pulls it harder

  Feedback

  Gravity pulls four times as hard, but the mass to be sped up is four times as big. The acceleration, and the period with it, do not change.

Once the prediction is locked in, the Scene changes

Other settings

Mass/nodes/pendulum/props/mass = 2

The Scene then shows: 2.01 s · The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

String length

1 mfixed

Gravity

1.62 m/s²fixed

Release angle

10 °fixed

Show energy and trace

Off

Other settings

Timer/nodes/pendulum/props/timer = false

Accepted within

± 2%

Computed from the Scene at this step’s setting

The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 12.17 s ± 5%

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• 3.88 s ± 3%

  Misconception: Leaves out the square root: takes the period to grow in proportion to the mass or the length, or to 1/k or 1/g.

  Feedback from the catalogue

  The period depends on the square root of the ratio: $T = 2\pi\sqrt{\dfrac{m}{k}}$$T =$$2\pi\sqrt{\dfrac{m}{k}}$⁠ for a spring and $T = 2\pi\sqrt{\dfrac{L}{g}}$$T =$$2\pi\sqrt{\dfrac{L}{g}}$⁠ for a pendulum. So four times the mass, or a string four times as long, gives a period only $\sqrt{4} = 2$⁠ times as long, and gravity six times weaker lengthens it $\sqrt{6} \approx 2.45$$\sqrt{6} \approx$$2.45$⁠ times, not six.

• 8 s ± 3%

  Misconception: Turns the ratio under the root upside down: 2π√(k/m) or 2π√(g/L).

  Feedback from the catalogue

  Check the direction with common sense: a heavier mass bounces more slowly, so $m$⁠ goes on top; a stiffer spring bounces faster, so $k$⁠ goes underneath. For a pendulum a long string slows it ($L$⁠ on top) and strong gravity speeds it up ($g$⁠ underneath). You can check with units too: $\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} = \mathrm{s}$$\sqrt{\dfrac{\mathrm{kg}}{\mathrm{N/m}}} =$$\mathrm{s}$⁠, while the other way up gives $\mathrm{1/s}$⁠, a unit of frequency, not of time.

Hint

Worked solution

The Scene at this step

String length

1 m

Gravity

1.62 m/s²fixed

Release angle

10 °fixed

Show energy and trace

Off

Read from the Scene at

The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

One setting that reaches it

String length

0.65 m

Hint

Worked solution

The Scene at this step

String length

1 mfixed

Gravity

9.81 m/s²fixed

Release angle

10 °

Show energy and trace

Off

Choices

• Correct It will get a little longer, by less than a fifth

• Wrong It will get about $8$⁠ times longer, because the arc is $8$⁠ times longer

  Misconception: Thinks a bigger swing takes longer, because it has farther to go.

  Feedback from the catalogue

  A bigger swing has farther to go, but it also moves faster: the restoring force grows in proportion to the distance from the middle, so the mass speeds up more. The two effects cancel exactly, and the period of a mass on a spring, $T = 2\pi\sqrt{m/k}$⁠, does not depend on the amplitude. For a pendulum it is almost true: at small angles the period hardly changes, and only at wide angles does it grow a little, by about $14\%$⁠ at $80^\circ$⁠, not several times over.

• Wrong It will not change at all: the period does not depend on the angle

  Feedback

  That is true only approximately, at small angles. The formula $2\pi\sqrt{L/g}$⁠ assumes $\sin\theta \approx \theta$$\sin\theta \approx$$\theta$⁠, which no longer holds at $80^\circ$⁠.

• Wrong It will get shorter

  Feedback

  At a wide angle the force along the arc, $mg\sin\theta$⁠, is weaker than a harmonic pendulum’s would be, so the bob is slower, not faster.

Once the prediction is locked in, the Scene changes

Release angle

80 °

The Scene then shows: 2.282 s · The free period the timer shows: 2π√(m/k), or a pendulum's exact period at its release angle/nodes/pendulum/outputs/period

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Push rate

0.25 Hz

Friction

0.03 damping

Hanging from

A stringfixed

Show energy and trace

Off

Accepted within

± 2%

Computed from the Scene at this step’s setting

Swings per second, 1 / period/nodes/swing/outputs/frequency

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 3.13 Hz ± 3%

  Misconception: Mixes up the frequency f with the angular frequency ω = 2πf.

  Feedback from the catalogue

  There are two ways to say how fast an oscillation repeats. The frequency $f$⁠ counts whole oscillations per second, in hertz. The angular frequency $\omega$⁠ counts the angle the point on the circle turns through per second, in radians per second. One oscillation is a full turn, $2\pi$⁠ radians, so $\omega = 2\pi f$⁠. The formulas $\omega = \sqrt{k/m}$⁠ and $v_{\max} = A\omega$⁠ use $\omega$⁠, not $f$⁠.

• 2.01 Hz ± 3%

  Misconception: Gives the period as the frequency, or confuses the two.

  Feedback from the catalogue

  The period $T$⁠ is how many seconds one oscillation takes. The frequency $f$⁠ is how many oscillations there are in one second, in hertz ($\mathrm{Hz}$⁠). Each is the inverse of the other: $f = \dfrac{1}{T}$⁠. An oscillation that takes 0.5 s repeats twice a second: 2 Hz.

Hint

Worked solution

The Scene at this step

Push rate

0.25 Hz

Friction

0.03 dampingfixed

Hanging from

A stringfixed

Show energy and trace

Off

Choices

• Wrong 0.25 Hz, a push every four seconds

  Feedback

  Pushes that come too slowly drift in and out of step with the swing, and some of them brake it. The pendulum mostly just moves along with the motor.

• Correct 0.50 Hz, the same rhythm as the pendulum

• Wrong 1.0 Hz, twice as many pushes

  Misconception: Thinks pushing more often always builds a bigger swing.

  Feedback from the catalogue

  A push adds energy only when it comes in the direction the object is already moving. Pushes that come too fast also arrive while the bob is coming back towards them, and then they brake it, so on average they give almost nothing. The biggest swing builds when the push rate equals the natural frequency.

Once the prediction is locked in, the Scene changes

Push rate

0.5 Hz

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Push rate

0.5 Hzfixed

Friction

0.03 dampingfixed

Hanging from

A springfixed

Show energy and trace

Off

Accepted within

± 3%

Computed from the Scene at this step’s setting

The size a driven swing settles to, from small-swing theory: the mass's bounce, or the bob's swing along its arc (0 when not driven)/nodes/swing/outputs/steadyAmplitude

Any unit of the same quantity is accepted; it is converted before checking.

Wrong answers it recognises

• 0.03 m ± 10%

  Misconception: Thinks the steady swing is as big as the motor's stroke.

  Feedback from the catalogue

  At the right rhythm every push adds a little more energy, and the swing grows far beyond the motor’s stroke, until the energy lost to friction in each cycle equals the energy the motor puts in. With a damping ratio $\zeta$⁠ that happens when the swing is $\dfrac{1}{2\zeta}$⁠ times the stroke. The swing equals the stroke only when you push very slowly, and the mass simply moves along with the motor.