C.1 Simple Harmonic Motion Questions

1C-1A-01

SHM: period & frequency·C.1 Simple harmonic motion

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksCalculate

An object oscillating with simple harmonic motion completes one full oscillation in 0.40 s.

What is its frequency?

Show mark scheme

Marking point	Mark	Notes
Step 1f = 1/T = 1/0.40 = 2.5 Hz.	✓ 1	Answer C

Answer: C · 1 stage of work, one mark

Every option, and why

AThis is T/2π: the period has been treated as if it were ω and converted with f = ω/2π.
BThis is the period itself. The frequency is the reciprocal of the period, not the period.
CCorrect: f = 1/T = 1/0.40 = 2.5 Hz.
DThis is the angular frequency ω = 2π/T in rad s⁻¹, not the frequency in Hz.

Syllabus understandingC.1 — the time period T as given by T = 1/f = 2π/ω Command term: Calculate

2C-1A-02

The defining equation a = −ω²x·C.1 Simple harmonic motion

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

A body oscillates with simple harmonic motion about its equilibrium position. Its displacement from the equilibrium position is x and its acceleration is a.

Which graph shows the variation of a with x during one complete oscillation?

Sketch graphs, not drawn to scale

Show mark scheme

Marking point	Mark	Notes
Step 1a = −ω²x, so a is directly proportional to x: the graph is a straight line through the origin.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2The minus sign means a always points opposite to x (towards equilibrium), so the gradient, −ω², is negative: graph A.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: a straight line through the origin with negative gradient −ω², as a = −ω²x requires.
BThe minus sign has been ignored: an acceleration in the same direction as the displacement would drive the body further from equilibrium.
CThis acceleration is always directed towards equilibrium but has a constant size. SHM also needs a to be proportional to x.
DThis confuses acceleration with speed: the speed is greatest at equilibrium, but the acceleration there is zero and is greatest at the extremes.

Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x Command term: Identify

3C-1A-03

Conditions for SHM·C.1 Simple harmonic motion

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

In which of the following is the motion simple harmonic?

Show mark scheme

Marking point	Mark	Notes
Step 1SHM requires an acceleration proportional to the displacement from a fixed equilibrium position and always directed towards that position.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Only the trolley meets this: each spring obeys Hooke's law, so the resultant force is proportional to the displacement and opposite to it.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: the two springs give a resultant force proportional to the displacement and directed towards the equilibrium position.
BThe motion is periodic, but between the walls there is no force at all, so there is no acceleration proportional to the displacement.
CAt 80° the restoring force is far from proportional to the angular displacement; a pendulum is simple harmonic only for small angles.
DThe motion is periodic, but between bounces the acceleration is the constant g, which does not depend on the displacement.

Syllabus understandingC.1 — conditions that lead to simple harmonic motion Command term: Deduce

4C-1A-04

The simple pendulum·C.1 Simple harmonic motion

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

The length of a simple pendulum is reduced to one quarter of its original value. The amplitude remains small.

What is (frequency after the change)/(frequency before the change)?

Show mark scheme

Marking point	Mark	Notes
Step 1T = 2π√(l/g), so T ∝ √l: the period is multiplied by √(1/4) = 1/2.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2f = 1/T, so the frequency is multiplied by 2.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThis treats the frequency as proportional to the length: both the square root and the reciprocal have been left out.
BThis is the factor for the period. The frequency is the reciprocal of the period, so it changes by the inverse factor.
CCorrect: the period halves because T ∝ √l, so the frequency doubles.
DThe reciprocal has been taken but not the square root: f ∝ 1/√l, not 1/l.

Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g) Command term: Determine

5C-1A-05

Energy in SHM·C.1 Simple harmonic motion

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksState

The bob of a simple pendulum is released from rest at one end of its swing and moves to the other end. Air resistance is negligible.

Which row describes the changes in the kinetic energy and in the gravitational potential energy of the bob during this motion?

Kinetic energyGravitational potential energy

Show mark scheme

Marking point	Mark	Notes
Step 1The bob is momentarily at rest at the highest points (each end of the swing) and moves fastest at the lowest point, so the kinetic energy rises then falls while the gravitational potential energy falls then rises.	✓ 1	Answer A

Answer: A · 1 stage of work, one mark

Every option, and why

ACorrect: kinetic energy is greatest at the lowest point and gravitational potential energy is greatest at the ends, with a constant sum.
BBoth energies cannot peak together. With no air resistance their sum is constant, so one rises while the other falls.
CBoth energies cannot fall together either: the bob speeds up as it descends, because gravitational potential energy is converted to kinetic energy.
DThe two positions are the wrong way round: at the ends of the swing the bob is at rest, so its kinetic energy is zero there, not a maximum.

Syllabus understandingC.1 — a qualitative approach to energy changes during one cycle of an oscillation Command term: State

6C-1A-06

Maximum speed & acceleration·C.1 Simple harmonic motion

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A particle oscillates with simple harmonic motion of amplitude x₀ and period T.

What is the magnitude of the maximum acceleration of the particle?

Show mark scheme

Marking point	Mark	Notes
Step 1The magnitude of a = −ω²x is greatest at maximum displacement, where x = x₀.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2With ω = 2π/T: a_max = ω²x₀ = 4π²x₀/T².	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThis is ωx₀: ω has not been squared, and the expression has the unit of a speed (m s⁻¹), not an acceleration.
BOnly the period has been squared. ω² = (2π/T)², so the factor 2π must be squared as well.
CCorrect: a_max = ω²x₀ = (2π/T)²x₀ = 4π²x₀/T².
DThe amplitude has been squared too. The acceleration is proportional to the displacement, not to its square.

Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x, with T = 1/f = 2π/ω Command term: Determine

7C-1A-07

The mass–spring oscillator·C.1 Simple harmonic motion

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A mass on a spring oscillates with period T. The mass is doubled and the spring is replaced by one with a spring constant four times as large.

What is the new period?

Show mark scheme

Marking point	Mark	Notes
Step 1T = 2π√(m/k), so T ∝ √(m/k).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2m/k becomes 2/4 = 1/2 of its original value, so the new period is T√(1/2) = T/√2.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThis divides by 4 rather than by √4 for the stiffer spring: only the square root of each factor reaches the period.
BCorrect: T ∝ √(m/k), and √(2/4) = 1/√2.
COnly the change of mass has been applied; the stiffer spring, which shortens the period, has been ignored.
DThis multiplies by √4 instead of dividing by it: a stiffer spring makes the oscillation faster, not slower.

Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k) Command term: Determine

8C-1A-08

The simple pendulum·C.1 Simple harmonic motion

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

On the Earth, a mass hanging from a spring and a simple pendulum both oscillate with period T. Both are taken to a planet where the gravitational field strength is one quarter of that on the Earth.

Which row gives the periods on the planet?

Mass–spring systemSimple pendulum

Show mark scheme

Marking point	Mark	Notes
Step 1T = 2π√(m/k) contains no g: a weaker field only moves the equilibrium position of the hanging mass, so its period stays T.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = 2π√(l/g) ∝ 1/√g: dividing g by 4 multiplies the pendulum's period by √4 = 2, giving 2T.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThe pendulum is right, but the mass–spring period does not depend on g; the field only changes the extension at equilibrium.
BThe mass–spring period is right, but the pendulum's dependence has been inverted: T ∝ 1/√g, so a weaker field gives a longer period.
CBoth errors: the mass–spring period is wrongly made to depend on g, and the pendulum's dependence on g is inverted.
DCorrect: the mass–spring period is independent of g, and the pendulum period doubles because T ∝ 1/√g.

Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k), and of a simple pendulum as given by T = 2π√(l/g) Command term: Deduce

9C-1B-01

The simple pendulum·C.1 Simple harmonic motion

Paper 1BHard11 marks

Data-based question11 steps to full marksDetermine

A student investigates how the period T of a pendulum depends on its length l. The bob is a steel nut tied to a light thread, and l is measured from the point of suspension to the centre of the nut with a metre rule. For each length the time for 10 oscillations, 10T, is measured with a hand-held stopwatch. The uncertainty in each value of 10T is ±0.2 s.

The student's hypothesis is that T = K lⁿ, where K and n are constants and n = 0.5. The table shows the data and the graph shows log T against log l for all six lengths.

l / m ± 0.002 m	10T / s ± 0.2 s	log( l / m )	log( T / s )
0.150	7.78	−0.824	−0.109
0.250	9.97	−0.602	−0.001
0.400	12.64	−0.398	0.102
0.600	15.52
0.850	18.57	−0.071	0.269
1.200	22.02	0.079	0.343

Graph drawn to scale

(a)

State the absolute uncertainty in the period T.

(1)

(b)

Complete the table for l = 0.600 m.

(2)

(c)

Explain why a graph of log T against log l is used to test the hypothesis.

(2)

(d)

Draw the line of best fit on the graph.

(1)

(e)

(i)

Determine the gradient of the line of best fit.

(2)

(ii)

State whether the data support the hypothesis.

(1)

(f)

The theory of the simple pendulum gives K = 2π/√g. Determine g using the line of best fit.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
±0.02 s	✓ 1	Accept ±0.02 s written with T, e.g. (1.55 ± 0.02) s.
Part (b)
log( l / m ) = log 0.600 = −0.222	✓ 1	Accept 2 or 4 decimal places.
T = 15.52/10 = 1.552 s AND log( T / s ) = 0.191	✓ 1	Accept 0.190 to 0.191. Both values needed for the second mark.
Part (c)
log T = n log l + log K	✓ 1	Taking logs of both sides of the hypothesis.
This has the form y = mx + c, so a straight line is expected, with gradient equal to n (and intercept log K)	✓ 1	OWTTE. Must link the gradient to n.
Part (d)
A single straight line with points distributed evenly on both sides, extended at least to log( l / m ) = 0	✓ 1	Do not accept a line joining the first and last points, or a curve.
Part (e)(i)
Uses two well-separated points on the drawn line (triangle covering more than half the line)	✓ 1
gradient = 0.50	✓ 1	Accept 0.48 to 0.52. No unit.
Part (e)(ii)
Yes: the gradient is equal to 0.5 to within the precision of the graph (and the line is straight)	✓ 1	Allow ECF from (e)(i): a gradient outside 0.47–0.53 leads to 'not supported'.
Part (f)
Intercept at log( l / m ) = 0 read as 0.303, so K = 10^0.303 = 2.01 s m^−0.5	✓ 1	Accept an intercept of 0.298 to 0.308.
g = 4π²/K² = 9.79 m s⁻²	✓ 1	Accept 9.6 to 10.0 m s⁻². Allow ECF from the intercept. Award [2] for CNA.

Answers: (a) ±0.02 s · (b) −0.222; 0.191 · (e)(i) 0.50 · (f) g ≈ 9.79 m s⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); Tool 3 — linearizing a power law with a log–log graph Command term: Determine

10C-1B-02

The mass–spring oscillator·C.1 Simple harmonic motion

Paper 1BHard11 marks

Data-based question11 steps to full marksDetermine

A spring hangs from a clamp and a mass hanger carrying slotted masses is attached to its lower end. The total mass m on the spring is measured with a top-pan balance. The mass is pulled down a short distance and released, and the time for 10 oscillations, 10T, is measured with a stopwatch. The uncertainty in each value of 10T is ±0.1 s.

The graph shows T² against m with error bars for T². The dashed lines are the lines of maximum and minimum gradient that pass through all the error bars.

m / kg	10T / s ± 0.1 s
0.100	4.27
0.150	5.14
0.200	5.80
0.250	6.48
0.300	7.06
0.350	7.64

Graph drawn to scale

(a)

For m = 0.200 kg, calculate T² and its absolute uncertainty.

(2)

(b)

Draw the line of best fit on the graph.

(1)

(c)

(i)

Show that the gradient of your line of best fit is about 1.6 s² kg⁻¹.

(1)

(ii)

Determine the spring constant k of the spring.

(2)

(d)

Using the dashed lines, determine the absolute uncertainty in k.

(2)

(e)

The manufacturer states that k = 25.0 N m⁻¹ ± 2 %. Comment on the student's result.

(1)

(f)

A computer fit shows that the line of best fit meets the m axis at m = −0.014 kg rather than at the origin. Suggest a reason for this, and explain why it does not affect the value of k found in (c)(ii).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T = 0.580 s, so T² = 0.336 s²	✓ 1
Δ(T²) = 2 × (0.01/0.580) × 0.336 = ±0.012 s²	✓ 1	Accept ±0.01 s². Uses the fractional uncertainty in T doubled.
Part (b)
Single straight line lying between the two dashed lines and passing through all error bars	✓ 1
Part (c)(i)
Gradient from two well-separated points on the line, e.g. (0.580 − 0.181)/(0.350 − 0.100) = 1.60	✓ 1	Must see the working AND an answer to at least 3 s.f. (between 1.55 and 1.67).
Part (c)(ii)
T² = (4π²/k) m, so gradient = 4π²/k	✓ 1
k = 4π²/1.60 = 24.7 N m⁻¹	✓ 1	Accept 23.9 to 25.5 N m⁻¹. Allow ECF from (c)(i). Award [2] for CNA.
Part (d)
Gradients of the dashed lines read as 1.69 and 1.51 s² kg⁻¹, giving k = 23.3 and 26.1 N m⁻¹	✓ 1	Accept ±0.03 on each gradient.
Δk = (26.1 − 23.3)/2 ≈ ±1.4 N m⁻¹, so k = (24.7 ± 1.4) N m⁻¹ OR (25 ± 1) N m⁻¹	✓ 1	Accept ±1.1 to ±1.7 N m⁻¹. Accept the largest difference from the best value.
Part (e)
The range 23.3–26.1 N m⁻¹ overlaps the manufacturer's range 24.5–25.5 N m⁻¹, so the result is consistent with the stated value	✓ 1	Allow ECF from (c)(ii) and (d). Must refer to both ranges or to the uncertainty.
Part (f)
The spring itself has mass, and part of it oscillates with the load, so the effective oscillating mass is greater than m	✓ 1	Accept 'the mass of the spring is not included in m'.
This adds the same mass to every reading, which shifts the line sideways but does not change its gradient	✓ 1	OWTTE. A systematic offset changes the intercept only.

Answers: (a) T² = (0.336 ± 0.012) s² · (c)(ii) k ≈ 24.7 N m⁻¹ · (d) Δk ≈ ±1.4 N m⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); Tool 3 — uncertainty in a gradient from lines of maximum and minimum gradient Command term: Determine

11C-1B-03

The defining equation a = −ω²x·C.1 Simple harmonic motion

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

A glider on a horizontal air track is attached between two stretched springs. A motion sensor measures the displacement x of the glider from its equilibrium position, and a wireless accelerometer fixed to the glider measures its acceleration a. A data-logger records both quantities at the same instants during one oscillation. Seven pairs of values are shown in the table and plotted on the graph.

x / m	a / m s⁻²
−0.075	+3.18
−0.050	+2.09
−0.025	+1.05
0.000	+0.05
+0.025	−0.91
+0.050	−1.98
+0.075	−3.00

Graph drawn to scale

(a)

Outline how the graph indicates that the glider moves with simple harmonic motion.

(2)

(b)

Draw the line of best fit on the graph.

(1)

(c)

Determine the angular frequency ω of the oscillation.

(3)

(d)

The student also times 20 oscillations with a stopwatch and records 19.6 s ± 0.3 s. Deduce whether this measurement is consistent with your answer to (c).

(2)

(e)

The line of best fit does not pass through the origin. Suggest a cause for this and state its effect on the value of ω.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
The graph is a straight line, so the (magnitude of the) acceleration is proportional to the displacement	✓ 1	Accept 'a varies linearly with x'.
The gradient is negative, so the acceleration is always directed opposite to the displacement, towards the equilibrium position	✓ 1	OWTTE.
Part (b)
Single straight line with points evenly distributed on both sides	✓ 1	Must not be forced through the origin.
Part (c)
Uses a triangle covering most of the line to find the gradient	✓ 1
gradient = −41 s⁻²	✓ 1	Accept −39 to −43 s⁻².
ω = √41 = 6.4 rad s⁻¹	✓ 1	Accept 6.2 to 6.6 rad s⁻¹. Allow ECF from the gradient. Award [3] for CNA with a gradient seen.
Part (d)
T = 19.6/20 = 0.980 s ± 0.015 s, i.e. between 0.965 s and 0.995 s	✓ 1	Uncertainty must also be divided by 20.
From (c), T = 2π/ω = 2π/6.4 = 0.98 s, which lies inside this range, so the measurements are consistent	✓ 1	Allow ECF from (c). Conclusion must be supported by comparison with the range.
Part (e)
The line crosses the a axis at about +0.07 m s⁻²: the accelerometer has a zero error (offset) and gives a non-zero reading when a = 0	✓ 1	Accept an intercept of +0.04 to +0.10 m s⁻². Accept any constant systematic offset of the accelerometer or a slightly tilted track.
The offset shifts every value of a by the same amount, so the gradient, and therefore ω, is unaffected	✓ 1	OWTTE.

Answers: (c) ω ≈ 6.4 rad s⁻¹ · (d) T = 0.980 ± 0.015 s; 2π/ω ≈ 0.98 s (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the defining equation of simple harmonic motion as given by a = −ω²x; the time period in terms of angular frequency as given by T = 2π/ω Command term: Determine

12C-2-01

Conditions for SHM·C.1 Simple harmonic motion

Paper 2Medium14 marks

Short answer & extended response14 steps to full marksExplain

A small puck moves without friction on a track made from two straight ramps. Each ramp is inclined at 6.0° to the horizontal, and the ramps meet at a smoothly rounded bottom of negligible length, as shown in Figure 1. The displacement x of the puck is measured along the track from the bottom, and is positive to the right.

The puck is released from rest at x = +0.40 m. Take g = 9.81 m s⁻².

Figure 1 — Diagram NOT to scale (the angle is exaggerated)

Figure 2 — for part (b)(ii)

(a)

State the two conditions on the acceleration of a body that are required for its motion to be simple harmonic.

(2)

(b)

(i)

Show that the magnitude of the acceleration of the puck on either ramp is about 1.0 m s⁻².

(1)

(ii)

Sketch, on the axes in Figure 2, a graph to show how the acceleration a of the puck varies with x for −0.40 m ≤ x ≤ +0.40 m.

(2)

(iii)

Hence explain why the motion of the puck is not simple harmonic.

(2)

(c)

Determine the period of the oscillation of the puck.

(3)

(d)

The ramps are replaced by a horizontal track on which the same puck, of mass 0.050 kg, is held between two stretched springs. The combined spring constant of the springs is 0.16 N m⁻¹.

(i)

Calculate the period of oscillation of the puck on the springs.

(1)

(ii)

Both oscillators are now started with an amplitude of 0.10 m instead of 0.40 m. Compare the periods of the two oscillators at this amplitude with their periods at the larger amplitude. Support your answer with a calculation.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the acceleration is «directly» proportional to the displacement from the equilibrium position	✓ 1	OWTTE. Accept a ∝ x
the acceleration is «always» directed towards the equilibrium position / opposite to the displacement	✓ 1	Award [2] for a = −ω²x with both the proportionality and the meaning of the minus sign explained
Part (b)(i)
a = g sin 6.0° = 9.81 × sin 6.0° «= 1.03 m s⁻²»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
horizontal line at a ≈ −1.0 m s⁻² for positive x AND horizontal line at a ≈ +1.0 m s⁻² for negative x, each extending to ±0.40 m	✓ 1	Allow ECF from (b)(i) for the size of a
a changes sign abruptly at x = 0	✓ 1	A vertical line at x = 0 is not required. Do not award MP2 for a sloping straight line through the origin
Part (b)(iii)
the magnitude of the acceleration is constant «≈ 1.0 m s⁻²» / does not depend on the displacement	✓ 1
so a is not proportional to x «the graph is not a straight line through the origin», which SHM requires, even though a is always directed towards the equilibrium position	✓ 1	OWTTE. Award [1] max for 'the period depends on the amplitude' with no reference to the acceleration
Part (c)
time to reach the bottom from s = ½at²: 0.40 = ½ × 1.03 × t²	✓ 1	Allow use of a = 1.0 m s⁻² throughout, giving T = 3.58 s
t = 0.883 s	✓ 1
one oscillation consists of four such journeys, so T = 4t = 3.53 s	✓ 1	Award [3] for a bald correct answer. Accept 3.5–3.6 s. Award [2] max for T = 2t
Part (d)(i)
T = 2π√(0.050/0.16) = 3.51 s	✓ 1	Accept 3.5 s
Part (d)(ii)
spring oscillator: the period is unchanged «at 3.51 s» because T = 2π√(m/k) does not depend on the amplitude	✓ 1
ramps: t ∝ √s, so when the distance is quartered the time «and the period» is halved	✓ 1	Accept a full recalculation: t = √(2 × 0.10/1.03) = 0.442 s
T = 1.77 s for the ramps «so the two oscillators no longer have the same period»	✓ 1	Allow ECF from (c). Accept 1.76–1.79 s

Answers: (b)(i) a = 1.03 m s⁻² · (c) T = 3.53 s · (d)(i) T = 3.51 s · (d)(ii) springs 3.51 s (unchanged); ramps 1.77 s (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — conditions that lead to simple harmonic motion, and the defining equation of simple harmonic motion as given by a = −ω²x (with A.1 — the equations of motion for uniformly accelerated motion) Command term: Explain

13C-2-02

The simple pendulum·C.1 Simple harmonic motion

Paper 2Hard15 marks

Short answer & extended response15 steps to full marksDetermine

A robotic lander on the surface of Mars carries a simple pendulum of length 0.500 m. A camera on the lander records that the pendulum makes 20 complete oscillations of small amplitude in 46.1 s.

(a)

(i)

Determine the gravitational field strength g_M at the surface of Mars.

(3)

(ii)

Outline why the time for 20 oscillations is measured rather than the time for one oscillation.

(1)

(b)

The radius of Mars is 3.39 × 10⁶ m.

(i)

Calculate the mass of Mars.

(2)

(ii)

The same pendulum is taken to the Earth. Calculate its period on the Earth.

(1)

(c)

Phobos, a moon of Mars, orbits at a distance of 9.38 × 10⁶ m from the centre of Mars.

(i)

Determine the gravitational field strength due to Mars at the orbit of Phobos.

(2)

(ii)

Determine the orbital period of Phobos, in hours.

(2)

(d)

The length of the pendulum is known to ±0.005 m and the time for 20 oscillations to ±0.2 s. Determine the absolute uncertainty in g_M, and state g_M with its uncertainty.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = 46.1/20 = 2.305 s	✓ 1
g_M = 4π²l/T² = 4π² × 0.500/2.305²	✓ 1
g_M = 3.72 m s⁻²	✓ 1	Award [3] for a bald correct answer. Accept 3.71–3.72 m s⁻². Award [2] max for using 46.1 s as the period
Part (a)(ii)
the «absolute» uncertainty in starting and stopping the timer is shared over 20 periods, so the percentage uncertainty in T is smaller	✓ 1	OWTTE. Do not accept 'more accurate' alone
Part (b)(i)
M = g_Mr²/G = 3.72 × (3.39 × 10⁶)²/6.67 × 10⁻¹¹	✓ 1	Allow ECF from (a)(i)
M = 6.40 × 10²³ kg	✓ 1	Accept 6.3–6.5 × 10²³ kg
Part (b)(ii)
T = 2π√(0.500/9.81) = 1.42 s	✓ 1
Part (c)(i)
g ∝ 1/r², so g = 3.72 × (3.39/9.38)²	✓ 1	ALTERNATIVE: g = GM/r² with M from (b)(i). Allow ECF
g = 0.485 N kg⁻¹	✓ 1	Accept 0.48–0.49 N kg⁻¹ or m s⁻²
Part (c)(ii)
g = v²/r so v = √(0.485 × 9.38 × 10⁶) = 2.13 × 10³ m s⁻¹	✓ 1	ALTERNATIVE: T = 2π√(r/g). Allow ECF from (c)(i)
T = 2πr/v = 2.76 × 10⁴ s = 7.7 h	✓ 1	Accept 7.6–7.7 h. Award [2] for CNA
Part (d)
fractional uncertainty in l = 0.005/0.500 = 0.010	✓ 1
fractional uncertainty in T² = 2 × 0.2/46.1 = 0.0087	✓ 1	Award this mark for 2 × the fractional uncertainty in T
Δg_M = (0.010 + 0.0087) × 3.72 = 0.069 m s⁻²	✓ 1	Allow ECF from (a)(i)
g_M = (3.72 ± 0.07) m s⁻²	✓ 1	MP4 is for matching the precision of the value and the uncertainty

Answers: (a)(i) g_M = 3.72 m s⁻² · (b)(i) 6.40 × 10²³ kg · (b)(ii) 1.42 s · (c)(i) 0.485 N kg⁻¹ · (c)(ii) 7.7 h · (d) (3.72 ± 0.07) m s⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g) (with D.1 — gravitational field strength at a point as given by g = F/m = GM/r², and the orbital motion of a satellite; Tool 3 — propagating uncertainties) Command term: Determine

14C-2-03

Energy in SHM·C.1 Simple harmonic motion

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksSketch

A long pendulum in a science museum consists of a bob of mass 28 kg on a steel wire. The distance from the point of suspension to the centre of the bob is 11.0 m.

The bob is pulled aside so that its centre rises 1.20 cm above its lowest position, and is released from rest at time t = 0. Treat the pendulum as a simple pendulum. Ignore air resistance except in part (d). Take g = 9.81 m s⁻².

Figure 1 — for part (c)

(a)

(i)

Show that the period of the pendulum is about 6.7 s.

(1)

(ii)

State the time t at which the speed of the bob is first a maximum.

(1)

(b)

Calculate the maximum speed of the bob.

(2)

(c)

Sketch, on the axes in Figure 1, graphs to show how the kinetic energy E_k and the gravitational potential energy E_p of the bob vary with time from t = 0 to t = T, where T is the period. E_p is measured from the lowest position of the bob. Label your graphs.

(3)

(d)

In practice the amplitude of the swing slowly decreases.

(i)

Outline why the amplitude decreases.

(1)

(ii)

An electromagnet under the floor gives the bob a small push once in each cycle so that the amplitude stays constant. Suggest where in the swing, and in which direction, the push should be given.

(2)

(e)

Suggest why the museum uses a heavy bob rather than a light bob of the same size.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = 2π√(11.0/9.81) «= 6.65 s»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
T/4 = 1.7 s	✓ 1	Accept 1.6–1.7 s. Allow ECF from (a)(i)
Part (b)
½mv² = mgh so v = √(2 × 9.81 × 0.0120)	✓ 1	Award [1] max if h = 1.20 is used «4.85 m s⁻¹»
v = 0.49 m s⁻¹	✓ 1	Accept 0.48–0.49 m s⁻¹. Award [2] for CNA
Part (c)
E_p = E at t = 0, T/2 and T, and E_p = 0 at T/4 and 3T/4	✓ 1	Labels required for MP1 and MP2
E_k = 0 when E_p = E, and E_k = E when E_p = 0	✓ 1
smooth curves «not straight lines» that are never negative, with E_k + E_p = E at every instant «the curves cross at E/2»	✓ 1	Do not award MP3 for curves with sharp corners at the maxima
Part (d)(i)
work is done against air resistance «and friction at the support», so energy of the oscillation is transferred to the surroundings as thermal energy	✓ 1	OWTTE
Part (d)(ii)
at «or near» the lowest point of the swing, in the direction in which the bob is moving	✓ 1	Accept 'in the direction of the velocity of the bob'
so the force does work on the bob / supplies energy in step with the motion, replacing the energy lost in each cycle	✓ 1	OWTTE. Do not accept a push against the motion
Part (e)
the resistive forces on bobs of the same size are about the same, but the heavier bob has more energy «and inertia», so a smaller fraction of its energy is lost in each swing «the amplitude decreases more slowly»	✓ 1	OWTTE

Answers: (a)(i) T = 6.65 s · (a)(ii) t = 1.7 s · (b) v = 0.49 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — a qualitative approach to energy changes during one cycle of an oscillation; the time period of a simple pendulum as given by T = 2π√(l/g) (with A.3 — the principle of the conservation of energy) Command term: Sketch

15C-1B-11

The mass–spring oscillator·C.1 Simple harmonic motion

Paper 1BEasy9 marks

Data-based question9 steps to full marksDetermine

A student determines the spring constant k of a spring by two methods.

Method 1: a mass of (0.300 ± 0.001) kg is hung from the spring. The position of the bottom of the spring is read on a millimetre scale before and after the mass is added. The extension is (9.6 ± 0.2) cm.

Method 2: the same mass is set oscillating vertically. The time for 20 oscillations is (12.5 ± 0.2) s.

Quantity	Measurement
mass m	(0.300 ± 0.001) kg
extension x (method 1)	(9.6 ± 0.2) cm
time for 20 oscillations (method 2)	(12.5 ± 0.2) s

(a)

Each reading of the scale has an uncertainty of ±0.1 cm. Outline why the uncertainty in the extension is ±0.2 cm.

(1)

(b)

Show that method 1 gives a value of k of about 31 N m⁻¹.

(1)

(c)

Calculate the percentage uncertainty in the value of k from method 1.

(2)

(d)

Determine the value of k from method 2 and its absolute uncertainty.

(3)

(e)

State whether the two methods agree.

(1)

(f)

Suggest one change to method 2 that would reduce the uncertainty in k.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
The extension is the difference of two readings, and the absolute uncertainties of the two readings add	✓ 1	OWTTE.
Part (b)
k = mg/x = 0.300 × 9.81/0.096 = 30.7 N m⁻¹	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (c)
Δx/x = 0.2/9.6 = 2.1 % and Δm/m = 0.33 %	✓ 1	Either fraction seen.
2.4 %	✓ 1	Accept 2 % (mass uncertainty ignored). Award [2] for CNA.
Part (d)
T = 12.5/20 = 0.625 s, so k = 4π²m/T² = 30.3 N m⁻¹	✓ 1
Fractional uncertainty = 2 × 0.2/12.5 + 0.001/0.300 = 0.035	✓ 1	Must double the fractional uncertainty in T. Accept 0.032 (mass ignored).
k = (30 ± 1) N m⁻¹	✓ 1	Accept (30.3 ± 1.1) N m⁻¹. MP3 is for matching the precision of value and uncertainty.
Part (e)
Method 1: 29.9–31.4 N m⁻¹; method 2: 29.2–31.4 N m⁻¹; the ranges overlap, so the methods agree	✓ 1	Allow ECF from (c) and (d). Must refer to the uncertainty ranges.
Part (f)
Time more oscillations (e.g. 50) in each run, OR use a fiducial marker at the equilibrium position to start and stop the timer, OR use a light gate/motion sensor	✓ 1	Do not accept 'repeat the experiment' without a specific reason. Do not accept 'use a more precise stopwatch'.

Answers: (b) k = 30.7 N m⁻¹ · (c) 2.4 % · (d) k = (30 ± 1) N m⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k); A.2 — Hooke's law as given by F_H = −kx Command term: Determine

16C-1B-12

The simple pendulum·C.1 Simple harmonic motion

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

A pendulum of length (0.800 ± 0.002) m swings through a light gate placed at the lowest point of the swing. A data-logger connected to the light gate displays the period T directly. The bob is released from rest at an angular amplitude θ₀, measured with a protractor, and the mean of the first five periods is recorded. The uncertainty in each mean period is ±0.004 s.

A student's hypothesis is that the period of a pendulum does not depend on its amplitude.

θ₀ / °	T / s
5	1.797
10	1.798
15	1.803
20	1.809
30	1.827
40	1.849
50	1.883
60	1.926

Graph drawn to scale

(a)

State the resolution of the period measurements.

(1)

(b)

Draw the curve of best fit on the graph.

(1)

(c)

(i)

Using the graph, estimate the largest amplitude for which the data support the hypothesis.

(1)

(ii)

Explain, with reference to the conditions for simple harmonic motion, why the period increases at large amplitudes.

(2)

(d)

Determine g, and its absolute uncertainty, using the period for θ₀ = 5°.

(3)

(e)

Comment on your answer to (d).

(1)

(f)

Suggest why only the first five periods after release were used for each amplitude.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.001 s	✓ 1	Do not accept ±0.004 s.
Part (b)
Smooth curve, almost horizontal at small amplitudes and rising increasingly steeply, passing through all error bars	✓ 1	Do not accept a straight line or dot-to-dot.
Part (c)(i)
About 10° to 15°: up to this amplitude the periods agree with the small-amplitude value within the uncertainty	✓ 1	Accept 10° to 15°.
Part (c)(ii)
SHM requires the restoring force (acceleration) to be proportional to the displacement; for a pendulum the restoring force is mg sin θ, which is proportional to θ only for small angles	✓ 1	Accept 'sin θ ≈ θ only for small angles'.
At large amplitudes sin θ < θ, so the restoring force and acceleration are smaller than SHM predicts and the bob takes longer to complete a cycle	✓ 1	OWTTE. Must link smaller acceleration to longer period.
Part (d)
g = 4π²l/T² = 4π² × 0.800/1.797² = 9.78 m s⁻²	✓ 1
Δg/g = Δl/l + 2ΔT/T = 0.002/0.800 + 2 × 0.004/1.797 = 0.0070	✓ 1	Must double the fractional uncertainty in T.
g = (9.78 ± 0.07) m s⁻²	✓ 1	MP3 is for matching the precision of value and uncertainty.
Part (e)
The accepted value 9.81 m s⁻² lies within the range 9.71–9.85 m s⁻², so the result is consistent with it	✓ 1	Allow ECF from (d).
Part (f)
The amplitude decreases with time because of air resistance (damping), so later periods would not correspond to the recorded amplitude	✓ 1	OWTTE.

Answers: (a) 0.001 s · (c)(i) about 10°–15° · (d) g = (9.78 ± 0.07) m s⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g); conditions that lead to simple harmonic motion Command term: Determine

17C-1B-13

The defining equation a = −ω²x·C.1 Simple harmonic motion

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

A plastic tube with a flat base is loaded with lead shot so that it floats upright in water. When the tube is pushed down a little and released, it oscillates vertically. A student varies the total mass m of the tube and shot, and uses a video recording to find the time for 10 oscillations, 10T.

The period is predicted to be T = 2π√(m/(ρgA)), where A is the cross-sectional area of the tube and ρ = 1000 kg m⁻³ is the density of water. The graph shows T² against m.

m / kg	10T / s	T² / s²
0.040	4.81	0.231
0.055	5.54	0.307
0.070	6.18
0.085	6.74	0.454
0.100	7.26	0.527
0.115	7.80	0.608

Graph drawn to scale

(a)

When the tube is pushed down by a distance y below its equilibrium position, the upward buoyancy force increases by ρgAy. Outline why the tube performs simple harmonic motion.

(2)

(b)

Show that T² is about 0.38 s² for m = 0.070 kg.

(1)

(c)

Draw the line of best fit on the graph.

(1)

(d)

Determine the gradient of the line of best fit.

(2)

(e)

Determine the external diameter of the tube.

(2)

(f)

The diameter measured with vernier calipers is (32.0 ± 0.1) mm. Comment on your answer to (e).

(1)

(g)

The line of best fit does not pass through the origin. Suggest a physical reason for this.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
The resultant force on the tube is ρgAy, which is proportional to the displacement y	✓ 1	Weight and equilibrium buoyancy cancel.
and is directed towards the equilibrium position (opposite to the displacement), so a = −(ρgA/m)y, the defining equation of SHM	✓ 1	OWTTE.
Part (b)
T = 6.18/10 = 0.618 s, T² = 0.3819 s²	✓ 1	Must see the substitution AND an answer to at least 3 s.f.
Part (c)
Single straight line with points evenly distributed on both sides, not forced through the origin	✓ 1
Part (d)
Uses a triangle covering more than half the line	✓ 1
gradient = 5.0 s² kg⁻¹	✓ 1	Accept 4.8 to 5.2 s² kg⁻¹.
Part (e)
Gradient = 4π²/(ρgA), so A = 4π²/(1000 × 9.81 × gradient) = 8.05 × 10⁻⁴ m²	✓ 1	Allow ECF from (d).
d = 2√(A/π) = 0.0320 m = 32.0 mm	✓ 1	Accept 31.4 to 32.7 mm.
Part (f)
The value from the graph agrees with the caliper value to within about 1 % (the graph method is much less precise), so the model is supported	✓ 1	Allow ECF. Accept a justified 'not consistent' only if the candidate's (e) is outside 31.4–32.7 mm.
Part (g)
Water around and below the tube moves with it, so the oscillating mass is larger than m by a constant amount	✓ 1	Accept 'the tube drags water with it'. Do not accept 'random error'.

Answers: (b) T² = 0.382 s² · (d) 5.0 s² kg⁻¹ · (e) d ≈ 32.0 mm (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — conditions that lead to simple harmonic motion; the defining equation of simple harmonic motion as given by a = −ω²x; Tool 3 — linearizing a relationship Command term: Determine

18C-2-16

The defining equation a = −ω²x·C.1 Simple harmonic motion

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine whether

A laboratory shaker has a horizontal platform that oscillates vertically with simple harmonic motion. A tray of mass 0.150 kg rests on the platform without being fixed to it. The platform oscillates with a frequency of 8.0 Hz and an amplitude of 2.0 mm. Take g = 9.81 m s⁻².

Figure 1 — Diagram NOT to scale

(a)

(i)

Show that the maximum acceleration of the platform is about 5 m s⁻².

(1)

(ii)

State the direction of the acceleration of the tray when the platform is at its highest point.

(1)

(b)

(i)

On Figure 1, draw and label arrows to represent the forces acting on the tray when the platform is at its highest point. The lengths of your arrows should show the relative sizes of the forces.

(2)

(ii)

Calculate the magnitude of the normal force on the tray at the highest point.

(2)

(c)

(i)

Determine the frequency at which the tray just loses contact with the platform when the amplitude is 2.0 mm.

(3)

(ii)

Explain why this frequency does not depend on the mass of the tray.

(1)

(d)

The frequency is returned to 8.0 Hz and the amplitude is increased to 4.0 mm. Determine whether the tray stays in contact with the platform throughout the oscillation.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
a_max = ω²A = (2π × 8.0)² × 2.0 × 10⁻³ «= 5.05 m s⁻²»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
downwards «towards the equilibrium position»	✓ 1
Part (b)(i)
two arrows only, starting on the tray: weight «W, mg, F_g» vertically downwards and normal force «N, R, F_N» vertically upwards, both labelled	✓ 1	Do not credit an extra 'restoring force' or 'acceleration' arrow
weight arrow clearly longer than the normal force arrow	✓ 1	MP2 only scores if MP1 scores. Accept lengths in the ratio of about 2 : 1
Part (b)(ii)
mg − N = ma, so N = 0.150 × (9.81 − 5.05)	✓ 1	Allow ECF from (a)(i); using a = 5 gives 0.72 N
N = 0.71 N	✓ 1	Accept 0.71–0.72 N
Part (c)(i)
contact is lost when N = 0 at the highest point, i.e. when the required downward acceleration ω²A equals g	✓ 1
ω = √(9.81/2.0 × 10⁻³) = 70.0 rad s⁻¹	✓ 1
f = ω/2π = 11.1 Hz	✓ 1	Award [3] for a bald correct answer. Accept 11 Hz
Part (c)(ii)
the weight and the resultant force needed for the acceleration are both proportional to the mass, so the condition ω²A = g does not involve m	✓ 1	OWTTE
Part (d)
ω²A = (2π × 8.0)² × 4.0 × 10⁻³ = 10.1 m s⁻²	✓ 1	ALTERNATIVE: largest amplitude for contact at 8.0 Hz = g/ω² = 3.88 mm
this is greater than g «= 9.81 m s⁻²»	✓ 1	ALTERNATIVE: 4.0 mm > 3.88 mm
so the tray does not stay in contact; it leaves the platform near the top of the motion	✓ 1	MP3 only scores if a comparison with g (or the limiting amplitude) is made

Answers: (a)(i) 5.05 m s⁻² · (b)(ii) N = 0.71 N · (c)(i) 11.1 Hz · (d) ω²A = 10.1 m s⁻² > g: contact is lost (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the defining equation of simple harmonic motion as given by a = −ω²x (with A.2 — free-body diagrams and Newton's second law of motion) Command term: Determine whether

19C-2-17

The simple pendulum·C.1 Simple harmonic motion

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksDetermine whether

A tall building sways from side to side in strong winds. The natural period of this sway is 6.0 s.

To reduce the sway, engineers hang a large steel ball from cables inside the top of the building. The ball and cables behave as a simple pendulum whose natural period is equal to that of the building. Hydraulic dampers connect the ball to the building. Take g = 9.81 m s⁻².

(a)

Show that the length of the pendulum must be about 9 m.

(1)

(b)

Explain, in terms of resonance, why gusts of wind that push on the building every 6.0 s are dangerous.

(3)

(c)

During a storm the ball swings with a horizontal amplitude of 1.5 m. The equation T = 2π√(l/g) is accurate to within 1 % for angular amplitudes up to about 23°. Determine whether the equation may be used for the pendulum during this storm.

(3)

(d)

The mass of the ball is 3.3 × 10⁵ kg. State and explain whether a ball of smaller mass, hung from cables of the same length, would have a different natural period.

(2)

(e)

The hydraulic dampers remove energy from the swinging ball. State the form of energy into which this energy is transferred.

(1)

(f)

Suggest one reason why the period of the real ball-and-cable system may differ from the value given by T = 2π√(l/g).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
l = gT²/4π² = 9.81 × 6.0²/4π² «= 8.95 m»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (b)
the driving frequency «1/6.0 s» equals the natural frequency of the building, so resonance occurs	✓ 1
energy is transferred to the building in every cycle «because each push is in step with the motion»	✓ 1	OWTTE
so the amplitude of the sway builds up to a large value that could damage the structure	✓ 1
Part (c)
sin θ = 1.5/8.95	✓ 1	Allow ECF from (a). Accept tan θ or the small-angle value 1.5/8.95 rad
θ = 9.7°	✓ 1	Accept 9.5°–9.8° «9.6° if l = 9.0 m is used»
this is less than 23°, so the equation may be used	✓ 1	MP3 only scores if a comparison is made
Part (d)
no, the period would be the same	✓ 1
T = 2π√(l/g) does not contain the mass, because the restoring force and the inertia are both proportional to the mass	✓ 1	OWTTE. Award [1] max for quoting the equation without explanation
Part (e)
thermal energy «internal energy» of the oil in the dampers and the surroundings	✓ 1	Accept 'heat'
Part (f)
the ball is large, not a point mass / the cables have mass / the dampers exert forces on the ball / the amplitude may not be small	✓ 1	Any one reasonable suggestion

Answers: (a) l = 8.95 m · (c) θ = 9.7° < 23°: yes (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a simple pendulum as given by T = 2π√(l/g) (with C.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency) Command term: Determine whether

20C-2-18

The mass–spring oscillator·C.1 Simple harmonic motion

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

An inertial balance consists of a tray clamped between two flexible horizontal steel strips. When the tray is pulled sideways and released, it oscillates horizontally with simple harmonic motion. An object of mass m clamped to the tray changes the period T according to

T = 2π√((m + m_t)/k)

where m_t is the mass of the tray and k is the effective spring constant of the strips. The table shows two calibration measurements.

Added mass m / kg	Time for 10 oscillations / s
0	4.20
0.200	6.30

Figure 1 — for part (c)

(a)

An inertial balance can be used to find the mass of an object on board an orbiting space station, where a spring balance (newton meter) cannot be used. Explain why.

(3)

(b)

(i)

Show that k is about 36 N m⁻¹.

(2)

(ii)

Determine m_t.

(2)

(c)

Draw, on the grid in Figure 1, the line that shows how T² varies with m for 0 ≤ m ≤ 0.30 kg.

(2)

(d)

The tray, carrying an unknown object, oscillates with a period of 0.560 s. Determine the mass of the object.

(2)

(e)

Suggest why the object must be clamped firmly to the tray.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
in orbit the station and everything in it are in free fall with the same acceleration	✓ 1	OWTTE
so no supporting force is needed and a spring balance reads zero «for any mass»	✓ 1
the inertial balance depends on the force needed to accelerate the mass «F = ma» / on the inertia of the mass, which does not depend on g	✓ 1	Do not accept 'there is no gravity in orbit'
Part (b)(i)
T₀ = 0.420 s and T₁ = 0.630 s, and subtracting: T₁² − T₀² = 4π² × 0.200/k	✓ 1	Must see the two periods used
k = 4π² × 0.200/(0.630² − 0.420²) «= 35.8 N m⁻¹»	✓ 1	Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
m_t = kT₀²/4π² = 35.8 × 0.420²/4π²	✓ 1	ALTERNATIVE: m_t/(m_t + 0.200) = 0.420²/0.630²
m_t = 0.160 kg	✓ 1	Accept 0.16 kg. Using k = 36 N m⁻¹ gives 0.161 kg
Part (c)
straight line with an intercept of T² = 0.176 s² at m = 0	✓ 1	Accept 0.17–0.18 s²
line passing through (0.200 kg, 0.397 s²) / gradient 1.10 s² kg⁻¹, drawn to m = 0.30 kg «T² = 0.507 s²»	✓ 1	Allow ECF from (b)
Part (d)
T² = 0.560² = 0.314 s²	✓ 1	Allow reading from the line in (c)
m = kT²/4π² − m_t = 0.124 kg	✓ 1	Accept 0.12–0.13 kg. Allow ECF from (b)
Part (e)
if it slides on the tray, not all of its mass is accelerated with the tray «and friction removes energy», so the period no longer corresponds to the full mass	✓ 1	OWTTE

Answers: (b)(i) k = 35.8 N m⁻¹ · (b)(ii) m_t = 0.160 kg · (d) m = 0.124 kg (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — the time period of a mass–spring system as given by T = 2π√(m/k) (with A.2 — Newton's second law of motion) Command term: Determine

21C-2-19

SHM: period & frequency·C.1 Simple harmonic motion

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine whether

A smartphone contains an accelerometer. Inside it, a tiny silicon block of mass 4.0 × 10⁻⁹ kg is held between thin silicon beams that act as a spring, so the block can move along one horizontal line. The block and beams behave as a mass on a spring. The displacement x of the block from its equilibrium position, relative to the case of the phone, is measured electrically.

In a test in which damping is negligible, the phone is tapped and the block oscillates with simple harmonic motion. The graph shows how x varies with time t after the tap.

Graph drawn to scale

(a)

Determine the frequency of the oscillation.

(2)

(b)

Show that the spring constant of the beams is about 2.5 N m⁻¹.

(1)

(c)

The phone is now given a constant acceleration of 3.0 m s⁻² along the line in which the block can move.

(i)

Explain why the block is displaced from its equilibrium position relative to the case of the phone.

(2)

(ii)

Calculate this displacement.

(2)

(d)

The electronics can detect a displacement of the block of 0.20 nm or more. Determine whether the accelerometer can detect a constant acceleration of 0.10 m s⁻².

(3)

(e)

The designer wants a larger displacement of the block for the same acceleration. Suggest, with reference to your answer to (c)(ii), how the natural frequency of the block and beams must change.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Four cycles take 1.00 ms, so T = 0.25 ms «2.5 × 10⁻⁴ s»	✓ 1	Accept T read from a single cycle.
f = 1/T = 4.0 × 10³ Hz	✓ 1	Accept 3.9–4.1 kHz. Allow ECF from an incorrect T. Award [2] for CNA.
Part (b)
k = 4π²m/T² = 4π² × 4.0 × 10⁻⁹/(2.5 × 10⁻⁴)² = 2.53 «N m⁻¹»	✓ 1	Must see full substitution OR answer to at least 3 s.f. Accept a rearrangement of T = 2π√(m/k).
Part (c)(i)
The block must accelerate with the phone, so a resultant force F = ma must act on it	✓ 1
This force can only be provided by the beams, which exert a force only when they are deformed «F = kx», so the block is displaced «towards the rear, opposite to the acceleration»	✓ 1	OWTTE. Do not accept 'the block is left behind' without reference to the force from the beams.
Part (c)(ii)
F = ma = 4.0 × 10⁻⁹ × 3.0 = 1.2 × 10⁻⁸ N	✓ 1	ALTERNATIVE: x = a/ω² with ω = 2π × 4.0 × 10³ rad s⁻¹.
x = F/k = 1.2 × 10⁻⁸/2.53 = 4.7 × 10⁻⁹ m	✓ 1	Accept 4.7–4.8 × 10⁻⁹ m «4.8 nm from k = 2.5 N m⁻¹». Award [2] for CNA.
Part (d)
x = ma/k = 4.0 × 10⁻⁹ × 0.10/2.53	✓ 1	ALTERNATIVE: smallest detectable acceleration = 0.20 × 10⁻⁹ × 2.53/4.0 × 10⁻⁹ = 0.13 m s⁻².
x = 1.6 × 10⁻¹⁰ m «0.16 nm»	✓ 1	Accept 1.6 × 10⁻¹⁰ m. Allow ECF from (b).
0.16 nm is less than 0.20 nm, so this acceleration cannot be detected	✓ 1	MP3 only scores if a displacement «or a smallest acceleration» has been calculated. Allow ECF for the conclusion.
Part (e)
For a given acceleration x = ma/k = a/ω² «= a/(2πf)²»	✓ 1	Allow ECF from (c)(ii).
so the natural frequency must be lower «e.g. a larger mass or less stiff beams»	✓ 1	MP2 needs a reason. OWTTE.

Answers: (a) f = 4.0 × 10³ Hz · (b) k = 2.53 N m⁻¹ · (c)(ii) 4.7 × 10⁻⁹ m · (d) 1.6 × 10⁻¹⁰ m — cannot be detected (the remaining parts are explanations — see the table above)

Syllabus understandingC.1 — a particle undergoing simple harmonic motion can be described using time period T, frequency f, angular frequency ω, amplitude, equilibrium position, and displacement; the time period in terms of frequency of oscillation and angular frequency as given by T = 1/f = 2π/ω; the time period of a mass–spring system as given by T = 2π√(m/k); A.2 — Newton's second law; the elastic restoring force given by Hooke's law F = −kx Command term: Determine whether

C.1 Simple harmonic motion: IB Physics SL exam-style questions

21 practice questions on C.1 Simple harmonic motion

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