AQA A-Level Physics Paper 1, June 2023: Question 6
9 marks · Hard difficulty · Extended Answer
Analyze simple harmonic motion and resonance of a floating pencil and ship in waves through equations, calculations, and explanations.
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Question text
06 A pencil is weighted with a thin coil of wire. The volume of the wire is negligible.
Figure 14 shows the pencil and wire floating in equilibrium in water.
Figure 14 Figure 15
In Figure 14 the combined weight of the pencil and wire is equal to an upwards force
called the buoyancy force. The length of the pencil that is submerged is l.
A student pushes the pencil down through a displacement y as shown in Figure 15.
The buoyancy force is now greater than the weight.
There is a resultant upward force F acting on the pencil when the student releases it.
The magnitude of F for any value of y is given by
F = Aρgy
where A is the cross-sectional area of the pencil
ρ is the density of water
g is the acceleration due to gravity.
The pencil is pushed down and released. The pencil then oscillates vertically about
the equilibrium position.
06.1 Show that the pencil moves with simple harmonic motion.
[2 marks]
06.2 The time period T of the vertical oscillations is given by
l
T = 2π
g
The measured value of l in Figure 15 is 85 mm.
The pencil is pushed down 5.0 mm and released.
Calculate the maximum acceleration of the pencil.
[2 marks]
22 −2
maximum acceleration = m s
A ship floating in the sea can be modelled by the pencil floating in water.
The ship can oscillate vertically. These oscillations are called heave oscillations.
Wave motion causes forced oscillations of the ship. Under certain conditions, heave
resonance may then occur.
06.3 Explain what is meant by resonance.
[2 marks]
06.4 Figure 16 shows a ship moving through continuous waves of wavelength 118 m
and velocity 14.2 m s−1.
The ship is moving steadily at 8.0 m s−1 relative to the seabed in the same direction as
the waves.
Figure 16
The natural frequency of heave oscillations of the ship is 0.13 Hz.
A crew member needs an emergency operation. The ship’s doctor is confident that
she can do the operation if the ship remains fairly steady.
There are two options:
• stop the ship’s motors and loosely anchor the ship to the seabed
• continue to sail the ship at 8.0 m s−1 in the same direction.
Deduce which is the better option.
Support your answer with a calculation.
[3 marks]
END OF SECTION A
Section B
Each of Questions 07 to 31 is followed by four responses, A, B, C and D.
For each question select the best response.
Only one answer per question is allowed.
For each question, completely fill in the circle alongside the appropriate answer.
CORRECT METHOD WRONG METHODS
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as shown.
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Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
06.1 Equates resultant force to ma In MP1: 2 AO1× 1
and shows a proportional to y, as Aρmg are all constant Condone upthrust/buoyancy force for AO2× 1
resultant force
F = ma = −Aρyg
Aρ yg
a = −
m
Condone missing minus signs in MP1.
Minus sign included and explained: In MP2:
(restoring) force/acceleration directed to centre of Minus because force/acceleration is in
oscillation opposite direction to y OWTTE
(hence SHM)
06.2 g −1 Alternative for MP1: 2 AO1 × 1
(T = 2π/ω) so ω = (= 10.74 rad s )
l calculates time (0.58(5) s) AND then uses ω AO2 × 1
g from this time
(a = − ω2y = × y = (9.81÷0.085) × 0.005 )
max max max
l
0.58 (m s−2) from some correct working MP2 for correct calculation of acceleration.
06.3 Idea that (at resonance) frequency of forced vibrations Accept fully labelled graph of amplitude vs 2 AO1 × 2
equals natural/resonant frequency 1 driving frequency with resonance frequency
clearly labelled1 and an amplitude peak. 2
Condone ‘wave frequency’ for ‘driving
Idea that amplitude (of vibrations/oscillations) is at a frequency’
maximum 2
Ignore references to phase
06.4 v 1 is for calculation of (driving) frequency 3 AO3× 3
stopped: wave frequency (= )= 0.12 Hz 1 when stopped. Condone reference to
λ
‘frequency of waves’.
If no reference to ship being stopped,
evidence can come from the substitution.
Reject simple “0.12 (Hz)”
2 is for a relevant comment about the
moving: when ship continues at 8 m s−1, forcing frequency
moving situation
will be further from resonant frequency 2
OR
calculation of forcing frequency with the ship
moving (giving 0.05 Hz)
For 2 accept incorrect calculation from
adding speeds provided comment that this
frequency is further from resonant frequency.
3 is for statement of why moving is the
better option
Moving option is better with reason,
24 Allow answer for 3 that mentions that
eg for stopped option wave/forcing frequency very close to damping will be highly likely, so amplitudes
natural frequency, (so amplitude of oscillations will be high) may not reach high enough values to prevent
operation
OR
for moving option resonance does not occur 3
Total 9
How to answer it
Simple Harmonic Motion and Resonance Study Guide
What this question tests
This exam question evaluates your ability to apply defining conditions for Simple Harmonic Motion (SHM) using given restoring force equations, perform maximum acceleration calculations using angular frequency relationships, define physical resonance phenomena clearly, and apply wave equation principles (v = f * lambda) to real-world context scenarios involving relative motion.
Proving Simple Harmonic Motion
💡 Key Knowledge
- The defining equation of SHM is a = -omega² * y or showing acceleration is directly proportional to displacement and directed towards equilibrium.
- Start from Newton's Second Law: F = m * a .
✅ Correct Answer & Derivation
- Equate resultant force to mass times acceleration: -A * p * g * y = m * a
- Rearrange for acceleration: a = -(A * p * g / m) * y
- A, p, g, and m are constants, proving a is proportional to -y .
- The minus sign indicates acceleration and displacement are in opposite directions (directed towards equilibrium).
Calculating Maximum Acceleration
📐 Step-by-Step Calculation
- Find angular frequency (omega): Use T = 2 * pi * sqrt(l / g) and omega = 2 * pi / T = sqrt(g / l) .
- Substitute values ( g = 9.81 m s⁻² , l = 0.085 m ): omega = sqrt(9.81 / 0.085) = 10.74 rad s⁻¹ .
- Calculate maximum acceleration: Use a_max = omega² * y_max .
- Substitute maximum displacement ( y_max = 5.0 mm = 5.0 × 10⁻³ m ): a_max = (10.74)² × 5.0 × 10⁻³ = 0.58 m s⁻² .
❌ Common Errors & Traps
- Unit conversion trap: Forgetting to convert millimeters ( 5.0 mm ) into meters ( 0.005 m ).
- Length conversion: Ensure length l is kept in standard SI units ( 0.085 m ).
Explaining Resonance
💡 Key Knowledge
- Resonance occurs when a system is driven at a frequency equal to its natural frequency.
- This results in a dramatic, sharp increase in the amplitude of oscillation.
✅ Expected Marking Points
- Point 1: The frequency of the forced vibrations equals the natural/resonant frequency of the system.
- Point 2: The amplitude of the oscillations reaches a maximum value.
Deducing the Best Emergency Option
🧠 Exam Technique & Analysis
- Calculate the driving frequency for the stopped option: f = v / lambda = 14.2 / 118 = 0.12 Hz .
- Compare this driving frequency ( 0.12 Hz ) with the ship's natural frequency ( 0.13 Hz ). They are extremely close, meaning resonance will occur (large amplitudes, dangerous for surgery).
- Calculate or deduce the frequency when moving at 8.0 m s⁻¹ in the same direction: relative speed v_rel = 14.2 - 8.0 = 6.2 m s⁻¹ , giving f = 6.2 / 118 = 0.05 Hz .
- Moving shifts the forcing frequency further away from the natural frequency ( 0.13 Hz ), preventing resonance.
✅ Final Conclusion
- Better option: Continue to sail the ship at 8.0 m s⁻¹ in the same direction.
- Reasoning: When stopped, the wave frequency ( 0.12 Hz ) is very close to the natural frequency ( 0.13 Hz ), leading to high-amplitude resonant oscillations. Continuing to move creates a lower forcing frequency further from resonance.
Topics
Physics · 3.6 Further mechanics and thermal physics (A-level only)
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.