Specification: AQA A-level Physics (7408) 3.6.1.3 | Paper: 1 | Data and formulae booklet: given in the exam
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Mass–spring system: a mass on a spring obeying Hooke’s law performs simple harmonic motion, because the restoring force is proportional to the displacement from equilibrium and directed towards the equilibrium position. Its period is T = 2π√(m/k), which is independent of the amplitude.
Spring constant: the force per unit extension, k = F ÷ ΔL, measured in N m⁻¹. A stiffer spring has a larger k and a shorter period.
Natural frequency: the frequency at which a system oscillates when it is displaced and then left to oscillate freely, f = 1 ÷ T. Resonance occurs when the driving frequency equals the natural frequency, giving a maximum amplitude (3.6.1.4).
The equation
time period = 2π × √(mass ÷ spring constant) T = 2π√(m/k) and frequency = 1 ÷ time period f = 1 ÷ T
| Quantity | Symbol | Unit |
|---|---|---|
| time period (time for one complete oscillation) | T | seconds (s) |
| oscillating mass | m | kilograms (kg) |
| spring constant | k | newtons per metre (N m⁻¹) |
| frequency | f | hertz (Hz), equivalent to s⁻¹ |
Rearranged: T² = 4π²m ÷ k | k = 4π²m ÷ T² | m = kT² ÷ 4π² | Square both sides first to remove the square root, and convert g to kg and kN m⁻¹ to N m⁻¹ before you substitute.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A 0.50 kg mass hangs on a spring of spring constant 20 N m⁻¹ and is set oscillating vertically. Calculate the time period of the oscillations, to 2 significant figures.
Show solution
F: T = 2π√(m/k)
I: T = 2π√(0.50 ÷ 20) = 2π√0.025
F: T is already the subject
A: 0.993… s = 0.99 s (2 s.f.)
Q2 (Core) A 0.20 kg mass oscillates on a spring of spring constant 45 N m⁻¹. Calculate (a) the time period, to 3 significant figures, and (b) the frequency of the oscillations, to 2 significant figures.
Show solution
F: T = 2π√(m/k)
I: T = 2π√(0.20 ÷ 45)
F: T is already the subject
A: (a) 0.4188… s = 0.419 s (3 s.f.)
(b) f = 1 ÷ T = 1 ÷ 0.4188… = 2.387… Hz = 2.4 Hz (2 s.f.)
Q3 (Core) A 250 g mass oscillates on a spring with a time period of 0.80 s. Calculate the spring constant, to 2 significant figures.
Show solution
Convert first: 250 g = 0.250 kg
F: T = 2π√(m/k), so T² = 4π²m ÷ k
I: 0.80² = 4π² × 0.250 ÷ k
F: k = 4π² × 0.250 ÷ 0.64
A: 15.42… N m⁻¹ = 15 N m⁻¹ (2 s.f.)
Watch out: forgetting to square T when rearranging. Square both sides before you make k the subject.
Q4 (Core) A spring of spring constant 1.2 kN m⁻¹ supports a mass that oscillates with a time period of 0.25 s. Calculate the mass, to 2 significant figures.
Show solution
Convert first: 1.2 kN m⁻¹ = 1200 N m⁻¹
F: T = 2π√(m/k), so T² = 4π²m ÷ k
I: 0.25² = 4π² × m ÷ 1200
F: m = kT² ÷ 4π² = 1200 × 0.0625 ÷ 4π²
A: 1.899… kg = 1.9 kg (2 s.f.)
Q5 (Exam) A 0.40 kg mass on a spring oscillates with a time period of 0.60 s. (a) The mass is doubled to 0.80 kg. Calculate the new time period, to 2 significant figures. (b) What mass on the same spring would give a time period of 0.30 s?
Show solution
F: T = 2π√(m/k), so T ∝ √m for the same spring
I: (a) Tnew = 0.60 × √(0.80 ÷ 0.40) = 0.60 × √2
F: Tnew is already the subject
A: 0.848… s = 0.85 s (2 s.f.)
(b) halving T means m must be multiplied by (½)² = ¼, so m = 0.40 ÷ 4 = 0.10 kg
Watch out: doubling the mass does not double the period. T is proportional to √m, so T only increases by a factor of √2 ≈ 1.41.
Q6 (Exam) In a required practical, a student hangs a 0.400 kg mass on a spring and uses a stopwatch to time 20 complete oscillations. The total time is 15.6 s and her reaction time uncertainty is ±0.2 s. (a) Calculate the spring constant, to 2 significant figures. (b) Explain, with calculations, why she timed 20 oscillations rather than one.
Show solution
Step 1 – time period
T = 15.6 ÷ 20 = 0.780 s
Step 2 – (a) spring constant
F: T² = 4π²m ÷ k
I: 0.780² = 4π² × 0.400 ÷ k
F: k = 4π² × 0.400 ÷ 0.780²
A: 25.95… N m⁻¹ = 26 N m⁻¹ (2 s.f.)
Step 3 – (b) percentage uncertainty
20 oscillations: 0.2 ÷ 15.6 × 100 = 1.3%; one oscillation: 0.2 ÷ 0.780 × 100 = 26%
A: the same ±0.2 s uncertainty is a much smaller fraction of a longer time, so timing many oscillations reduces the percentage uncertainty in T (from about 26% to 1.3%)
Watch out: dividing by 20 does not remove the reaction time error, it shares it out. Also time from the equilibrium position, where the mass moves fastest, using a fiducial marker.
Q7 (Exam) When a 300 g mass is hung on a spring, the spring extends by 4.9 cm and the mass comes to rest. The mass is then pulled down a little and released. Calculate (a) the spring constant and (b) the time period of the oscillations, both to 2 significant figures. (g = 9.81 m s⁻²)
Show solution
Convert first: 300 g = 0.300 kg; 4.9 cm = 0.049 m
Step 1 – (a) spring constant from Hooke’s law
at equilibrium, kΔL = mg, so k = 0.300 × 9.81 ÷ 0.049 = 2.943 ÷ 0.049 = 60.06… N m⁻¹
A: 60 N m⁻¹ (2 s.f.)
Step 2 – (b) time period
F: T = 2π√(m/k)
I: T = 2π√(0.300 ÷ 60.06…)
F: T is already the subject
A: 0.444… s = 0.44 s (2 s.f.)
Q8 (Exam) A 0.75 kg mass on a spring of spring constant 120 N m⁻¹ is attached to a vibration generator. The frequency of the generator is slowly increased from 0.5 Hz. Calculate the frequency at which the amplitude of the mass is a maximum, to 2 significant figures.
Show solution
Step 1 – natural period
F: T = 2π√(m/k)
I: T = 2π√(0.75 ÷ 120)
F: T is already the subject
A: T = 0.4967… s
Step 2 – natural frequency
f = 1 ÷ T = 1 ÷ 0.4967… = 2.013… Hz
A: resonance happens when the driving frequency equals the natural frequency, so the amplitude is a maximum at 2.0 Hz (2 s.f.)
Watch out: quoting the period (0.50 s) when a frequency is asked for. Always finish with f = 1 ÷ T.
Q9 (Stretch) Two identical light springs each have a spring constant of 20 N m⁻¹. A 0.60 kg mass is hung from (a) one spring, (b) the two springs in series (one below the other), and (c) the two springs in parallel (side by side). Calculate the time period in each case, to 2 significant figures.
Show solution
Step 1 – effective spring constants
series: 1 ÷ k = 1 ÷ 20 + 1 ÷ 20, so k = 10 N m⁻¹; parallel: k = 20 + 20 = 40 N m⁻¹
Step 2 – time periods
F: T = 2π√(m/k)
I: (a) T = 2π√(0.60 ÷ 20); (b) T = 2π√(0.60 ÷ 10); (c) T = 2π√(0.60 ÷ 40)
F: T is already the subject
A: (a) 1.088… s = 1.1 s; (b) 1.539… s = 1.5 s; (c) 0.7695… s = 0.77 s (all 2 s.f.)
the series arrangement has a quarter of the parallel spring constant, so its period is √4 = 2 times longer
Watch out: springs combine the opposite way to resistors. In series the extensions add, so the combination is less stiff.
Q10 (Stretch) A 0.80 kg mass oscillates on a spring of spring constant 50 N m⁻¹ with an amplitude of 4.0 cm. (a) Calculate the time period, to 3 significant figures. (b) Calculate the maximum speed of the mass, to 2 significant figures. (c) Show that the maximum kinetic energy equals the maximum elastic potential energy stored, ½kA².
Show solution
Convert first: 4.0 cm = 0.040 m
Step 1 – (a) time period
F: T = 2π√(m/k)
I: T = 2π√(0.80 ÷ 50) = 2π√0.016
A: 0.7947… s = 0.795 s (3 s.f.)
Step 2 – (b) maximum speed
ω = 2π ÷ T = √(k/m) = √(50 ÷ 0.80) = 7.905… rad s⁻¹
vmax = ωA = 7.905… × 0.040 = 0.3162… m s⁻¹
A: 0.32 m s⁻¹ (2 s.f.)
Step 3 – (c) energy check
maximum kinetic energy = ½mvmax² = 0.5 × 0.80 × 0.3162…² = 0.040 J
½kA² = 0.5 × 50 × 0.040² = 0.040 J, so the two are equal
A: 0.040 J in both cases
Watch out: leaving the amplitude in cm. 4.0 instead of 0.040 makes the energy 10 000 times too big.
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