Mass-Spring Systems and Pendulums

The two most important physical systems that exhibit SHM at A-Level are the mass-spring system and the simple pendulum. Each has its own period equation, and understanding where these equations come from — and their limitations — is essential for exam success.

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The Mass-Spring System

Deriving the Period

For a mass m attached to a spring of spring constant k, Hooke's law gives the restoring force:

F = −kx

By Newton's second law, F = ma, so:

ma = −kx

a = −(k/m)x

Comparing this with the SHM equation a = −ω²x:

ω² = k/m

ω = √(k/m)

Since T = 2π/ω:

T = 2π√(m/k)

Key Features

• The period increases with mass (heavier objects oscillate more slowly)
• The period decreases with spring constant (stiffer springs give faster oscillations)
• The period does not depend on amplitude (isochronicity)
• The period does not depend on gravitational field strength — a mass-spring system would have the same period on the Moon as on Earth

Assumptions

The equation T = 2π√(m/k) assumes:

• The spring obeys Hooke's law (force proportional to extension)
• The spring is massless (all inertia comes from the attached mass)
• There is no damping (no friction or air resistance)
• Oscillations are along the spring's axis (no lateral motion)

For a vertical mass-spring system, the equilibrium position is shifted downward by the weight of the mass, but the period equation is the same. The gravitational force provides a constant offset but does not affect the restoring force, which depends only on displacement from the new equilibrium.

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The Simple Pendulum

Deriving the Period

A simple pendulum consists of a point mass (bob) m suspended by a light, inextensible string of length L. When displaced by a small angle θ from vertical, the component of gravitational force along the arc is:

F = −mg sin θ

For the arc length s = Lθ (where θ is in radians), the acceleration along the arc is:

a = −g sin θ

For small angles, sin θ ≈ θ (in radians), so:

a ≈ −gθ = −(g/L)s

This is SHM with ω² = g/L. Therefore:

T = 2π√(L/g)

Key Features

• The period increases with length (longer pendulums swing more slowly)
• The period decreases with gravitational field strength (a pendulum on the Moon has a longer period than on Earth)
• The period does not depend on the mass of the bob
• The period does not depend on amplitude (for small oscillations)

The Small-Angle Approximation

The SHM treatment of a pendulum relies on sin θ ≈ θ. How accurate is this?

Angle (°)	sin θ	θ (rad)	% error in sin θ ≈ θ
5	0.0872	0.0873	0.1%
10	0.1736	0.1745	0.5%
15	0.2588	0.2618	1.2%
20	0.3420	0.3491	2.1%
30	0.5000	0.5236	4.7%
45	0.7071	0.7854	11.1%

For angles up to about 10°, the approximation introduces less than 1% error, and the motion is very close to SHM. Beyond 15°, the error grows significantly, and the period begins to depend on amplitude.

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Using SHM to Measure g

A simple pendulum provides an elegant method to determine the local gravitational field strength g:

1. Measure the period T for different string lengths L
2. Plot T² against L — this should give a straight line through the origin
3. The gradient of the line is 4π²/g
4. Calculate g from the gradient

Since T = 2π√(L/g), squaring gives T² = 4π²L/g. Comparing with y = mx, the gradient is 4π²/g.

Sources of Uncertainty

• Timing many oscillations and dividing (reduces percentage uncertainty in T)
• Measuring L from the pivot to the centre of mass of the bob (not the bottom)
• Ensuring small amplitude oscillations (so the small-angle approximation holds)
• Minimising air resistance and friction at the pivot
• Starting and stopping the timer at the same point in the cycle (e.g., as the bob passes through the centre)

Worked Example: Determining g

A student obtains the following data:

L (m)	T for 20 oscillations (s)	T (s)	T² (s²)
0.300	22.0	1.10	1.21
0.500	28.4	1.42	2.02
0.700	33.6	1.68	2.82
0.900	38.0	1.90	3.61
1.100	42.0	2.10	4.41

Gradient of T² vs L = (4.41 − 1.21)/(1.100 − 0.300) = 3.20/0.800 = 4.00 s² m⁻¹

g = 4π²/gradient = 39.48/4.00 = 9.87 m s⁻²

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Using SHM to Measure k

Similarly, a mass-spring system can determine the spring constant k:

1. Measure the period T for different masses m
2. Plot T² against m — this gives a straight line through the origin
3. The gradient is 4π²/k
4. Calculate k from the gradient

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Combinations of Springs

Springs in Parallel

When two springs (constants k₁ and k₂) support a mass side by side:

k_effective = k₁ + k₂

The system is stiffer, so the period is shorter.

Springs in Series

When two springs are connected end to end:

1/k_effective = 1/k₁ + 1/k₂

k_effective = k₁k₂/(k₁ + k₂)

The system is less stiff, so the period is longer.

Worked Example: Springs in Series and Parallel

Two identical springs each have k = 50 N m⁻¹. A 0.2 kg mass is attached.

In parallel: k_eff = 50 + 50 = 100 N m⁻¹ T = 2π√(0.2/100) = 2π√(0.002) = 2π × 0.0447 = 0.281 s

In series: k_eff = (50 × 50)/(50 + 50) = 2500/100 = 25 N m⁻¹ T = 2π√(0.2/25) = 2π√(0.008) = 2π × 0.0894 = 0.562 s

The series arrangement gives exactly double the period of the parallel arrangement.

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Comparing Mass-Spring and Pendulum Systems

Property	Mass-Spring	Pendulum
Period equation	T = 2π√(m/k)	T = 2π√(L/g)
Period depends on	m, k	L, g
Period independent of	A, g	A, m
Works on the Moon?	Same period	Longer period (smaller g)
Works in zero gravity?	Yes (same period)	No (no restoring force)
Energy stored as	Elastic PE	Gravitational PE
Restoring force	Spring force (−kx)	Component of gravity (−mg sin θ)
Small-angle approximation needed?	No	Yes
Practical applications	Vehicle suspension, seismometers	Clocks, measuring g

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Worked Example: Vertical Mass-Spring System

A vertical spring has a natural length of 0.30 m. When a 0.15 kg mass is hung from it, the spring extends to 0.34 m at static equilibrium.

(a) Find the spring constant.

At equilibrium: mg = kΔx k = mg/Δx = 0.15 × 9.81/0.04 = 36.8 N m⁻¹

(b) Find the period of vertical oscillations.

T = 2π√(m/k) = 2π√(0.15/36.8) = 2π√(0.00408) = 2π × 0.0639 = 0.401 s

(c) The mass is pulled down a further 3 cm and released. Find the maximum speed.

A = 0.03 m, ω = 2π/T = 2π/0.401 = 15.67 rad s⁻¹

v_max = Aω = 0.03 × 15.67 = 0.470 m s⁻¹

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Worked Example: Pendulum on Different Planets

A pendulum has a period of 2.00 s on Earth (g = 9.81 m s⁻²). Find its period on:

(a) The Moon (g = 1.63 m s⁻²)

T_Moon/T_Earth = √(g_Earth/g_Moon) = √(9.81/1.63) = √6.02 = 2.45

T_Moon = 2.00 × 2.45 = 4.90 s

(b) Jupiter (g = 24.8 m s⁻²)

T_Jupiter/T_Earth = √(9.81/24.8) = √0.396 = 0.629

T_Jupiter = 2.00 × 0.629 = 1.26 s

(c) Inside the ISS (g ≈ 0)

The pendulum would not oscillate at all — there is no gravitational restoring force. The bob would float at whatever angle it was released. This is why mass-spring systems are used to measure mass in microgravity.

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Real-World Application: Measuring Mass in Space

On the International Space Station, astronauts cannot use conventional scales (which rely on weight). Instead, the Body Mass Measurement Device (BMMD) uses a spring-based oscillation system. The astronaut sits on a chair attached to a spring, and the period of oscillation is measured. Since T = 2π√(m/k) and k is known, the mass can be calculated:

m = kT²/(4π²)

This technique works perfectly in microgravity because the mass-spring period depends on inertial mass, not weight.

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A-Level Deep Dive: Mass-Spring Systems and Pendulums

Spec mapping

Edexcel 9PH0 specification Topic 13 — Oscillations covers simple harmonic motion, the mass-spring oscillator, the simple pendulum, and the conditions under which each obeys SHM (refer to the official specification document for exact wording). Mass-spring and pendulum systems are the canonical concrete realisations of the abstract SHM equation a = -ω²x. They are examined principally in 9PH0 Paper 2 (Advanced Physics II), where Topic 13 sits alongside thermodynamics, gravitation and astrophysics, but synoptic links push them into Paper 1 (forces, energy) and Paper 3 (general and practical). Required-practical CP9 (simple pendulum) and CP12 (mass-spring SHM) are also explicitly assessable, including in Paper 3 short-answer practical questions.

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Worked example with full mark scheme

Question (8 marks):

(a) A trolley of mass m is attached between two identical springs, each of spring constant k, on a frictionless horizontal track. By considering the net force when the trolley is displaced by x from equilibrium, derive an expression for the period T of small oscillations. (5)

(b) The trolley has m = 0.40 kg and each spring has k = 25 N m⁻¹. Calculate T, and state how T would change if m were doubled. (3)

Solution with mark scheme:

(a) Step 1 — write the restoring force.

When the trolley is displaced by x, one spring is stretched by x and the other compressed by x. Both springs pull (or push) the trolley back toward equilibrium, so the net restoring force is:

$F_{\text{net}} = -kx - kx = -2kx$Fnet​=−kx−kx=−2kx

M1 — recognising that both springs contribute to the restoring force (the most common error is using only one spring constant). The negative sign indicates the force opposes displacement.

Step 2 — apply Newton's second law.

$ma = -2kx \implies a = -\dfrac{2k}{m}x$ma=−2kx⟹a=−m2k​x

M1 — correctly applying F = ma to obtain a in terms of x.

Step 3 — identify the SHM form.

Comparing with the defining SHM equation a = -ω²x:

$\omega^2 = \dfrac{2k}{m} \implies \omega = \sqrt{\dfrac{2k}{m}}$ω2=m2k​⟹ω=m2k​​

A1 — correct identification of ω². Candidates who write ω = 2k/m (forgetting the square root) lose this mark.

Step 4 — convert ω to period.

Using T = 2π/ω:

$T = 2\pi\sqrt{\dfrac{m}{2k}}$T=2π2km​​

M1 A1 — correct manipulation and final expression. The "m on top, k on bottom inside the root" pattern is the universal mass-spring fingerprint.

(b) Step 1 — substitute values. With m = 0.40 kg and k = 25 N m⁻¹ (so 2k = 50 N m⁻¹):

$T = 2\pi\sqrt{\dfrac{0.40}{50}} = 2\pi\sqrt{0.008} = 2\pi \times 0.0894 \approx 0.562 \text{ s}$T=2π500.40​​=2π0.008​=2π×0.0894≈0.562 s

M1 — correct substitution. A1 — T ≈ 0.56 s (2 s.f.) or 0.562 s (3 s.f.).

Step 2 — effect of doubling m. Since T ∝ √m, doubling m multiplies T by √2 ≈ 1.41. B1 — quoted as a factor of √2, not simply "doubles".

Total: 8 marks (M4 A3 B1).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (7 marks): A simple pendulum of length L = 1.20 m hangs from the ceiling of a stationary lift. The bob has mass 0.15 kg.

(a) Calculate the period of small oscillations. Use g = 9.81 N kg⁻¹. (2)

(b) The lift now accelerates upward at 2.50 m s⁻². Without re-deriving the period formula, explain qualitatively how T changes, and calculate the new period. (3)

(c) The amplitude is 8.0°. State, with a reason, whether the small-angle approximation remains valid. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO2) — substituting into T = 2π√(L/g): T = 2π√(1.20/9.81).
• A1 (AO2) — T ≈ 2.20 s.

(b)

• M1 (AO1) — recognising that in the lift's accelerating frame, the effective gravitational field strength is g_eff = g + a (apparent weight increases).
• M1 (AO2) — substituting g_eff = 9.81 + 2.50 = 12.31 N kg⁻¹ into T = 2π√(L/g_eff).
• A1 (AO2) — T_new ≈ 2π√(1.20/12.31) ≈ 1.96 s, shorter than before.

(c)

• B1 (AO3) — stating that 8° in radians is ≈ 0.140 rad, and sin(0.140) ≈ 0.1395, a discrepancy of about 0.3%.
• B1 (AO3) — concluding that the small-angle approximation is valid (typical A-Level threshold is amplitudes below ~10–15°).

Total: 7 marks split AO1 = 1, AO2 = 4, AO3 = 2. Part (b) is the synoptic crux — it links Topic 13 oscillations to Topic 5 (forces) via the equivalence principle, a classically Edexcel-style cross-topic test.

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Synoptic links

Connects to:

1. Topic 13 SHM equations: the mass-spring derivation produces a = -(k/m)x, exactly the canonical SHM form a = -ω²x. Without confidence in the abstract SHM framework, the mass-spring and pendulum cases are isolated formulas; with it, they are two instances of one principle.

2. Topic 4 — Hooke's law (F = -kx): the very property that makes a spring obey SHM is its linear restoring force. Hooke's law is the cause of the SHM behaviour, and its breakdown (springs stretched beyond the elastic limit) is exactly the breakdown of SHM.