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Two specific oscillators dominate the AQA A-Level Physics treatment of simple harmonic motion: the mass-on-a-spring (horizontal or vertical) and the simple pendulum. The first is the textbook archetype because its restoring force, F = −kx, is exactly linear; the second is a near-miss, linear only in the small-angle limit, but is the basis for centuries of timekeeping. Both feature in Required Practical 10, where students measure period as a function of mass (spring) or length (pendulum) to test the predicted T(m) and T(L) relationships. This lesson develops both systems from first principles, derives their period formulas, and works through the laboratory investigation that anchors them.
Spec mapping: This lesson sits under AQA 7408 section 3.6.1.2 and provides the foundation for Required Practical 10. It covers the derivation of T = 2π√(m/k) for a mass-spring oscillator from Hooke's law and Newton's second law, the small-amplitude derivation of T = 2π√(L/g) for the simple pendulum from sin θ ≈ θ, the limits of validity of both formulas, and the laboratory procedures used to measure T as a function of m or L. The course's RP10 anchor lesson (lesson 7) deepens the experimental treatment. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.6.1.2 (SHM defining condition): the systems here are the canonical examples of the a = −ω²x equation, and the algebraic templates used to derive their periods generalise to any linear restoring system.
- Section 3.4 (energy conservation in mechanics): the energy bookkeeping for a vertical mass-spring (where both elastic and gravitational PE are involved) is more delicate than for the horizontal case and is a routine synoptic target.
- Section 3.7.2 (gravitational fields — g variation): the pendulum period's dependence on g means a pendulum clock runs slightly differently at different latitudes and altitudes. Foucault's pendulum, the seconds pendulum (designed for T = 2 s, requiring L ≈ 0.994 m), and the use of pendulums to map local g are excellent synoptic applications.
Consider a block of mass m attached to one end of a light spring whose other end is fixed. The spring has natural length and spring constant k. Place the block on a frictionless horizontal surface and define x as the displacement of the block from its equilibrium position (where x = 0 means the spring is at its natural length and exerts no force on the block).
When the block is at displacement x, the spring's restoring force on the block is given by Hooke's law:
F_spring = −kx
The minus sign is critical: when x is positive (spring stretched), the spring pulls the block back towards equilibrium; when x is negative (spring compressed), the spring pushes the block forward towards equilibrium. The restoring force is always opposite to the displacement.
By Newton's second law, the block's acceleration is
a = F/m = −kx/m = −(k/m)x.
Comparing with the defining condition for SHM (a = −ω²x):
ω² = k/m → ω = √(k/m)
The block performs SHM with angular frequency √(k/m). The period is
T = 2π/ω = 2π√(m/k)
and the frequency is
f = 1/T = (1/2π)√(k/m).
Notice two important features:
When two springs of spring constants k₁ and k₂ are used together, the effective spring constant depends on how they are connected.
Parallel (both springs share the displacement): the springs act independently and contribute additively to the restoring force. The block experiences a total restoring force F_total = −(k₁ + k₂)x, so
k_eff (parallel) = k₁ + k₂
The system is stiffer in parallel; the period drops.
Series (the springs share the load and add their extensions): the same force passes through both springs and their extensions add. The effective spring constant is the harmonic-mean-style combination
1/k_eff (series) = 1/k₁ + 1/k₂ → k_eff = k₁k₂/(k₁ + k₂)
The system is less stiff in series; the period rises.
A 0.30 kg block sits on a frictionless track, attached to a spring of spring constant 120 N m⁻¹. The block is pulled 0.040 m from equilibrium and released. Calculate (a) the angular frequency, (b) the period, (c) the maximum speed, (d) the speed at x = 0.020 m, (e) the maximum acceleration.
(a) ω = √(k/m) = √(120/0.30) = √400 = 20.0 rad s⁻¹.
(b) T = 2π/ω = 2π/20.0 = 0.314 s.
(c) v_max = Aω = 0.040 × 20.0 = 0.800 m s⁻¹.
(d) v = ω√(A² − x²) = 20.0 × √(0.0016 − 0.0004) = 20.0 × √0.0012 = 20.0 × 0.0346 = 0.693 m s⁻¹.
(e) a_max = Aω² = 0.040 × 400 = 16.0 m s⁻².
When a spring hangs vertically with a mass attached, gravity plays a role. At the natural length of the spring, the only force on the mass is its weight mg — there is a net downward force, so the mass is not at equilibrium. Equilibrium is at the stretched position where the spring's upward force balances gravity:
kx_eq = mg → x_eq = mg/k.
Now define the variable y as the displacement of the mass from this dynamical equilibrium. The forces on the mass are: weight mg downward, and spring force k(x_eq + y) upward (because the total stretch is x_eq + y). The net force is
F_net = k(x_eq + y) − mg, taking up as positive in the spring expression.
Wait — let's redo with clean sign conventions. Take downward as positive. The mass is at position y below the equilibrium (y positive means stretched further). The forces are:
Net force: mg − k(x_eq + y) = mg − kx_eq − ky = 0 − ky = −ky (because at equilibrium kx_eq = mg).
The mg and kx_eq cancel exactly. The net force is just −ky, exactly the restoring force you would have on a horizontal spring with the equilibrium shifted to y = 0.
The vertical mass-spring system performs SHM about the stretched equilibrium with the same period T = 2π√(m/k) as the horizontal case.
Gravity does not change the period. It only shifts the equilibrium position. This is a routine exam result.
A 0.50 kg mass hangs in equilibrium on a vertical spring, stretching it by 0.10 m. The mass is then pulled down by an additional 0.030 m and released. Calculate (a) the spring constant, (b) the period of oscillation, (c) the maximum speed.
(a) At equilibrium: kx_eq = mg → k = mg/x_eq = 0.50 × 9.81/0.10 = 49.05 N m⁻¹.
(b) T = 2π√(m/k) = 2π√(0.50/49.05) = 2π × 0.101 = 0.634 s.
(c) ω = 2π/T = 9.90 rad s⁻¹. v_max = Aω = 0.030 × 9.90 = 0.297 m s⁻¹.
Note that A is the amplitude of the SHM about the stretched equilibrium, not the original stretch x_eq.
A simple pendulum consists of a small (point-like) bob of mass m attached to a light, inextensible string of length L. The other end of the string is fixed. When the bob is displaced through an angle θ from the vertical, gravity exerts a torque that pulls it back towards equilibrium.
Take the equilibrium position as θ = 0 (bob directly below the pivot). The bob hangs at the end of the string of length L. Its position along the arc, measured from equilibrium, is
s = Lθ (arc length, by definition of radian measure).
The forces on the bob are: weight mg downward, and tension T along the string upward through the pivot. The tension has no component along the arc (it is purely radial); the weight does have a tangential component.
Resolving the weight along the arc direction (tangent to the swing), the tangential component is
F_tangential = −mg sin θ
The minus sign indicates that the tangential force is opposite to the displacement s = Lθ: when the bob is to the right of vertical (θ > 0), the tangential gravity component points to the left, back towards equilibrium. The magnitude is mg sin θ.
The tangential acceleration of the bob is a_tan = dv_tan/dt = L(d²θ/dt²) (because v_tan = L(dθ/dt) and the arc radius L is constant).
By Newton's second law along the arc:
m × L(d²θ/dt²) = −mg sin θ → d²θ/dt² = −(g/L) sin θ.
This is not the SHM equation; it is the pendulum equation, which contains sin θ rather than θ. For large θ, the pendulum equation is non-linear and the motion is not SHM.
For small angles (typically θ less than about 10° or 0.17 rad), the Taylor expansion of sin θ gives
sin θ = θ − θ³/6 + θ⁵/120 − ... ≈ θ (for small θ).
In this limit the pendulum equation linearises to
d²θ/dt² ≈ −(g/L)θ → a_angular = −(g/L)θ.
Comparing with the SHM defining condition (in angular form) a_angular = −ω²θ:
ω² = g/L → ω = √(g/L)
The period is
T = 2π/ω = 2π√(L/g).
This is the small-amplitude period of a simple pendulum. It depends on L and g — and crucially, like all SHM, not on the amplitude. To the accuracy of the small-angle approximation, a pendulum is isochronous.
Notice also that T does not depend on the mass of the bob. A pendulum's period is set entirely by its length and the local gravitational field strength. This is why pendulums have historically been used to measure g: timing the period and inverting the formula gives g = 4π²L/T².
The small-angle approximation sin θ ≈ θ is accurate to within about 0.5% for θ ≤ 10°, and within about 4% at 30°. The period of a real pendulum grows slowly with amplitude — at θ_max = 30°, T is about 1.7% longer than the small-angle prediction. The exact period requires a complete elliptic integral and cannot be written in closed form, but the leading correction is
T(θ_max) ≈ 2π√(L/g) × (1 + θ_max²/16 + ...)
For exam purposes, the small-angle formula T = 2π√(L/g) is accepted whenever amplitudes are stated as "small" or less than about 10°.
A grandfather clock pendulum has length 1.00 m. Calculate (a) its period at the surface of the Earth (g = 9.81 m s⁻²), (b) its frequency, (c) the period if the clock is taken to the equator where g = 9.78 m s⁻².
(a) T = 2π√(L/g) = 2π√(1.00/9.81) = 2π × 0.3193 = 2.006 s.
(b) f = 1/T = 0.499 Hz.
(c) T' = 2π√(1.00/9.78) = 2π × 0.3198 = 2.010 s.
The pendulum runs about 0.2% slower at the equator than at higher latitudes, which corresponds to about 1.7 minutes per day. Pre-electronic clocks had to be calibrated for their installation latitude.
Required Practical 10 in AQA A-Level Physics asks students to investigate the factors that affect the period of a mass-spring system or pendulum and to verify the theoretical predictions.
Hypothesis: T = 2π√(m/k), so T² = (4π²/k) × m. A graph of T² against m should be a straight line through the origin with gradient 4π²/k.
Procedure (summary): suspend a vertical spring from a clamp, attach a mass hanger, add masses in regular increments (e.g. 100, 200, 300, ..., 800 g). For each mass, pull the hanger down by a small amount (about 2 cm — the amplitude should be small compared with the natural length to avoid coil overlap) and release. Time 10 complete oscillations with a digital stopwatch; divide by 10 to get T. Repeat each measurement three times and take the mean. Plot T² against m and read off k from the gradient.
Sources of uncertainty: stopwatch reaction time (typically ±0.2 s on a single oscillation, reduced by a factor of 10 if 10 oscillations are timed), starting the timer accurately at the extreme, parallax in reading the displacement amplitude. The "many-oscillation" trick (timing 10 cycles rather than 1) is the most important uncertainty-reduction technique.
Hypothesis: T = 2π√(L/g), so T² = (4π²/g) × L. A graph of T² against L should be a straight line through the origin with gradient 4π²/g.
Procedure (summary): suspend a pendulum bob from a clamp using a light string. Measure the length L from the suspension point to the centre of mass of the bob (using calipers or a ruler with the bob lifted into measurement position). Displace the bob through a small angle (less than 10°) and release. Time 10 complete oscillations and divide by 10 to get T. Repeat for L values across a wide range (e.g. 0.20, 0.40, 0.60, 0.80, 1.00 m). Plot T² against L and read off g.
Sources of uncertainty: measuring L to the centre of mass rather than the surface of the bob is the most common error (introduces a systematic offset). The small-angle requirement should be checked — too-large amplitudes give a period that is biased upward by a few per cent.
(Lesson 7 develops RP10 in full detail, including a complete write-up of uncertainty analysis, CPAC competencies and improvements.)
flowchart TD
A["RP10: investigate factors<br/>affecting period of SHM"] --> B{"Choose system"}
B -->|"Mass-spring"| C["Vary m, fix k<br/>Plot T² vs m"]
B -->|"Pendulum"| D["Vary L, fix g<br/>Plot T² vs L"]
C --> E["Gradient = 4π²/k<br/>→ k = 4π²/gradient"]
D --> F["Gradient = 4π²/g<br/>→ g = 4π²/gradient"]
E --> G["Time 10 oscillations<br/>to reduce uncertainty"]
F --> G
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