SHM is the heartbeat of oscillating systems — from pendulum clocks to atomic vibrations. Master the defining equation a = −ω²x and explore how energy flows between kinetic and potential forms.
📐State the defining condition a = −ω²x and explain the significance of the negative sign
🔢Use x = A cos(ωt) and v = −Aω sin(ωt) to find displacement and velocity at any time
⏱️Apply T = 2π√(m/k) for mass-spring and T = 2π√(L/g) for simple pendulum
⚡Describe the energy exchange between KE and PE in SHM
🔬Determine g and k from practical oscillation experiments
📊Sketch displacement, velocity and acceleration graphs and state their phase relationships
Defining Simple Harmonic Motion
An object executes Simple Harmonic Motion (SHM) when its acceleration is proportional to its displacement from the equilibrium position and always directed toward that equilibrium:
a = −ω²x
where x is displacement (m) and ω is angular frequency (rad s⁻¹). The negative sign is crucial — it shows the restoring nature: when x is positive (displaced right), a is negative (acceleration toward left/equilibrium). This makes the motion oscillatory.
SHM condition: a ∝ −x. The force must be a linear restoring force: F = −kx (Hooke's Law for a spring, or the restoring component of gravity for a pendulum at small angles).
The angular frequency ω is related to the ordinary frequency f and period T:
ω = 2πf = 2π/T
Maximum acceleration occurs at maximum displacement (the amplitude A): a_max = ω²A
Displacement, Velocity and Acceleration Equations
Starting from maximum positive displacement at t = 0, the displacement equation is:
x = A cos(ωt)
Differentiating to find velocity:
v = −Aω sin(ωt)
Differentiating again for acceleration:
a = −Aω² cos(ωt) = −ω²x ✓
The velocity as a function of displacement (eliminating t using sin²+cos²=1):
v = ±ω√(A² − x²)
Maximum velocity occurs at x = 0 (equilibrium): v_max = ωA
Phase relationships: displacement and acceleration are in antiphase (180° apart). Velocity leads displacement by 90° (π/2 rad). At maximum displacement, velocity is zero; at equilibrium, velocity is maximum.
Position
Displacement x
Velocity v
Acceleration a
Equilibrium
0
±ωA (maximum)
0
Amplitude
±A
0
∓ω²A (maximum magnitude)
Periods of Common SHM Systems
Mass-spring system: A mass m on a spring of spring constant k (N m⁻¹). The restoring force is F = −kx (Hooke's Law). Comparing with F = ma = m(−ω²x): k = mω², so ω = √(k/m).
T = 2π√(m/k)
Note: T is independent of amplitude for ideal SHM. T increases with greater mass (more inertia) and decreases with greater k (stiffer spring, stronger restoring force).
Simple pendulum: A point mass on a light inextensible string of length L. For small angles (θ < ~15°), the restoring force component along the arc ≈ −mg sinθ ≈ −mgθ = −(mg/L)x. This gives ω² = g/L:
T = 2π√(L/g)
Again, T is independent of amplitude (for small angles) and, remarkably, independent of the mass of the bob. T only depends on L and g — this is exploited in pendulum clocks and in measurements of g.
⚠️ The simple pendulum formula T = 2π√(L/g) only holds for small angles (sinθ ≈ θ in radians). For large amplitudes, the period increases slightly — this is a deviation from ideal SHM.
Energy in SHM
Energy continually transforms between kinetic energy (KE) and potential energy (PE) — elastic PE for a spring, gravitational PE for a pendulum. The total mechanical energy remains constant (no damping):
E_total = ½mω²A² = ½kA² (constant)
Kinetic energy at displacement x:
KE = ½mv² = ½mω²(A² − x²)
Potential energy at displacement x:
PE = ½mω²x²
Check: KE + PE = ½mω²(A²−x²) + ½mω²x² = ½mω²A² ✓
At the equilibrium position (x = 0): all energy is kinetic. At maximum displacement (x = ±A): all energy is potential. The energy–displacement graph shows KE as an inverted parabola and PE as an upright parabola, with their sum a horizontal line.
The frequency of energy oscillation is twice the frequency of displacement — PE and KE each go through two complete cycles per oscillation of x (think: both extremes are positions of maximum PE).
Isochronous Oscillations and Amplitude Independence
A key property of ideal SHM is isochronism — the period is independent of amplitude. This makes SHM systems useful as timekeepers (pendulums, quartz crystal oscillators). Isochronism arises because, at larger amplitudes, both the restoring force and the distance to be covered scale proportionally, keeping the period constant.
In practice, real systems deviate from ideal SHM at large amplitudes. A pendulum with θ > 15° oscillates with a slightly longer period. A spring oscillated beyond its elastic limit no longer obeys Hooke's Law. Both are departures from the a ∝ −x condition.
Isochronous: Having equal time — the property that the period of oscillation does not depend on amplitude, characteristic of ideal SHM systems.
A mass of 0.40 kg oscillates on a spring of spring constant 25 N m⁻¹ with amplitude 0.05 m. Find (a) the angular frequency, (b) the period, (c) the maximum velocity, (d) the velocity when x = 0.03 m.
L ≈ 1.00 m on Earth; T_Moon ≈ 4.92 s (longer period because gravity is weaker)
An object in SHM has displacement x = 0.08 cos(5πt) m. Find (a) the amplitude, (b) the period, (c) the displacement at t = 0.15 s, (d) the maximum acceleration.
1From x = A cos(ωt): A = 0.08 m and ω = 5π rad s⁻¹
2Period: T = 2π/ω = 2π/(5π) = 2/5 = 0.40 s
3At t = 0.15 s: x = 0.08 cos(5π × 0.15) = 0.08 cos(0.75π) = 0.08 cos(135°) = 0.08 × (−0.7071) = −0.0566 m
A = 0.08 m; T = 0.40 s; x(0.15s) = −0.057 m; a_max = 19.7 m s⁻²
A mass-spring system has total energy 0.18 J, spring constant 50 N m⁻¹. Find (a) the amplitude, (b) the KE when displacement is half the amplitude.
1Total energy E = ½kA²: A² = 2E/k = 2×0.18/50 = 0.0072 → A = 0.0849 m ≈ 0.085 m
2At x = A/2: KE = E − PE = ½kA² − ½k(A/2)² = ½kA²(1 − ¼) = ¾ × E = ¾ × 0.18 = 0.135 J
A = 0.085 m; KE at x = A/2 is 0.135 J (75% of total energy)
Q1. Which of the following is the correct definition of SHM?
Q2. A mass of 0.20 kg on a spring (k = 80 N m⁻¹) oscillates in SHM. What is the period?
Q3. At which position in SHM is the kinetic energy at a maximum?
Q4. A pendulum of length 0.25 m swings on Earth. What is its period?
Q5. An SHM oscillator has amplitude 0.12 m and ω = 10 rad s⁻¹. What is the maximum acceleration?
Challenge 1. A 0.30 kg mass oscillates on a spring (k = 48 N m⁻¹) with amplitude 0.10 m. (a) Write the equation for displacement if x = +A at t = 0. (b) At what time does the mass first reach x = +0.05 m (moving away from equilibrium)? (c) What is the KE at that point?
✓ ω = √(k/m) = √(48/0.30) = √160 = 12.65 rad s⁻¹. (a) x = 0.10 cos(12.65t) m. (b) 0.05 = 0.10 cos(12.65t) → cos(12.65t) = 0.5 → 12.65t = π/3 → t = π/(3×12.65) = 0.0826 s. (c) KE = ½mω²(A²−x²) = ½×0.30×160×(0.01−0.0025) = 0.5×0.30×160×0.0075 = 0.18 J. Alternatively: E_total = ½kA² = ½×48×0.01 = 0.24 J; PE = ½k(0.05)² = ½×48×0.0025 = 0.06 J; KE = 0.24−0.06 = 0.18 J.
Challenge 2. A student measures the period of a simple pendulum for different lengths and plots T² against L. The gradient is found to be 4.02 s² m⁻¹. (a) What value of g does this give? (b) The accepted value is 9.81 m s⁻². Calculate the percentage error. (c) Suggest two improvements to reduce systematic error.
✓ (a) From T² = (4π²/g)L, the gradient = 4π²/g → g = 4π²/gradient = 4π²/4.02 = 39.48/4.02 = 9.82 m s⁻². (b) % error = |9.82−9.81|/9.81 × 100% = 0.10%. Excellent result! (c) Improvements: (i) Measure from the pivot to the centre of mass of the bob (not the top of the string); (ii) Time at least 20 oscillations per measurement to reduce timing percentage uncertainty; (iii) Use a fiducial marker at the equilibrium position and start timing when the bob passes it (reduces reaction time error); (iv) Keep angle below 5° to ensure true SHM conditions.
Challenge 3 (Synoptic). Derive the expression T = 2π√(m/k) for a mass on a spring from Newton's second law and the SHM condition. Clearly state any assumptions made.
✓ Assumptions: spring is ideal (obeys Hooke's Law throughout motion), mass is a point mass, no damping (no friction or air resistance), oscillation is vertical or spring is horizontal. Derivation: By Hooke's Law, restoring force F = −kx (negative because it opposes displacement). By Newton's 2nd law: F = ma, so ma = −kx, giving a = −(k/m)x. Comparing with the SHM condition a = −ω²x: ω² = k/m → ω = √(k/m). Since T = 2π/ω: T = 2π/√(k/m) = 2π√(m/k). QED.
Challenge 4. Two identical springs (k = 30 N m⁻¹) are connected (a) in series, (b) in parallel, with a 0.5 kg mass. For each case, find the effective spring constant and the period of oscillation.
✓ (a) Series: 1/k_eff = 1/k₁ + 1/k₂ = 1/30 + 1/30 = 2/30 → k_eff = 15 N m⁻¹. T = 2π√(0.5/15) = 2π√0.0333 = 2π×0.1826 = 1.15 s. (b) Parallel: k_eff = k₁ + k₂ = 30+30 = 60 N m⁻¹. T = 2π√(0.5/60) = 2π√0.00833 = 2π×0.0913 = 0.574 s. Note: Springs in series are softer (lower k, longer period); springs in parallel are stiffer (higher k, shorter period).
Required Practical
Determining g and k from Oscillation Experiments
Part A: Simple Pendulum — Determining g
Apparatus: Clamp and stand, string (varying lengths 0.3–1.2 m), metal bob, split-cork pivot, metre rule, stopwatch, protractor.
Method:
Suspend pendulum from split cork. Measure L from pivot to centre of bob.
Displace bob by less than 5° from vertical. Release and time 20 complete oscillations.
Calculate T = total time / 20. Repeat for different lengths (at least 6 values).
Plot T² (y-axis) against L (x-axis). The graph should be a straight line through the origin.
Gradient = 4π²/g → g = 4π²/gradient.
Part B: Mass-Spring System — Determining k
Apparatus: Spring, stand and clamp, masses (varying 50–300 g), ruler/metre rule, stopwatch.
Method:
Hang spring vertically. Add a mass m and measure the extension e to find k from F = ke → k = mg/e (static method). This gives a preliminary estimate of k.
Dynamic method: displace mass slightly downward and release. Time 20 oscillations. Calculate T.
Repeat for at least 5 different masses. Plot T² against m.
Gradient = 4π²/k → k = 4π²/gradient.
Analysis and Error
Source of Error
Effect
Minimisation
Measuring L to centre of bob
Systematic under/overestimate of L
Use callipers to find bob diameter; add radius to string length
Large amplitude pendulum
Period longer than T = 2π√(L/g)
Keep angle < 5°; use protractor
Reaction time in timing
Random error in T
Time 20 oscillations; use fiducial marker at centre
Spring oscillating with mass of spring
Effective mass > m
Add m_spring/3 to hanging mass in analysis (advanced)
⚠️ For the mass-spring system, do not exceed the elastic limit of the spring — oscillations must be small. Ensure the retort stand is clamped firmly to the bench or a heavy base.