Simple Harmonic Motion (SHM) is the motion that results when an object experiences a restoring force PROPORTIONAL to its displacement: F = −kx. The negative sign means the force always points back toward equilibrium. The further you pull it, the harder it pulls back. Mass-spring systems are the cleanest example. Pendulums are SHM for small angles.
Key Properties
Restoring Force
F = −kx. Linear in displacement.
Amplitude (A)
Maximum displacement.
Position vs Time
Sinusoidal: x(t) = A·cos(ωt + φ).
Pendulum Caveat
SHM only for small angles (< ~15°).
Example
Spring force at various displacements
A spring has constant k = 50 N/m. When stretched 0.1 m, the force is F = −kx = −(50)(0.1) = −5 N (pulling back toward equilibrium).
When stretched 0.2 m, F = −(50)(0.2) = −10 N. Double the stretch → double the restoring force. This linear relationship is what makes the motion simple harmonic.
Topic 7.2
Frequency and Period of SHM
Period depends on the system, not on amplitude
The Big Idea
The two formulas you must memorize: T_spring = 2π√(m/k) and T_pendulum = 2π√(L/g). Both have a remarkable property: amplitude does NOT appear. Pull a spring back twice as far, and it still oscillates at the same period. The frequency is f = 1/T, in hertz.
Key Equations
Mass-Spring
T = 2π√(m/k).
Pendulum (small angle)
T = 2π√(L/g).
Frequency
f = 1/T (Hz).
Angular Frequency
ω = 2π/T = 2πf.
Example
Period of a mass on a spring
A 2 kg block is attached to a spring with constant k = 200 N/m. What is the period?
T = 2π√(m/k) = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.63 s
If you DOUBLE the amplitude (say, from 0.05 m to 0.10 m), the period stays at 0.63 s. If you double the mass to 4 kg, T grows by √2: T = 0.89 s. If you double k to 400 N/m, T shrinks by √2: T = 0.44 s.
Topic 7.3
Representing and Analyzing SHM
Position, velocity, and acceleration in sync
The Big Idea
In SHM, position traces out a sine/cosine curve in time: x(t) = A·cos(ωt + φ). Velocity is the derivative — also sinusoidal, but shifted 90°: v(t) = −Aω·sin(ωt + φ). Acceleration is shifted another 90° from velocity: a(t) = −Aω²·cos(ωt + φ) = −ω²·x. Key takeaway: v is max where x is zero; a is max where x is max (but opposite direction).
Where Things Are Max and Min
x_max = A
At turning points. v = 0, |a| = max.
v_max = Aω
At equilibrium (x = 0). a = 0.
a_max = Aω²
At turning points (|x| = A).
Phase Relations
x leads v by 90°; v leads a by 90°.
Example
Maximum speed of a mass on a spring
A 1 kg block on a spring (k = 100 N/m) oscillates with amplitude A = 0.05 m. Find the maximum speed.
ω = √(k/m) = √(100/1) = 10 rad/s
v_max = Aω = (0.05)(10) = 0.5 m/s
The maximum acceleration: a_max = Aω² = (0.05)(100) = 5 m/s² (at the turning points).
Topic 7.4
Energy of Simple Harmonic Oscillators
PE and KE trade off; total stays constant
The Big Idea
The total mechanical energy of an SHM oscillator is E = ½kA². It's constant — energy conservation. The energy swaps continuously between PE (½kx², stored in the spring) and KE (½mv², the moving mass). At the turning points, all the energy is PE. At equilibrium, all the energy is KE. In between, it's split.
Energy at Key Positions
At Turning Points (x = ±A)
U = ½kA²; K = 0. ALL potential.
At Equilibrium (x = 0)
U = 0; K = ½kA². ALL kinetic.
In Between
½mv² + ½kx² = ½kA².
Double the Amplitude
→ Quadruple the energy.
Example
Speed at half-amplitude
A 0.5 kg block on a spring (k = 200 N/m) has amplitude A = 0.10 m. Find the speed when x = 0.05 m (half-amplitude).
Energy conservation: ½kA² = ½kx² + ½mv²
½(200)(0.10)² = ½(200)(0.05)² + ½(0.5)v²
1.0 = 0.25 + 0.25v² → 0.75 = 0.25v² → v² = 3 → v ≈ 1.73 m/s
Compare to v_max = A·√(k/m) = 0.1·√400 = 2 m/s. At x = A/2, speed is about 87% of max.
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How to use this visual review
Take 1–2 minutes per slide. Read the Big Idea first, then definitions, then the worked example. Four slides total — one per topic.
The two formulas you must master: T_spring = 2π√(m/k) and T_pendulum = 2π√(L/g). Most Unit 7 problems start from one of these.