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A Level Physics - How to Solve: Simple Harmonic Motion (SHM) – Complete Guide

How to Solve: Simple Harmonic Motion (SHM) – Complete Guide

For GCSE & A-Level Physics (Edexcel, AQA, OCR, IB)

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Introduction

"Mastering Simple Harmonic Motion (SHM) unlocks 10–15% of your mechanics exam marks—think pendulum clocks, car suspension, and even molecular vibrations in chemistry. One question on displacement, velocity, or acceleration could be the difference between a 6 and a 9 at GCSE, or an A and an A at A-Level."

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WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Graphs of trigonometric functions (sine, cosine) – SHM is just a sine/cosine wave in motion.
2. Basic calculus (differentiation) – Velocity is the derivative of displacement; acceleration is the derivative of velocity. (A-Level only)
3. Hooke’s Law – Springs obey F = -kx, where k is the spring constant.

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KEY TERMS & FORMULAS

Key Terms

Term	Definition
Amplitude (A)	Maximum displacement from equilibrium (metres, m).
Period (T)	Time for one complete oscillation (seconds, s).
Frequency (f)	Number of oscillations per second (Hertz, Hz). f = 1/T
Angular frequency (ω)	ω = 2πf or ω = 2π/T (radians per second, rad/s).
Phase difference (φ)	How "out of sync" two oscillations are (radians).
Equilibrium position	Point where net force is zero (displacement = 0).

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Formulas

1. Displacement (x)

MEMORISE THIS x = A cos(ωt + φ) - x = displacement (m) - A = amplitude (m) - ω = angular frequency (rad/s) - t = time (s) - φ = phase difference (rad) (usually 0 unless stated)

Alternative (if starting at max displacement): x = A sin(ωt) (if released from equilibrium at t=0)

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2. Velocity (v)

MEMORISE THIS (A-Level) / Given on sheet (GCSE) v = -Aω sin(ωt + φ) - v = velocity (m/s) - Negative sign = direction (velocity is max at equilibrium, zero at max displacement).

Maximum velocity: v_max = Aω (occurs at equilibrium)

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3. Acceleration (a)

MEMORISE THIS (A-Level) / Given on sheet (GCSE) a = -Aω² cos(ωt + φ) - a = acceleration (m/s²) - Negative sign = acceleration is always towards equilibrium.

Maximum acceleration: a_max = Aω² (occurs at max displacement)

Key relationship: a = -ω²x (acceleration is proportional to displacement but opposite in direction)

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4. Period of a Spring (T)

MEMORISE THIS T = 2π√(m/k) - T = period (s) - m = mass (kg) - k = spring constant (N/m)

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5. Period of a Pendulum (T)

MEMORISE THIS T = 2π√(L/g) - T = period (s) - L = length of pendulum (m) - g = acceleration due to gravity (9.81 m/s²)

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STEP-BY-STEP METHOD

Step 1: Identify the System

• Spring? Use T = 2π√(m/k).
• Pendulum? Use T = 2π√(L/g).
• Given displacement graph? Read amplitude (A) and period (T) directly.

Step 2: Calculate Angular Frequency (ω)

• ω = 2π/T or ω = 2πf.
• If given k and m (spring), ω = √(k/m).
• If given L and g (pendulum), ω = √(g/L).

Step 3: Write the Displacement Equation

• If released from max displacement at t=0: x = A cos(ωt).
• If released from equilibrium at t=0: x = A sin(ωt).
• If phase difference (φ) is given: x = A cos(ωt + φ).

Step 4: Find Velocity or Acceleration

• Velocity: Differentiate displacement: v = dx/dt = -Aω sin(ωt).
• Acceleration: Differentiate velocity: a = dv/dt = -Aω² cos(ωt).
• OR use a = -ω²x (faster for A-Level).

Step 5: Plug in Values

• Substitute t, A, ω into the equation.
• For max/min values, use sin or cos = ±1.

Step 6: Check Units & Direction

• Displacement (x) → metres (m).
• Velocity (v) → m/s.
• Acceleration (a) → m/s².
• Direction matters! Negative sign = opposite direction.

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WORKED EXAMPLES

Example 1 – Basic (GCSE)

A mass on a spring oscillates with amplitude 0.2 m and period 3 s. Write the displacement equation and find the displacement at t = 1 s.

Step 1: System = spring → use T = 2π√(m/k) (but we don’t need m or k here). Step 2: ω = 2π/T = 2π/3 ≈ 2.09 rad/s. Step 3: Released from max displacement → x = A cos(ωt) = 0.2 cos(2.09t). Step 4: At t = 1 s, x = 0.2 cos(2.09 × 1) ≈ 0.2 × (-0.5) = -0.1 m. Answer: x = 0.2 cos(2.09t); x = -0.1 m at t = 1 s.

What we did and why: - Used ω = 2π/T to find angular frequency. - Chose cos because the mass starts at max displacement. - Substituted t = 1 s to find displacement at that time.

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Example 2 – Medium (A-Level)

A pendulum has length 1.5 m. Calculate its period and maximum velocity if released from 0.1 m displacement. (g = 9.81 m/s²)

Step 1: System = pendulum → T = 2π√(L/g). Step 2: T = 2π√(1.5/9.81) ≈ 2.46 s. Step 3: ω = 2π/T ≈ 2.55 rad/s. Step 4: Max velocity = v_max = Aω = 0.1 × 2.55 ≈ 0.255 m/s. Answer: T = 2.46 s; v_max = 0.255 m/s.

What we did and why: - Used pendulum period formula to find T. - Calculated ω from T. - Used v_max = Aω because velocity is max at equilibrium.

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Example 3 – Exam-Style (A-Level)

A spring-mass system has k = 50 N/m and m = 0.5 kg. The mass is pulled 0.3 m from equilibrium and released. Find: a) The period of oscillation. b) The displacement at t = 0.4 s. c) The acceleration at t = 0.4 s.

Step 1: System = spring → T = 2π√(m/k). Step 2: T = 2π√(0.5/50) ≈ 0.628 s. Step 3: ω = 2π/T ≈ 10 rad/s. Step 4: Released from max displacement → x = 0.3 cos(10t). Step 5: At t = 0.4 s, x = 0.3 cos(10 × 0.4) = 0.3 cos(4) ≈ 0.3 × (-0.653) ≈ -0.196 m. Step 6: a = -ω²x = -(10)² × (-0.196) ≈ 19.6 m/s². Answer: a) T = 0.628 s b) x = -0.196 m at t = 0.4 s c) a = 19.6 m/s² at t = 0.4 s

What we did and why: - Used spring period formula for T. - Calculated ω from T. - Used x = A cos(ωt) for displacement. - Used a = -ω²x for acceleration (faster than differentiating).

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COMMON MISTAKES

MISTAKE	WHY IT HAPPENS	CORRECT APPROACH
Forgetting the negative sign in a = -ω²x	Thinks acceleration is in the same direction as displacement.	Acceleration is always towards equilibrium (opposite to displacement).
Mixing up sin and cos for displacement	Doesn’t check initial conditions (max displacement vs. equilibrium).	cos = starts at max displacement; sin = starts at equilibrium.
Using f instead of ω in equations	Confuses frequency (f) with angular frequency (ω).	ω = 2πf – always convert f to ω first.
Ignoring units in T = 2π√(m/k)	Uses grams instead of kg or N/cm instead of N/m.	Convert m to kg and k to N/m before plugging in.
Assuming v_max = A or a_max = A	Forgets v_max = Aω and a_max = Aω².	Velocity and acceleration depend on ω, not just amplitude.

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EXAM TRAPS

TRAP	HOW TO SPOT IT	HOW TO AVOID IT
Phase difference (φ) in displacement equation	Question says "released 0.1 s after t=0" or "out of phase by π/2".	Add φ to the argument: x = A cos(ωt + φ).
Non-SHM motion (e.g., large angle pendulum)	Pendulum angle > 10° or spring stretched beyond elastic limit.	State: "Assumes small angle approximation (θ < 10°)."
Graph interpretation (displacement vs. time)	Asks for velocity/acceleration from a x-t graph.	Velocity = gradient of x-t graph; acceleration = gradient of v-t graph.

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1-MINUTE RECAP (Night Before Exam)

"Right, listen up—this is SHM in 60 seconds. First, identify the system: spring or pendulum? Springs use T = 2π√(m/k), pendulums use T = 2π√(L/g). Next, find ω: ω = 2π/T or ω = √(k/m) for springs. Displacement is x = A cos(ωt) if released from max, x = A sin(ωt) if from equilibrium. Velocity is v = -Aω sin(ωt), acceleration is a = -Aω² cos(ωt)—or just a = -ω²x for a shortcut. Max velocity is Aω, max acceleration is Aω². Watch the signs—negative means opposite direction. Graphs: displacement is a cosine wave, velocity is a sine wave, acceleration is a negative cosine wave. Common traps: forgetting ω = 2πf, mixing up sin and cos, and ignoring units. Now go smash that exam!"

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