Simple Harmonic Motion

Understanding the mathematics of oscillations: displacement, velocity, and acceleration

SPEC 3.6.1.2 • PERIODIC MOTION

What is an Oscillation?

An oscillation is a repetitive motion about an equilibrium position.

Equilibrium Position

The position where the net force is zero. When displaced, a restoring force acts to return the object to this position.

Examples:

A pendulum swinging

A mass on a spring

A vibrating guitar string

Simple Pendulum

A and x measured horizontally

Key Definitions

QuantitySymbolDefinitionUnit

DisplacementxDistance from equilibrium (with direction)m

AmplitudeAMaximum displacement from equilibriumm

PeriodTTime for one complete oscillations

FrequencyfNumber of oscillations per secondHz

Angular FrequencyωRate of change of phase anglerad s⁻¹

Key Relationships

T = \frac{1}{f}

ω = 2 π f

ω = \frac{2 π}{T}

The Defining Condition for SHM

For motion to be simple harmonic motion (SHM), it must satisfy:

Definition of SHM

Acceleration is directly proportional to displacement from equilibrium

Acceleration is always directed towards equilibrium (opposite to displacement)

Mathematical Statement

a \propto - x

or

a = - ω^{2} x

Critical

The negative sign is essential! It shows acceleration is opposite to displacement — a restoring acceleration.

The Acceleration Equation

Fundamental Equation

a = - ω^{2} x

a = acceleration (m s⁻²) | ω = angular frequency (rad s⁻¹) | x = displacement (m)

Straight line through origin →

ω = - gradient

01Worked Example

An object oscillates with SHM with T = 0.50 s and A = 0.080 m. Calculate:

(a) angular frequency (b) maximum acceleration (c) acceleration when x = 0.040 m

a)

ω = \frac{2 π}{T} = \frac{2 π}{0.50} = 4 π

= 12.6 rad s⁻¹

b)

a_{max} = ω^{2} A = (4 π)^{2} \times 0.080

= 12.6 m s⁻²

c)

a = - ω^{2} x = - (4 π)^{2} \times 0.040

= −6.3 m s⁻²

Displacement Equation

Displacement varies sinusoidally. The equation depends on starting position:

Starting at x = +A

x = A cos (ω t)

x = A when t = 0

Starting at x = 0 (moving +ve)

x = A sin (ω t)

x = 0 when t = 0

Exam Tip

Timing usually starts at amplitude, so use x = A cos(ωt) as default. Always check initial conditions!

02Worked Example

A particle starts at x = +0.12 m (maximum) with f = 2.0 Hz.

(a) Write the displacement equation. (b) Find x at t = 0.10 s.

a)

Starts at +A → cosine: ω = 2πf = 4π rad s⁻¹

$x = 0.12 cos (4 π t)$

b)

x = 0.12 cos (4 π \times 0.10) = 0.12 cos (0.4 π) = 0.12 \times 0.309

= 0.037 m

Velocity in SHM

Velocity can be expressed as a function of time or displacement:

Velocity as Function of Time

If

x = A cos (ω t)

:

v = \frac{d x}{d t} = - ω A sin (ω t)

If

x = A sin (ω t)

, then

v = + ω A cos (ω t)

Velocity as Function of Displacement

v = \pm ω A^{2} - x^{2}

± shows velocity can be in either direction

Velocity-Displacement Graph

The equation

v = \pm ω A^{2} - x^{2}

describes an ellipse:

At x = 0

v = \pm ω A

— Maximum speed

At x = ±A

v = 0

— Momentarily stationary

Worked Example03

A mass oscillates with A = 0.15 m and f = 1.5 Hz. Find the speed when x = 0.10 m.

1

ω = 2 π \times 1.5 = 3 π

rad s⁻¹

2

v = \pm ω A^{2} - x^{2} = \pm 3 π 0.1 5^{2} - 0.1 0^{2}

3

v = \pm 3 π 0.0125 = \pm 3 π \times 0.1118 = \pm 1.05

m s⁻¹

Speed = 1.05 m s⁻¹

Maximum Velocity and Acceleration

Maximum Velocity

v_{max} = ω A

At equilibrium (x = 0)

Maximum Acceleration

a_{max} = ω^{2} A

At amplitude (x = ±A)

Alternative Forms

v_{max} = 2 π f A = \frac{2 π A}{T}

a_{max} = (2 π f)^{2} A = \frac{4 π ^{2} A}{T ^{2}}

Memory Aid

At equilibrium: fastest, zero acceleration. At amplitude: stationary, maximum acceleration.

Worked Example04

An object has A = 0.20 m and $v_{max}$ = 0.80 m s⁻¹. Find: (a) period (b) maximum acceleration

a)

ω = \frac{v _{max}}{A} = \frac{0.80}{0.20} = 4.0

rad s⁻¹

T = \frac{2 π}{ω} = \frac{2 π}{4.0}

= 1.57 s

b)

a_{max} = ω^{2} A = 4. 0^{2} \times 0.20

= 3.2 m s⁻²

SHM Equations: The Complete Set

The three key equations are linked by differentiation:

Differentiation Chain

x = A cos (ω t)

↓ differentiate:

v = \frac{d x}{d t}

v = - ω A sin (ω t)

↓ differentiate:

a = \frac{d v}{d t}

a = - ω^{2} A cos (ω t) = - ω^{2} x

Key Insight

Acceleration is always proportional to displacement and opposite in direction — this confirms SHM!

Worked Example05

A particle starts at x = +0.050 m (maximum) and oscillates at 4.0 Hz. Find the time to first reach x = +0.025 m.

1

ω = 2 π \times 4.0 = 8 π

rad s⁻¹

2

0.025 = 0.050 cos (8 π t)

→

cos (8 π t) = 0.5

3

8 π t = \frac{π}{3}

→

t = \frac{1}{24}

= 0.042 s

Graphical Representations

For motion starting at x = A (t = 0). Each graph is the gradient of the one above:

Phase Relationships

Phase Differences

Velocity leads displacement by π/2 (90° or T/4)

Acceleration leads velocity by π/2 (90° or T/4)

Acceleration is in antiphase with displacement (π or 180°)

Timexva

t = 0+A (max)0−ω²A (max −ve)

t = T/40−ωA (max −ve)0

t = T/2−A (max −ve)0+ω²A (max +ve)

t = 3T/40+ωA (max +ve)0

Worked Example06

An object oscillates at 5.0 Hz. When x = +0.030 m, v = −0.70 m s⁻¹. Find the amplitude.

1

ω = 2 π f = 2 π \times 5.0 = 10 π

rad s⁻¹

2

Using

v^{2} = ω^{2} (A^{2} - x^{2})

:

3

0.7 0^{2} = (10 π)^{2} (A^{2} - 0.03 0^{2})

4

0.49 = 986.96 (A^{2} - 0.0009)

5

A^{2} = \frac{0.49}{986.96} + 0.0009 = 0.001397

A = 0.037 m

Worked Example07

$a_{max}$ = 8.0 m s⁻² and $v_{max}$ = 0.40 m s⁻¹. Find the amplitude A and period T.

1

Using

a_{max} = ω^{2} A

and

v_{max} = ω A

:

2

Dividing:

\frac{a _{max}}{v _{max}} = \frac{ω ^{2} A}{ω A} = ω

3

ω = \frac{8.0}{0.40} = 20

rad s⁻¹

4

A = \frac{v _{max}}{ω} = \frac{0.40}{20}

= 0.020 m

5

T = \frac{2 π}{ω} = \frac{2 π}{20}

= 0.31 s

Practice

Question 1: A mass oscillates with f = 2.5 Hz and A = 0.060 m. Calculate:

(a) maximum velocity (b) maximum acceleration

1

ω = 2 π \times 2.5 = 5 π

= 15.7 rad s⁻¹

a)

v_{max} = ω A = 5 π \times 0.060

= 0.94 m s⁻¹

b)

a_{max} = ω^{2} A = (5 π)^{2} \times 0.060

= 14.8 m s⁻²

Practice

Question 2:

1

A = 0.08 m, T = 0.4 s,

f = 1/0.4

= 2.5 Hz

2

Starts at +A → cosine:

ω = 5 π

rad s⁻¹ → $x = 0.08 cos (5 π t)$

Practice

Question 3: An object starts at x = +A with T = 2.0 s and A = 0.10 m. Find x and v at t = 0.25 s.

1

ω = \frac{2 π}{2.0} = π

rad s⁻¹

2

x = 0.10 cos (π \times 0.25) = 0.10 \times 0.707

= 0.071 m

3

v = - π \times 0.10 \times sin (0.25 π) = - 0.10 π \times 0.707

= −0.22 m s⁻¹

Summary

Key Equations for SHM

QuantityEquationNotes

Defining condition

a = - ω^{2} x

a ∝ −x

Displacement

x = A cos (ω t)

Starts at x = A

Velocity (time)

v = - ω A sin (ω t)

Derivative of x

Velocity (displacement)

v = \pm ω A^{2} - x^{2}

Very useful!

Acceleration (time)

a = - ω^{2} A cos (ω t)

= −ω²x

Max velocity

v_{max} = ω A

At x = 0

Max acceleration

a_{max} = ω^{2} A

At x = ±A

Exam Skills

v = 0 at amplitude; v = max at equilibrium

a = max at amplitude; a = 0 at equilibrium

a and x are always in antiphase • Gradient of a-x graph = −ω²