SHM Systems

SPEC 3.6.1.3-4 • FURTHER MECHANICS

Simple Harmonic Systems

Mass-spring systems, pendulums, energy, damping, and resonance

Two key systems demonstrate simple harmonic motion in the AQA specification:

Mass-Spring System

A mass attached to a spring oscillates when displaced from equilibrium.

Restoring force: F = -kx

Simple Pendulum

A mass on a string oscillates when displaced from vertical.

Restoring force: F = -mg sin θ

Key Requirement

Both systems require a restoring force proportional to displacement for SHM to occur.

Mass-Spring System

Restoring Force

F = - k x

Force F always points opposite to displacement x (towards equilibrium)

Mass-Spring Period

Formula

T = 2 π \frac{m}{k}

SymbolQuantityUnit

TPeriods

mMasskg

kSpring constantN m⁻¹

Important

Period is independent of amplitude - a defining feature of SHM.

Vertical vs Horizontal Springs

Energy conversion:

KE ↔ Elastic PE only

No change in height, so no gravitational PE involved.

Energy conversion:

KE ↔ Elastic PE + Gravitational PE

Height changes during oscillation.

Same Period Formula

Both orientations use

T = 2 π \frac{m}{k}

- the equilibrium position just shifts for vertical springs.

Angular Frequency Relationship

For a vertical mass-spring system at equilibrium, the spring extension e satisfies:

At Equilibrium

m g = k e

Weight = Spring force at equilibrium extension e

Since

ω^{2} = \frac{k}{m}

, we can substitute

k = \frac{m g}{e}

:

Key Result

ω^{2} = \frac{k}{m} = \frac{g}{e}

Alternative Period Formula

This gives:

T = 2 π \frac{e}{g}

Useful when you know the equilibrium extension.

Physical Meaning

A spring that stretches more at equilibrium (larger e) will have a longer period.

Worked Example01

A 0.50 kg mass is attached to a spring with spring constant 200 N m⁻¹. Calculate the period and frequency of oscillation.

1

Identify values: m = 0.50 kg, k = 200 N m⁻¹

2

Calculate period:

T = 2 π \frac{m}{k} = 2 π \frac{0.50}{200} = 2 π 0.0025 = 2 π \times 0.05 = 0.314

s

3

Calculate frequency:

f = \frac{1}{T} = \frac{1}{0.314} = 3.18

Hz

AnswerT = 0.31 s, f = 3.2 Hz

Worked Example02

A mass-spring system oscillates with period 0.80 s. If the mass is 0.32 kg, calculate the spring constant.

1

Rearrange formula:

T = 2 π \frac{m}{k}

→

T^{2} = 4 π^{2} \frac{m}{k}

→

k = \frac{4 π ^{2} m}{T ^{2}}

2

Substitute:

k = \frac{4 π ^{2} \times 0.32}{0.8 0 ^{2}} = \frac{4 \times 9.87 \times 0.32}{0.64}

3

Calculate:

k = \frac{12.6}{0.64} = 19.7

N m⁻¹

Answerk = 20 N m⁻¹ (2 s.f.)

Simple Pendulum

Setup

Mass m on string of length L

Displaced by angle θ from vertical

Restoring Force

F = - m g sin θ

Small Angle (θ < 10°)

sin θ ≈ θ (radians)

F ≈ −mgθ → F ∝ −x (SHM)

Pendulum Period

Formula

T = 2 π \frac{L}{g}

SymbolQuantityUnit

TPeriods

LLength (pivot to centre of mass)m

gGravitational field strengthm s⁻²

Key Features

Period is independent of mass and independent of amplitude (for small angles).

Worked Example03

A simple pendulum has length 0.80 m. Calculate its period on Earth (g = 9.81 m s⁻²) and on the Moon (g = 1.62 m s⁻²).

Formula:

T = 2 π \frac{L}{g}

Earth (g = 9.81 m s⁻²)

T = 2 π \frac{0.80}{9.81}

T = 1.80 s

Moon (g = 1.62 m s⁻²)

T = 2 π \frac{0.80}{1.62}

T = 4.41 s

Lower g → Longer period (Moon period is 2.45× Earth period)

Energy in SHM

In both systems, energy continuously transfers between kinetic and potential forms:

Mass-Spring

KE:

\frac{1}{2} m v^{2}

Elastic PE:

\frac{1}{2} k x^{2}

Total:

E = \frac{1}{2} k A^{2}

Pendulum

KE:

\frac{1}{2} m v^{2}

Gravitational PE:

m g h

Total:

E = m g h_{ma x}

Energy Conservation

In an undamped system, total mechanical energy is constant:

KE + PE = constant

Energy Using ω²

Alternative energy formulas use angular frequency ω:

Key Formulas

Kinetic Energy

K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})

Potential Energy

PE = \frac{1}{2} m ω^{2} x^{2}

Total Energy

E_{t o t a l} = \frac{1}{2} m ω^{2} A^{2}

Why ω² Formulas?

These are equivalent to the k-based formulas since

k = m ω^{2}

for a mass-spring system.

At x = 0: KE = ½mω²A² (maximum), PE = 0

At x = ±A: KE = 0, PE = ½mω²A² (maximum)

Total Energy

Adding KE + PE at any position:

E = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) + \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} A^{2}

✓

Energy vs Displacement

At x = 0 (Equilibrium)

PE = 0, KE = maximum

Object at maximum speed v = ωA

At x = ±A (Extremes)

PE = maximum, KE = 0

Object momentarily stationary

Energy vs Time

Energy Oscillates at Double Frequency

Both KE and PE complete two full cycles during one period T of oscillation.

This is because energy depends on x² and v² — squaring sin(ωt) or cos(ωt) doubles the frequency.

Conservation of Energy

At every instant: KE + PE = E_T (constant). When PE is maximum, KE is zero and vice versa.

Worked Example04

A 0.25 kg mass on a spring (k = 100 N m⁻¹) oscillates with amplitude 0.08 m. Calculate: (a) the total energy, (b) the maximum speed, (c) the speed when displacement is 0.04 m.

1

Total energy:

E = \frac{1}{2} k A^{2} = \frac{1}{2} \times 100 \times 0.0 8^{2} = 0.32

J

2

Maximum speed (at x = 0, all energy is KE):

\frac{1}{2} m v_{ma x}^{2} = 0.32

→

v_{ma x} = \frac{2 \times 0.32}{0.25} = 1.6

m s⁻¹

3

Speed at x = 0.04 m: PE = ½ × 100 × 0.04² = 0.08 J, KE = 0.32 - 0.08 = 0.24 J

v = \frac{2 \times 0.24}{0.25} = 1.39

m s⁻¹

Answer(a) 0.32 J, (b) 1.6 m s⁻¹, (c) 1.4 m s⁻¹

Free vs Forced Vibrations

Free Vibrations

System oscillates at its natural frequency (f₀) after being displaced and released.

No external driving force

Frequency determined by system properties

Amplitude decreases due to damping

Example: Plucked guitar string

Forced Vibrations

System is driven by an external periodic force at the driving frequency (f).

Continuous energy input

Frequency set by driver

Amplitude depends on f vs f₀

Example: Pushing a child on swing

Natural Frequency

f_{0} = \frac{1}{2 π} \frac{k}{m}

(mass-spring) or

f_{0} = \frac{1}{2 π} \frac{g}{L}

(pendulum)

Free Oscillations

Ideal Free Oscillation

In the absence of damping, amplitude remains constant.

System oscillates at natural frequency f₀ indefinitely.

In Practice

Real systems always have some energy loss (damping), so amplitude gradually decreases.

Damped Oscillations

Damping Ratio ζ

The damping ratio characterises how quickly oscillations decay:ζ < 1

Underdamped

Oscillates, amplitude decreasesζ = 1

Critical

Fastest return, no overshootζ > 1

Overdamped

Slow return, no oscillation

Worked Example05

Classify each as free or forced vibration: (a) A tuning fork struck and left to vibrate, (b) A building swaying due to wind gusts, (c) A guitar string plucked, (d) A washing machine drum during spin cycle.

a

Tuning fork struck: FREE - Initial energy input, then oscillates at natural frequency with no continuous driving force.

b

Building in wind: FORCED - Wind provides continuous periodic driving force at wind's frequency.

c

Guitar string plucked: FREE - Initial displacement, then oscillates at natural frequency (determines pitch).

d

Washing machine: FORCED - Motor continuously drives rotation at set frequency.

Key distinction: Free = one-off energy input; Forced = continuous energy input

Damping Types

Damping is the loss of energy from an oscillating system, usually due to resistive forces.

Underdamped (ζ < 1)

Amplitude decreases gradually over many oscillations.Examples:

Swinging pendulum in air

Guitar string vibrating

Diving board after jump

Critical (ζ = 1)

Returns to equilibrium in the shortest time without oscillating.Examples:

Car shock absorbers

Door closers

Weighing scales

Overdamped (ζ > 1)

Returns slowly to equilibrium without oscillating.Examples:

Heavy door with strong closer

Pendulum in thick oil

Old car suspensions

Exam Tip

Critical damping is optimal when you want the fastest return to equilibrium without overshoot.

Damping Comparison

Critical Damping is Optimal

Returns to equilibrium in the shortest time without overshooting.

Exam Tip

Be able to sketch and identify these three curves from their shapes.

Types of Damping

Natural Damping

Energy loss due to inherent resistive forces in the system:

Air resistance - pendulum in air

Internal friction - molecular forces in spring

Fluid viscosity - oscillating in water

Always present to some degree

Artificial Damping

Deliberately added damping mechanisms:

Shock absorbers - car suspension

Electromagnetic braking - galvanometers

Building dampers - skyscraper stabilisation

Engineered for specific behaviour

Engineering Application

Artificial damping is designed to achieve critical damping where rapid settling without oscillation is required.

Worked Example06

A car's suspension system uses dampers (shock absorbers). Explain why critical damping is preferred over light or heavy damping.

1

Light damping would be problematic: After hitting a bump, the car would oscillate up and down many times, causing discomfort and loss of tyre contact with road.

2

Heavy damping would also be problematic: The suspension would be too stiff and return to equilibrium very slowly, failing to absorb the next bump effectively.

3

Critical damping is ideal: The suspension returns to equilibrium in the shortest time without overshooting. This maximises comfort and maintains good road contact.

AnswerCritical damping allows fastest return to equilibrium without oscillation, maintaining ride comfort and tyre-road contact.

Resonance

Definition

Resonance occurs when the driving frequency equals the natural frequency of the system. This results in maximum energy transfer and maximum amplitude.

Condition for Resonance

f = f₀

When driving frequency (f) matches natural frequency (f₀), the system absorbs maximum energy from the driver.

Useful Resonance

Musical instruments, radio tuning, MRI scanners

Destructive Resonance

Bridge collapse (Tacoma Narrows), building damage in earthquakes

Resonance Curves

Effect of Damping on Resonance

More damping → lower peak amplitude

More damping → broader peak (less sharp)

Heavy damping → resonant frequency shifts slightly lower

Phase at Resonance

The phase difference between driver and oscillator changes with frequency:

ConditionPhase DifferenceMeaning

f < f₀≈ 0Oscillator follows driver (in phase)

f = f₀π/2 (90°)Oscillator lags by quarter cycle

f > f₀≈ π (180°)Oscillator opposes driver (antiphase)

Worked Example07

A child on a swing (natural frequency 0.5 Hz) is pushed periodically. Describe what happens to the amplitude when pushed at: (a) 0.3 Hz, (b) 0.5 Hz, (c) 0.8 Hz.

a

At 0.3 Hz (below f₀): Amplitude builds up but remains relatively small. Energy transfer is inefficient. Swing nearly in phase with pushes.

b

At 0.5 Hz (at f₀): RESONANCE - maximum amplitude achieved. Efficient energy transfer on each push. Swing lags pushes by 90°.

c

At 0.8 Hz (above f₀): Amplitude remains small. Pushes fight against swing motion (out of phase). Energy transfer inefficient.

AnswerMaximum amplitude at f = f₀ = 0.5 Hz (resonance). Lower amplitudes when f ≠ f₀.

Worked Example08

A building has natural frequency 2.0 Hz. An earthquake produces ground oscillations at 2.0 Hz. (a) What phenomenon occurs? (b) How can engineers protect the building? (c) What would happen if dampers were added?

a

Resonance occurs because the driving frequency (earthquake, 2.0 Hz) matches the building's natural frequency (2.0 Hz), causing potentially destructive large amplitude oscillations.

b

Engineering solutions: Change natural frequency (different mass/stiffness), add dampers to reduce peak amplitude, base isolation to reduce energy transfer from ground.

c

Effect of dampers: Reduce the peak amplitude at resonance and broaden the resonance curve, making the building less sensitive to the specific earthquake frequency.

Answer(a) Resonance; (b) Change f₀ or add damping; (c) Reduced peak amplitude, broader response

Comparing SHM Systems

Mass-Spring

T = 2 π \frac{m}{k}

Pendulum

T = 2 π \frac{L}{g}

PropertyMass-SpringPendulum

Depends on mass?Yes (T ∝ √m)No

Restoring forceF = −kxF ≈ −(mg/L)x

Energy typeElastic PEGravitational PE

Common Feature

Both have period independent of amplitude — the hallmark of SHM.

Practice

Question 1: A mass of 0.40 kg attached to a spring oscillates with period 0.60 s. Calculate the spring constant.

Rearrange:

k = \frac{4 π ^{2} m}{T ^{2}} = \frac{4 π ^{2} \times 0.40}{0.6 0 ^{2}} = \frac{15.8}{0.36} = 43.9

N m⁻¹

Answerk = 44 N m⁻¹

Practice

Question 2: Calculate the length of a simple pendulum with period 2.0 s on Earth (g = 9.81 m s⁻²).

Rearrange:

L = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.81 \times 2. 0 ^{2}}{4 π ^{2}} = \frac{39.24}{39.48} = 0.994

m

AnswerL = 0.99 m ≈ 1.0 m

Practice

Question 3: A 0.15 kg mass on a spring (k = 60 N m⁻¹) oscillates with amplitude 0.05 m. Calculate (a) the total energy and (b) the maximum speed.

(a) Total energy:

E = \frac{1}{2} k A^{2} = \frac{1}{2} \times 60 \times 0.0 5^{2} = 0.075

J

(b) Maximum speed: All energy is KE at equilibrium:

\frac{1}{2} m v_{ma x}^{2} = 0.075

→

v_{ma x} = \frac{2 \times 0.075}{0.15} = 1.0

m s⁻¹

Answer(a) 0.075 J, (b) 1.0 m s⁻¹

Summary

Key Formulas

SystemPeriodTotal Energy

Mass-Spring

T = 2 π \frac{m}{k}

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}

Pendulum

T = 2 π \frac{L}{g}

E = m g h_{ma x}

Exam Checklist

Period is independent of amplitude for SHM

Pendulum formula only valid for small angles (θ < 10°)

Resonance: maximum amplitude when f = f₀

Damping reduces resonance amplitude and broadens peak

Energy oscillates at 2f (twice the oscillation frequency)

Critical damping (ζ = 1): fastest return without oscillation