DP Physics HL C.1 Simple Harmonic Motion

DP Physics HL C.1 Simple Harmonic Motion - Full notes

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DP Physics HL | C.1 Simple Harmonic Motion

DP PHYSICS · THEME C: WAVE BEHAVIOUR

C.1

Simple Harmonic Motion

Full notes for HL · IB-style

Contents

1	Describing oscillations
2	Mass–spring systems and pendulums
3	Graphs of SHM
4	Energy in SHM
5	Mathematics of SHM
6	Damping and resonance
7	Exam technique
8	Data-based questions and further practice
9	Common misconceptions and links
10	Exam-style question with markscheme
11	Check your understanding
12	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define period, frequency, angular frequency and amplitude
state and use the defining equation a = −ω²x
calculate periods of mass–spring systems and pendulums
interpret x–t, v–t, a–t and a–x graphs and phase differences
describe energy changes in SHM
describe damping, forced oscillations and resonance
use x = x₀ sin(ωt + φ) and v = ±ω√(x₀² − x²)
use energy equations for SHM

Sections marked (HL) are Additional Higher Level content: displacement, velocity and energy equations. Every other section is common to SL and HL.

Guiding questions

What makes the harmonic oscillator model applicable to a wide range of physical phenomena? Why must the defining equation of SHM take the form it does? How can the energy and motion of an oscillation be analysed both graphically and algebraically?

How this topic is examined

C.1 appears in Papers 1 and 2. Expect calculations of period, maximum speed and acceleration, interpretation of SHM graphs, energy and resonance; HL adds the displacement, velocity and energy equations.

Use radians in all SHM calculations.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Describing oscillations

Motion that repeats.

Think about it

A child on a swing, a guitar string and the atoms in a crystal all oscillate. What do they have in common?

Oscillations are repeated back-and-forth motions about an equilibrium position. Simple harmonic motion (SHM) is the most important type: it models pendulums, springs, vibrating molecules and the motion behind waves.

1.1 Key terms

Displacement x is the distance from equilibrium in a stated direction. Amplitude x₀ is the maximum displacement. Period T is the time for one complete oscillation; frequency f = 1/T is the number of oscillations per second (Hz). Angular frequency ω = 2π/T = 2πf.

WORKED EXAMPLE 1 Pendulum (basic)

A pendulum makes 20 oscillations in 32 s. Find T, f and ω.

Solution

T = 32/20 = 1.6 s; f = 0.625 HzA1

ω = 2π/1.6 = 3.9 rad s⁻¹A1

1.2 Conditions for oscillation

Oscillation occurs when a body is displaced from stable equilibrium and a restoring force acts back towards the equilibrium position. Inertia carries it past equilibrium, so it overshoots, and the cycle repeats.

Quick check

1. Why does a pendulum overshoot the equilibrium position?

Answers: 1. it has kinetic energy (momentum) there; the restoring force is zero at equilibrium

1.3 The defining equation of SHM

SHM occurs when the acceleration is proportional to the displacement and directed towards the equilibrium position: a = −ω²x. Equivalently, the restoring force F = −kx. The period is then independent of the amplitude (isochronous).

Figure 1. Acceleration against displacement for SHM: a straight line through the origin with negative gradient.

Key point

Two features define SHM: acceleration ∝ displacement, and acceleration always points towards equilibrium (the minus sign).

WORKED EXAMPLE 2 Testing for SHM (standard)

An object has a = −25x. Show this is SHM and find its period.

Solution

Of the form a = −ω²x with ω² = 25R1

ω = 5.0 rad s⁻¹; T = 2π/5.0 = 1.26 sA1

1.4 Restoring force graphs

For SHM, a graph of restoring force against displacement is a straight line through the origin with negative gradient −k. A non-linear graph means the motion is only approximately simple harmonic for small displacements.

Figure 2. Restoring force against displacement for a spring: F = −kx.

Quick check

1. What does the gradient of an F–x graph give?

Answers: 1. −k

1.5 Examples of SHM

The two standard examples are a mass on a spring and a simple pendulum with small swings. Many other systems (molecules, floating objects, a liquid in a U-tube, tuning forks) are approximately simple harmonic for small displacements.

Figure 3. A mass on a spring and a simple pendulum.

2 Mass–spring systems and pendulums

Calculating periods.

2.1 The mass–spring system

For a mass m on a spring of constant k: F = −kx gives a = −(k/m)x, so ω² = k/m and T = 2π√(m/k). The period does not depend on g or on the amplitude, so it is the same for vertical and horizontal springs.

WORKED EXAMPLE 3 Spring period (standard)

A 0.50 kg mass oscillates on a spring of k = 200 N m⁻¹. Find T.

Solution

T = 2π√(0.50/200)M1

= 0.31 sA1

WORKED EXAMPLE 4 Changing mass (standard)

The mass is quadrupled. Find the new period.

Solution

T ∝ √mM1

Doubles to 0.63 sA1

2.2 The simple pendulum

For small angles (less than about 10°), the restoring force on a pendulum bob is approximately −(mg/L)x, giving T = 2π√(L/g). The period depends only on length and g, not on the mass of the bob.

WORKED EXAMPLE 5 Pendulum (standard)

Find the length of a pendulum with a period of 2.0 s.

Solution

L = gT²/4π² = 9.81 × 4/4π²M1

= 0.99 mA1

WORKED EXAMPLE 6 On the Moon (exam level)

The pendulum is taken to the Moon (g = 1.62 N kg⁻¹). Find its period.

Solution

T = 2π√(0.99/1.62)M1

= 4.9 sA1

Common mistake

Using the pendulum formula for large swings. It is only accurate for small angles.

2.3 Vertical springs

When a mass hangs on a spring, it stretches to a new equilibrium where ke = mg. Oscillations take place about this new equilibrium position, with the same period T = 2π√(m/k) as horizontally; gravity only shifts the equilibrium.

Figure 4. A vertical spring: natural length, new equilibrium and displaced position.

WORKED EXAMPLE 7 Static extension (exam level)

A mass stretches a spring by 6.2 cm at equilibrium. Find the period of vertical oscillations.

Solution

k/m = g/e, so T = 2π√(e/g) = 2π√(0.062/9.81)M1

= 0.50 sA1

2.4 Large-amplitude pendulums

The pendulum formula assumes sin θ ≈ θ, valid for small angles. At larger amplitudes the period increases slightly, so a pendulum is no longer exactly simple harmonic.

Figure 5. Period of a pendulum relative to the small-angle value as amplitude increases.

WORKED EXAMPLE 8 Error (standard)

Use the graph to estimate the percentage error in the period at 30° amplitude.

Solution

Ratio ≈ 1.017M1

≈ 1.7% longerA1

2.5 Measuring g with a pendulum

Time 20 oscillations for several lengths and plot T² against L. The graph is a straight line through the origin with gradient 4π²/g.

Figure 6. T² against length for a simple pendulum.

WORKED EXAMPLE 9 From the gradient (exam level)

Find g from the graph.

Solution

Gradient = 4.02 s² m⁻¹M1

g = 4π²/4.02 = 9.82 m s⁻²A1

Try this

Time 20 swings from a fiducial marker at the centre of the swing; measure L to the centre of the bob.

2.6 Springs in combination

Springs in parallel share the load (keff = k₁ + k₂), so the period decreases; springs in series stretch more (1/keff = 1/k₁ + 1/k₂), so the period increases.

WORKED EXAMPLE 10 Two springs (exam level)

A mass has period 0.80 s on one spring. Find the period on two identical springs (a) in parallel; (b) in series.

Solution

(a) k doubles: T = 0.80/√2 = 0.57 sA1

(b) k halves: T = 0.80√2 = 1.13 sA1

2.7 Other simple harmonic systems

A floating test tube pushed down and released, liquid in a U-tube and a ball rolling in a bowl all oscillate with SHM for small displacements, because their restoring forces are proportional to displacement.

WORKED EXAMPLE 11 Floating cylinder (exam level)

A cylinder of mass 0.20 kg and cross-sectional area 5.0 × 10⁻⁴ m² floats in water. Show its restoring force is −(ρAg)x and find its period.

Solution

Extra upthrust when pushed down x: ρAgx, so k = ρAg = 4.9 N m⁻¹M1

T = 2π√(0.20/4.9) = 1.27 sA1

2.8 Liquid in a U-tube

If liquid in a U-tube is displaced so one side is x above equilibrium and the other x below, the unbalanced column of height 2x provides a restoring force proportional to x. The liquid oscillates with SHM of period T = 2π√(L/2g), where L is the total length of the liquid column.

Figure 7. Liquid oscillating in a U-tube.

WORKED EXAMPLE 12 U-tube (exam level)

A liquid column of total length 0.40 m oscillates in a U-tube. Find its period.

Solution

T = 2π√(0.40/(2 × 9.81))M1

= 0.90 sA1

3 Graphs of SHM

Displacement, velocity and acceleration.

3.1 Displacement, velocity and acceleration against time

If the oscillator is released from maximum displacement at t = 0, x = x₀ cos ωt. Velocity is the gradient of the x–t graph and acceleration is the gradient of the v–t graph. Velocity is a quarter cycle (π/2) ahead of displacement; acceleration is in antiphase with displacement.

•At maximum displacement: v = 0, |a| maximum.
•At equilibrium: |v| maximum, a = 0.
•Maximum speed vmax = ωx₀; maximum acceleration amax = ω²x₀.

Figure 8. Displacement, velocity and acceleration against time for x₀ = 0.10 m, T = 2.0 s.

WORKED EXAMPLE 13 Maxima (standard)

For the oscillator in the figure, find the maximum speed and maximum acceleration.

Solution

ω = π rad s⁻¹; vmax = 0.10π = 0.31 m s⁻¹A1

amax = 0.10π² = 0.99 m s⁻²A1

3.2 Reading SHM graphs

From an x–t graph, read the amplitude (peak height) and period (time for one cycle). Use ω = 2π/T to find maximum speed and acceleration. On a v–t graph, the area under one half-cycle equals 2x₀.

WORKED EXAMPLE 14 From a graph (exam level)

An x–t graph shows amplitude 4.0 cm and period 0.50 s. Find vmax and amax.

Solution

ω = 4π = 12.6 rad s⁻¹M1

vmax = 0.50 m s⁻¹; amax = 6.3 m s⁻²A1

3.3 Different starting points

The equation for x depends on where timing starts. Released from maximum displacement: x = x₀ cos ωt. Starting from equilibrium moving in the positive direction: x = x₀ sin ωt. Both have the same shape, shifted by a quarter period.

Figure 9. Displacement–time graphs for two different starting conditions.

Quick check

1. If x = x₀ cos ωt, where is the oscillator at t = T/4?

Answers: 1. at equilibrium

3.4 Velocity against displacement

Plotting v against x for SHM gives an ellipse: v = 0 at x = ±x₀ and |v| maximum at x = 0. The motion goes round the ellipse once per period.

Figure 10. Velocity against displacement for SHM (x₀ = 0.10 m, ω = π rad s⁻¹).

Quick check

1. What are the semi-axes of the ellipse?

Answers: 1. x₀ and ωx₀

3.5 Phase and phase difference

Two oscillators with the same frequency may be out of step. The phase difference φ (in radians) describes how far one leads or lags the other: φ = 2πΔt/T, where Δt is the time offset between corresponding points.

Figure 11. Two oscillators with a phase difference of π/2.

WORKED EXAMPLE 15 Phase difference (standard)

Two pendulums of period 1.2 s pass through equilibrium 0.30 s apart (moving the same way). Find the phase difference.

Solution

φ = 2π × 0.30/1.2M1

= π/2 radA1

Quick check

1. What is the phase difference between displacement and acceleration in SHM?

Answers: 1. π (antiphase)

3.6 Comparing SHM quantities

A summary of how each quantity varies through one cycle.

Position	Displacement	Velocity	Acceleration	Kinetic energy	Potential energy
x = +x₀	maximum (+)	zero	maximum (−)	zero	maximum
x = 0 (moving −)	zero	maximum (−)	zero	maximum	zero
x = −x₀	maximum (−)	zero	maximum (+)	zero	maximum
x = 0 (moving +)	zero	maximum (+)	zero	maximum	zero

Quick check

1. At which positions is the acceleration zero?

Answers: 1. at equilibrium (x = 0)

3.7 SHM and circular motion

The projection onto a diameter of a point moving round a circle at constant angular velocity ω moves with SHM of the same ω. This is why the same symbol ω appears in both topics and why SHM equations use sines and cosines.

Figure 12. The projection of uniform circular motion is simple harmonic.

4 Energy in SHM

Kinetic and potential energy exchange.

4.1 Energy changes during an oscillation

In an undamped oscillator, energy continually transfers between kinetic and potential forms while the total stays constant. Kinetic energy is maximum at equilibrium; potential energy is maximum at the extremes.

Figure 13. Energy against displacement for a 0.50 kg oscillator (x₀ = 0.10 m, T = 2.0 s).

Figure 14. Kinetic and potential energy against time: each varies at twice the oscillation frequency.

WORKED EXAMPLE 16 Total energy (standard)

Find the total energy of the oscillator in the figure.

Solution

ET = ½mω²x₀² = ½(0.50)(π²)(0.10)²M1

= 0.025 J (25 mJ)A1

Quick check

1. How often does the kinetic energy reach a maximum during one period?

Answers: 1. twice

4.2 Energy of a mass–spring system

For a spring, the total energy is ½kx₀². At displacement x the elastic potential energy is ½kx², so the kinetic energy is ½k(x₀² − x²).

WORKED EXAMPLE 17 Speed at a point (exam level)

A 0.40 kg mass on a spring (k = 160 N m⁻¹) oscillates with amplitude 5.0 cm. Find its speed at x = 3.0 cm.

Solution

½mv² = ½k(x₀² − x²) = ½(160)(0.0025 − 0.0009)M1

v = 0.80 m s⁻¹A1

4.3 Energy and amplitude

The total energy of an oscillator is proportional to the square of its amplitude. Doubling the amplitude quadruples the energy (and doubles the maximum speed).

WORKED EXAMPLE 18 Amplitude change (standard)

An oscillator's amplitude falls from 6.0 cm to 3.0 cm. What fraction of its energy remains?

Solution

E ∝ x₀²M1

(3/6)² = one quarterA1

4.4 Energy at different positions

Bar charts show how the fixed total energy is shared at different positions. At x = x₀/2, potential energy is one quarter of the total and kinetic energy three quarters.

Figure 15. Energy shares at three positions for a 25 mJ oscillator.

Quick check

1. What fraction of energy is kinetic at x = x₀/2?

Answers: 1. 3/4

4.5 Energy in a damped oscillator

With damping, the total energy decreases with time as work is done against resistive forces, transferring energy to the surroundings as internal energy. For light damping the decrease is approximately exponential.

Figure 16. Total energy of a lightly damped oscillator against time.

WORKED EXAMPLE 19 Energy decay (exam level)

Use the graph to find the time for the energy to halve.

Solution

E falls from 25 to 12.5 mJM1

t ≈ 2.3 sA1

5 Mathematics of SHM

Equations for displacement, velocity and energy.

5.1 Displacement and velocity equations (HL)

Solutions of a = −ω²x are x = x₀ sin(ωt + φ) and v = ωx₀ cos(ωt + φ), where φ is the initial phase. Starting at equilibrium, φ = 0; starting at maximum displacement, φ = π/2 (so x = x₀ cos ωt).

WORKED EXAMPLE 20 Position at a time (exam level)

An oscillator starts at equilibrium moving in the positive direction with x₀ = 0.080 m and T = 1.2 s. Find x and v at t = 0.20 s.

Solution

ω = 5.24 rad s⁻¹; ωt = 1.047 radM1

x = 0.080 sin 1.047 = 0.069 m; v = 0.419 cos 1.047 = 0.21 m s⁻¹A1 A1

5.2 Velocity in terms of displacement (HL)

Eliminating time gives v = ±ω√(x₀² − x²). The speed is greatest at x = 0 and zero at x = ±x₀.

WORKED EXAMPLE 21 Speed at half amplitude (exam level)

Find the speed at x = x₀/2 as a fraction of vmax.

Solution

v = ω√(x₀² − x₀²/4) = ωx₀√3/2M1

= 0.87 vmaxA1

5.3 Energy equations (HL)

Ek = ½mω²(x₀² − x²), Ep = ½mω²x² and ET = ½mω²x₀².

WORKED EXAMPLE 22 Equal energies (exam level)

At what displacement are kinetic and potential energies equal?

Solution

½mω²x² = ½mω²(x₀² − x²)M1

x = x₀/√2 = 0.71x₀A1

5.4 Time to reach a position (HL)

Rearranging x = x₀ cos ωt (or sin) gives the time to reach a given displacement.

WORKED EXAMPLE 23 Time (exam level)

A mass released from x₀ = 0.10 m with T = 2.0 s. How long does it take to reach x = 0.050 m for the first time?

Solution

0.050 = 0.10 cos(πt): πt = π/3M1

t = 0.33 sA1

6 Damping and resonance

Real oscillators lose energy.

6.1 Damping

Damping is the loss of energy from an oscillator due to resistive forces (friction, drag), so the amplitude decreases with time. Light damping: amplitude decays slowly and the period is almost unchanged. Critical damping: the system returns to equilibrium in the shortest time without oscillating. Heavy (over) damping: slow return without oscillation.

Figure 17. Displacement against time for three degrees of damping.

In the real world

Car suspensions are designed to be close to critically damped, so the car body returns smoothly to its level after a bump without bouncing.

6.2 Light damping in detail

In light damping, the amplitude decreases by the same fraction each cycle (exponential decay). The energy lost per cycle is proportional to the square of the amplitude.

Figure 18. A lightly damped oscillation with its exponential envelope.

WORKED EXAMPLE 24 Energy loss (exam level)

The amplitude of a lightly damped oscillator falls from 10 cm to 8.0 cm in one cycle. Find the fraction of energy lost.

Solution

E ∝ A²: remaining (0.80)² = 0.64M1

Lost 36%A1

6.3 Forced oscillations and resonance

A periodic driving force makes a system oscillate at the driving frequency. When the driving frequency equals the natural frequency, the amplitude is greatest: resonance. Damping reduces the peak amplitude and broadens the resonance curve; light damping gives a sharp, high peak.

Figure 19. Amplitude of a forced oscillator against driving frequency for different damping.

WORKED EXAMPLE 25 Resonance (standard)

A wobbly washing machine shakes violently at one particular spin speed. Explain.

Solution

At that speed the driving frequency equals a natural frequency of the machineR1

Resonance: energy transfer is most efficient, giving large amplitudeR1

6.4 Uses and problems of resonance

Useful: musical instruments, microwave ovens (water molecules), MRI, radio tuning circuits, quartz clocks. Problems: bridges and buildings in wind or earthquakes, vibrating machinery, rattling car panels. Engineers add damping or change natural frequencies to avoid destructive resonance.

In the real world

The Millennium Bridge in London swayed on its opening day in 2000 because pedestrians' footsteps synchronised with a natural frequency of the bridge; dampers were added to fix it.

6.5 Natural frequencies of molecules

Atoms in molecules behave like masses on springs and vibrate at natural frequencies in the infrared range. When infrared radiation matches these frequencies, it is absorbed by resonance: this is the mechanism of the greenhouse effect (B.2) and infrared spectroscopy.

WORKED EXAMPLE 26 Molecular vibration (exam level)

A CO₂ bending vibration has natural frequency 2.0 × 10¹³ Hz. Find the wavelength of radiation that resonates with it.

Solution

λ = c/f = 3.0 × 10⁸/2.0 × 10¹³M1

= 15 μm (infrared)A1

6.6 Avoiding destructive resonance

Engineers prevent dangerous resonance by: adding damping (tuned mass dampers in skyscrapers, shock absorbers in cars); stiffening structures to raise natural frequencies above likely driving frequencies; or changing mass to move natural frequencies away from driving frequencies.

In the real world

Taipei 101 has a 660-tonne steel pendulum (tuned mass damper) near its top that swings against the building's motion, reducing sway in typhoons and earthquakes.

6.7 Barton's pendulums

Several pendulums of different lengths hang from a string driven by a heavy driver pendulum. Only the pendulum of the same length as the driver (same natural frequency) oscillates with large amplitude; it lags the driver by π/2 at resonance.

Try this

Make Barton's pendulums with paper cones on threads hung from a horizontal string. Swing a heavy driver and watch which cone responds most.

Quick check

1. Which pendulum resonates in Barton's pendulums?

Answers: 1. the one with the same length (natural frequency) as the driver

7 Exam technique

Getting SHM questions right.

7.1 Method checklist

•Identify x₀ and T from the graph or data; calculate ω = 2π/T first.
•Use vmax = ωx₀ and amax = ω²x₀.
•For energy, use ET = ½mω²x₀² (or ½kx₀² for springs).
•Check calculator mode: radians.
•In definitions, state both proportionality and direction (towards equilibrium).

7.2 Multi-step problem: car suspension

WORKED EXAMPLE 27 Suspension (exam level)

A car body of mass 1000 kg rests on four springs, each of k = 2.5 × 10⁴ N m⁻¹ (in parallel). Find (a) the period; (b) the speed at which a road with bumps 12 m apart would cause resonance.

Solution

(a) k = 1.0 × 10⁵ N m⁻¹; T = 2π√(1000/10⁵) = 0.63 sM1 A1

(b) v = 12/0.63 = 19 m s⁻¹M1 A1

7.3 Multi-step problem: graph analysis

WORKED EXAMPLE 28 From an a–x graph (exam level)

An a–x graph for a 0.20 kg oscillator is a straight line through (−0.05 m, 2.0 m s⁻²) and (0.05 m, −2.0 m s⁻²). Find ω, T and the total energy if x₀ = 0.05 m.

Solution

ω² = 2.0/0.05 = 40: ω = 6.3 rad s⁻¹M1 A1

T = 1.0 sA1

E = ½(0.20)(40)(0.05)² = 0.010 JA1

8 Data-based questions and further practice

Applying SHM.

8.1 Experimental data: mass–spring

A student measures the period for different masses on a spring.

m / kg	0.10	0.20	0.30	0.40	0.50
T / s	0.29	0.40	0.49	0.57	0.63

Figure 20. T² against mass for the spring data.

WORKED EXAMPLE 29 Spring constant (exam level)

Use the graph to find k.

Solution

Gradient = 4π²/k = 0.79 s² kg⁻¹M1

k = 4π²/0.79 = 50 N m⁻¹A1

8.2 Graded practice

WORKED EXAMPLE 30 Tuning fork (standard)

A tuning fork prong vibrates at 440 Hz with amplitude 0.50 mm. Find its maximum acceleration.

Solution

ω = 2π × 440 = 2760 rad s⁻¹M1

amax = ω²x₀ = 3.8 × 10³ m s⁻²A1

WORKED EXAMPLE 31 Bouncing car (exam level)

A 1200 kg car sinks 2.0 cm when 80 kg of passengers get in. Find the natural frequency of the loaded car on its springs.

Solution

k = 80 × 9.81/0.020 = 3.9 × 10⁴ N m⁻¹M1

f = (1/2π)√(k/m) = 0.88 HzA1

WORKED EXAMPLE 32 Clock (exam level)

A pendulum clock with a 0.994 m pendulum (T = 2.0 s at g = 9.81) is taken where g = 9.78 N kg⁻¹. Does it gain or lose time?

Solution

Smaller g: longer periodR1

The clock loses timeA1

8.3 Graded practice: periods

WORKED EXAMPLE 33 Pendulum length change (standard)

A pendulum of length 0.50 m is lengthened to 2.0 m. By what factor does the period change?

Solution

T ∝ √LM1

√4 = 2A1

WORKED EXAMPLE 34 Spring constant from frequency (standard)

A 0.30 kg mass oscillates at 2.5 Hz on a spring. Find k.

Solution

k = m(2πf)² = 0.30 × (15.7)²M1

= 74 N m⁻¹A1

WORKED EXAMPLE 35 Seconds pendulum on Mars (exam level)

Find the length of a pendulum with a 2.0 s period on Mars (g = 3.71 N kg⁻¹).

Solution

L = gT²/4π² = 3.71 × 4/39.5M1

= 0.38 mA1

8.4 Graded practice: graphs and energy

WORKED EXAMPLE 36 Energy at a point (standard)

An oscillator has total energy 40 mJ and amplitude 8.0 cm. Find its kinetic energy at x = 4.0 cm.

Solution

Ep = 40 × (4/8)² = 10 mJM1

Ek = 30 mJA1

WORKED EXAMPLE 37 Maximum speed from energy (standard)

A 0.25 kg oscillator has total energy 0.050 J. Find its maximum speed.

Solution

½mv² = 0.050M1

v = 0.63 m s⁻¹A1

WORKED EXAMPLE 38 Reading a v–t graph (exam level)

A v–t graph shows a sine curve with peak 0.60 m s⁻¹ and period 0.40 s. Find the amplitude.

Solution

ω = 2π/0.40 = 15.7 rad s⁻¹M1

x₀ = vmax/ω = 3.8 cmA1

8.5 Graded practice: damping and resonance

WORKED EXAMPLE 39 Driving a swing (standard)

Explain how a parent pushing a swing at the right moments builds up a large amplitude.

Solution

Pushes at the swing's natural frequency (once per cycle)R1

Energy is added in phase each cycle: resonanceR1

WORKED EXAMPLE 40 Amplitude decay (exam level)

A lightly damped oscillator loses 10% of its amplitude each cycle. Find the amplitude after 5 cycles as a fraction of the start, and the fraction of energy remaining.

Solution

A: 0.90⁵ = 0.59A1

E: 0.59² = 0.35A1

8.6 Key terms

Term	Meaning
amplitude x₀	maximum displacement from equilibrium
period T	time for one complete oscillation
angular frequency ω	2π/T
SHM	a = −ω²x: acceleration ∝ displacement, towards equilibrium
phase difference	fraction of a cycle (in radians) between two oscillations
damping	loss of energy to resistive forces
natural frequency	frequency of free oscillation
resonance	maximum amplitude when driving frequency = natural frequency

8.7 Self-test

Try these without notes, then check the answers.

Quick check

1. State the defining equation of SHM.

2. Find T for a 0.20 kg mass on a 50 N m⁻¹ spring.

3. Where is speed maximum in SHM?

4. What happens to the period of a pendulum if its mass doubles?

5. If the amplitude halves, what happens to the total energy?

6. What type of damping returns a system to equilibrium fastest without oscillating?

Answers: 1. a = −ω²x 2. 0.40 s 3. at equilibrium 4. nothing 5. it falls to one quarter 6. critical damping

8.8 Common exam errors

•Omitting the minus sign or the 'towards equilibrium' condition when defining SHM.
•Using degrees instead of radians in x = x₀ sin(ωt + φ).
•Claiming the pendulum period depends on the bob's mass.
•Thinking resonance needs a large driving force (it needs matching frequency).
•Confusing frequency with angular frequency.

9 Common misconceptions and links

Students often think…	Actually…
the period depends on amplitude	SHM is isochronous
acceleration is maximum at equilibrium	it is zero there and maximum at the extremes
a pendulum's period depends on its mass	it depends only on L and g
resonance needs a large force	it needs matching frequency
damping changes the period greatly	light damping barely changes it

How C.1 connects to the rest of the course

Topic	Connection
A.2 Forces and momentum	restoring forces, F = ma
C.2 Wave model	oscillating particles carry waves
C.3 Wave phenomena (resonance)	standing waves and resonance
A.2 Circular motion	ω and the reference circle

10 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Figure 21. Displacement, velocity and acceleration of an oscillator.

A 0.50 kg mass oscillates on a spring with SHM. Its displacement varies as shown (amplitude 0.10 m, period 2.0 s).

(a) State the amplitude and period. [1]

(b) Calculate the spring constant. [2]

(d) Calculate the total energy of the oscillation. [2]

(e) Describe and explain how the graph would change if the system were lightly damped. [2]

(f) Calculate the speed when the displacement is 0.060 m. [2]

Markscheme

Part	Answer	Marks
(a)	0.10 m and 2.0 s	A1
(b)	k = 4π²m/T² = 4π²(0.50)/4.0	M1
	4.9 N m⁻¹	A1
(c)	vmax = ωx₀ = π × 0.10	M1
	0.31 m s⁻¹	A1
(d)	E = ½kx₀² = ½(4.93)(0.010)	M1
	0.025 J	A1
(e)	amplitude decreases gradually (exponentially)	R1
	period almost unchanged; energy lost to resistive forces	R1
(f)	v = ω√(x₀² − x²) = π√(0.0100 − 0.0036)	M1
	0.25 m s⁻¹	A1

Examiner's comments. In (e), many answers stated the period decreases; light damping leaves it almost unchanged.

11 Check your understanding

Section A: multiple choice

1. In SHM, acceleration is

A. proportional to displacement and opposite in direction B. constant C. proportional to velocity D. maximum at equilibrium

2. The period of a mass–spring system depends on

A. amplitude B. g C. m and k D. length of spring only

3. At maximum displacement in SHM

A. acceleration is zero B. velocity is zero C. kinetic energy is maximum D. velocity is maximum

4. Doubling the length of a pendulum multiplies its period by

A. 2 B. 4 C. 1/2 D. √2

5. Velocity leads displacement by

A. π B. π/2 C. 2π D. zero

6. Total energy of an oscillator is proportional to

A. amplitude squared B. amplitude C. period D. frequency

7. Resonance occurs when

A. damping is heavy B. amplitude is small C. driving force is large D. driving frequency equals natural frequency

8. Critical damping

A. causes oscillation forever B. increases amplitude C. returns the system to equilibrium fastest without oscillation D. has no effect

9. At x = x₀/2, the speed is

A. vmax/2 B. vmax C. zero D. about 0.87 vmax

Section B: structured

10. Show that a simple pendulum undergoes SHM for small displacements. [3]

11. A mass on a spring has T = 0.80 s and x₀ = 4.0 cm. Find vmax and amax. [3]

12. Sketch energy–displacement graphs for an undamped oscillator. [3]

13. Explain why soldiers break step when crossing a bridge. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C 9. D

10. Restoring force −mg sin θ ≈ −(mg/L)x for small θ: a = −(g/L)x, so SHM.

11. ω = 7.85 rad s⁻¹; vmax = 0.31 m s⁻¹; amax = 2.5 m s⁻².

12. Ep parabola from zero at x = 0; Ek inverted parabola; total horizontal line.

13. To avoid driving the bridge at its natural frequency and causing resonance.

12 Summary and topic map

Idea	Remember
Definitions	T, f, ω = 2πf, x₀
SHM	a = −ω²x
Periods	T = 2π√(m/k); T = 2π√(L/g)
Maxima	vmax = ωx₀; amax = ω²x₀
Energy	ET = ½mω²x₀²
Damping and resonance	amplitude decays; peak at f = f₀

Before the exam, make sure you…

•use radians
•identify maxima and zeros on graphs
•use the correct period formula
•explain resonance by matching frequencies

Simple Harmonic Motion on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 22. Topic map for Simple Harmonic Motion.

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