C. Wave behaviour

Start with the restoring force

Simple harmonic motion occurs when the restoring force is proportional to displacement from equilibrium and always points back toward equilibrium: $F\propto -x$. For constant mass this gives $a\propto -x$.

Locate equilibrium

Measure $x$ from the equilibrium position, not from an arbitrary origin. At equilibrium $x=0$, so the restoring force and acceleration are zero; away from equilibrium, the acceleration points opposite to the displacement.

Check the model

A spring–mass system is an SHM model when the spring force is linear. A simple pendulum approximates SHM only for small angular displacements, where $\sin\theta\approx\theta$ in radians.

Common trap

“Periodic” motion alone is not enough. The defining condition is the restoring acceleration $a=-\omega^2x$, including the opposite direction and proportional dependence.

Write the definition

Simple harmonic motion is defined by $a=-\omega^2x$, where $x$ is displacement from equilibrium and $\omega$ is angular frequency. The acceleration is proportional to displacement and points in the opposite direction.

Interpret the minus sign

If the particle is displaced to positive $x$, acceleration is negative; if it is displaced to negative $x$, acceleration is positive. At equilibrium, $x=0$ and $a=0$, although the particle may have maximum speed there.

Connect frequency to acceleration

A larger $\omega$ gives a larger acceleration for the same displacement. The equation also shows why increasing amplitude does not change the period of ideal SHM: acceleration scales with the displacement.

Common trap

Do not write $a=+\omega^2x$ or measure $x$ from an arbitrary origin. The displacement must be relative to equilibrium, and the sign must restore the particle toward it.

Name each quantity

The equilibrium position is the central position where the resultant restoring force is zero. Displacement $x$ is the signed distance from equilibrium. Amplitude $x_0$ is the maximum magnitude of displacement.

Connect time measures

The period $T$ is the time for one complete cycle. Frequency $f$ is the number of cycles per second, so $T=1/f$. Angular frequency is $\omega=2\pi f=2\pi/T$, measured in radians per second.

Keep amplitude and displacement distinct

Amplitude is a non-negative fixed maximum for an ideal oscillation; displacement changes continuously between $-x_0$ and $+x_0$. The sign of displacement identifies the side of equilibrium.

Common trap

Do not call the distance travelled in one cycle the amplitude. Amplitude is measured from equilibrium to an extreme position, not from one extreme to the other.

Three equivalent measures

$$T=\frac1f=\frac{2\pi}{\omega},\qquad \omega=2\pi f$$

Use seconds for $T$, hertz for $f$, and radians per second for $\omega$.

Worked example from the mapped local textbook

A guitar-string point oscillates at $f=196\,\mathrm{Hz}$.

$$T=\frac1{196}=5.10\times10^{-3}\,\mathrm s$$

$$\omega=2\pi(196)=1.23\times10^3\,\mathrm{rad\,s^{-1}}$$

Common trap

Do not mix this general conversion objective with the separate spring and pendulum period models. Also, $\omega$ is $2\pi$ times $f$, not $f/(2\pi)$.

Mass–spring period

For an ideal mass $m$ attached to a linear spring of spring constant $k$,

$$T=2\pi\sqrt{\frac{m}{k}}$$

Use $m$ in kilograms and $k$ in $\mathrm{N\,m^{-1}}$ to obtain $T$ in seconds.

Read the dependence

$T\propto\sqrt m$: more mass increases the period. $T\propto1/\sqrt k$: a stiffer spring decreases the period. The ideal period is independent of amplitude while Hooke's law remains valid.

Worked example from local textbook question 6

For $T=1.00\,\mathrm s$ and $k=84\,\mathrm{N\,m^{-1}}$,

$$m=k\left(\frac{T}{2\pi}\right)^2=84\left(\frac{1.00}{2\pi}\right)^2=2.13\,\mathrm{kg}$$

Boundary

This model assumes a linear spring and that the stated oscillating mass includes any effective mass the question requires. Do not substitute amplitude for $m$.

Simple-pendulum period

For a pendulum of length $l$ undergoing small-angle oscillations,

$$T=2\pi\sqrt{\frac{l}{g}}$$

Measure $l$ from the pivot to the bob's centre of mass and use $g$ in $\mathrm{m\,s^{-2}}$.

Read the dependence

$T\propto\sqrt l$ and $T\propto1/\sqrt g$. Bob mass does not appear, so changing mass alone does not change the ideal period.

Worked example from local practice question 9

Changing $M$ to $4M$ has no effect. Changing $l$ to $0.25l$ gives

$$T'=2\pi\sqrt{\frac{0.25l}{g}}=0.5T$$

Boundary

The equation is the small-angle approximation, where $\sin\theta\approx\theta$ with $\theta$ in radians. Large amplitudes do not follow this period exactly.

Describe one complete cycle

In ideal SHM, total mechanical energy is constant. As the particle moves from an extreme position to equilibrium, stored potential energy changes into kinetic energy; from equilibrium to the opposite extreme, kinetic energy changes back into potential energy.

Use quarter-cycle checkpoints

At an extreme, speed and kinetic energy are zero while the relevant potential energy is maximum. At equilibrium, speed and kinetic energy are maximum while that potential energy is minimum. The energy pattern repeats every cycle.

Connect the model

A circular-motion picture can help visualize phase: the projected coordinate oscillates between two extremes and passes equilibrium twice per cycle. Use it as a representation of the oscillation, while the energy explanation remains based on the SHM position and speed.

Common trap

Do not claim that the energy transfer stops at equilibrium. The particle has maximum speed there, so the transfer reverses direction as it continues toward the next extreme.

Define phase angle

The phase angle $\phi$ specifies where an oscillator is within its cycle relative to a reference sine wave. It is an angular quantity measured in radians; one complete cycle is $2\pi$ radians.

Include the initial condition

For the chosen sine convention, displacement is $x=x_0\sin(\omega t+\phi)$, and velocity is $v=\omega x_0\cos(\omega t+\phi)$. The value of $\phi$ sets the displacement and direction of motion at $t=0$.

Compare oscillations

A phase difference of $\pi/2$ means one oscillation is a quarter-cycle ahead of the other; $\pi$ means they are in antiphase. Choose the smallest signed or positive phase difference required by the question.

Common trap

Do not mix degrees and radians in the equations, and do not infer phase from amplitude. Phase describes timing within the cycle, not the size of the oscillation.

HL motion equations

$$x=x_0\sin(\omega t+\phi)$$
$$v=\omega x_0\cos(\omega t+\phi)$$
$$v=\pm\omega\sqrt{x_0^2-x^2}$$

The sign of $v$ records direction; radians are used for phase.

HL energy equations

For an ideal oscillator,

$$E_T=\frac12m\omega^2x_0^2,\qquad E_p=\frac12m\omega^2x^2,\qquad E_k=E_T-E_p$$

At $|x|=x_0$, $E_p=E_T$; at $x=0$, $E_k=E_T$.

Stable solution order

Convert all quantities to SI units, find $\omega=2\pi f$, evaluate the phase in radians, substitute, and retain the velocity sign. Check that $|x|\le x_0$ and that each energy lies between zero and $E_T$.

Common trap

Do not omit $\omega$ from the velocity equation, confuse $x$ with amplitude $x_0$, or apply the quantitative energy equations to SL-only work.

Follow one particle, not the drawn shape

At a fixed position, a medium particle oscillates with time about equilibrium. In a position snapshot, different particles have different displacements at the same instant. The travelling pattern and energy move through the medium; the particles do not travel with the pattern.

Classification rule

Compare the particle or field oscillation direction with the propagation direction. A sinusoidal-looking graph alone does not determine whether a wave is transverse or longitudinal.

Define the quantities

Wavelength $\lambda$ is the shortest distance between points in phase. Frequency $f$ is cycles per second, period $T$ is the time for one cycle, and amplitude is maximum displacement from equilibrium. Wave speed $v$ is the speed at which the disturbance and energy propagate.

Connect time and space

For a travelling wave, $v=f\lambda=\lambda/T$. Use a spatial wavelength measured in metres and a temporal frequency measured in hertz; the result is in metres per second.

Read the same ideas in both wave types

For transverse waves, wavelength can be measured crest-to-crest or trough-to-trough. For longitudinal waves, measure compression-to-compression or rarefaction-to-rarefaction. In both cases the points are in phase.

Common trap

Do not use the distance from a crest to the next trough as one wavelength; that is half a wavelength. Do not confuse the speed of the medium particles with the propagation speed $v$.

Sound needs a mechanical medium

Sound is produced by a vibrating source and travels through matter as a mechanical wave. In air it is longitudinal: air molecules oscillate back and forth parallel to the direction in which the disturbance and energy propagate.

Compressions and rarefactions

A compression is a region of higher particle density and pressure; a rarefaction is a region of lower density and pressure. One wavelength is the distance between neighbouring compressions or neighbouring rarefactions.

Exam-language calibration from local practice question 2

A complete description connects all four ideas: longitudinal particle motion, propagation through air, alternating compressions/rarefactions, and energy transfer away from the source.

Boundary

The air molecules oscillate locally; they are not carried from loudspeaker to listener. Sound cannot propagate through a vacuum because there are no particles to sustain the mechanical disturbance.

Oscillating fields

An electromagnetic wave consists of oscillating electric and magnetic fields. The two fields are perpendicular to each other and both are perpendicular to the direction of propagation and energy transfer, so the wave is transverse.

No material medium is required

Electromagnetic fields can propagate through a vacuum. Every electromagnetic wave travels in vacuum at $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$, with $c=f\lambda$.

One spectrum, approximate regions

Radio, microwave, infrared, visible, ultraviolet, X-ray and gamma radiation are all electromagnetic waves. Use the approximate wavelength orders of magnitude supplied in the Physics data booklet; the named regions do not have perfectly sharp physical boundaries.

Common trap

Different spectrum regions do not have different vacuum speeds. They differ in frequency and wavelength while satisfying the same value of $c$.

Energy moves without net medium displacement

In a travelling mechanical wave, particles oscillate about equilibrium and pass the disturbance onward, so energy moves even though the medium has no resultant displacement after a complete cycle. In an electromagnetic wave, oscillating fields carry energy through space.

Use the same relationship carefully

Both models obey $v=f\lambda$, but $v$ is set by the relevant medium or, for electromagnetic waves in vacuum, by $c$. The source frequency links the spatial pattern to how rapidly the local oscillation repeats.

Common trap

“Transfers energy” does not mean matter must travel from source to receiver. It also does not mean mechanical and electromagnetic waves have the same physical oscillator.

Define a wavefront

A wavefront is a line or surface joining points that are in phase. Adjacent wavefronts are separated by one wavelength. In a uniform medium, the wavefronts are perpendicular to the direction of propagation.

Use rays to show propagation

A ray is a line showing the direction in which the wave transfers energy. Draw rays perpendicular to the local wavefronts; straight, parallel wavefronts give parallel rays, while circular wavefronts from a point source give radial rays.

Read the geometry

When a wavefront diagram changes direction at a boundary, compare the ray direction and the spacing of wavefronts on each side. The wavefront construction helps distinguish a change in speed from a change in frequency.

Common trap

Do not draw rays parallel to wavefronts. A ray follows energy propagation and is normal to the wavefront at each point.

Write the boundary relationship

For a ray crossing from medium 1 to medium 2, Snell’s law is $n_1\sin\theta_1=n_2\sin\theta_2$. Each angle is measured between the ray and the normal, not between the ray and the surface.

Solve the geometry first

Draw or identify the normal at the boundary, label incident and refracted angles, then substitute the refractive indices and sines. If the question asks for speed, combine the result with $n=c/v$.

Check the bend

Entering a higher-index medium decreases speed and bends the ray toward the normal. Entering a lower-index medium increases speed and bends it away from the normal; the frequency remains fixed at a stationary boundary.

Common trap

Do not use the angle between a wavefront and the normal as if it were the ray angle. A wavefront is perpendicular to the ray, so convert the angle when the diagram labels wavefronts.

Define refractive index

The refractive index of a medium is $n=c/v$, where $c$ is the speed of light in vacuum and $v$ is its speed in the medium. A larger $n$ means a lower light speed in that medium.

Relate speed and wavelength

At a stationary boundary the frequency is unchanged. Since $v=f\lambda$, a lower speed means a shorter wavelength. For two media, $n_2/n_1=\lambda_1/\lambda_2$ when the frequency is common.

Use a graph or measurement

If a graph’s gradient represents $n$, state that interpretation before reading the value. For uncertainty, use the spread from suitable maximum and minimum lines or the specified uncertainty method.

Common trap

Do not use $n=v/c$, and do not assume wavelength stays fixed when light enters a different medium. Frequency is the quantity that remains fixed at a stationary boundary.

Check the two conditions

Total internal reflection can occur only when a wave travels from a higher-index medium to a lower-index medium, and the incidence angle is greater than the critical angle. At the critical angle, the refracted ray travels along the boundary: $\theta_2=90^\circ$.

Calculate the critical angle

From Snell’s law, $\sin\theta_c=n_2/n_1$ for $n_1>n_2$. For a dense medium to air, $n_2\approx1$, so $\sin\theta_c=1/n_1$.

Use the boundary picture

For incidence below $\theta_c$, there is a refracted ray. At $\theta_c$, it grazes the boundary. Above $\theta_c$, no refracted ray propagates into the lower-index medium and all the light is reflected back into the denser medium.

Common trap

Do not use the critical-angle equation when light travels from lower to higher refractive index, and do not measure the critical angle from the surface rather than the normal.

Add overlapping displacements

When waves overlap, the resultant displacement at a point is the algebraic sum of the individual displacements: $y_{\mathrm{resultant}}=y_1+y_2$. The waves then continue propagating after the overlap.

Keep the signs

Displacements on the same side of equilibrium add; opposite displacements partially or completely cancel. Equal opposite pulses can produce zero displacement at an instant without destroying either wave.

Connect to interference

Repeated superposition of coherent waves can create stable maxima and minima. A diffraction pattern extending beyond a geometrical shadow is evidence that wave overlap and interference are involved.

Common trap

Do not add amplitudes as positive magnitudes only, and do not treat destructive interference as permanent disappearance of the waves.

Define coherence

Two waves are coherent if they have the same frequency and a constant phase difference. The phase relationship does not drift with time.

Connect coherence to a pattern

When coherent waves overlap, the locations of constructive and destructive interference remain fixed, producing a stable interference pattern. An ordinary pair of independent light sources usually has a changing phase relationship and does not produce a stable pattern.

Use one source when needed

A single source split into two paths can provide a common frequency and phase relationship. The resulting secondary sources can then act coherently for a double-source interference experiment.

Common trap

Same frequency alone is not enough. The phase difference must also remain constant; otherwise bright and dark locations move or wash out over time.

Define path difference

Path difference is the difference between the distances travelled by two waves from their sources to the same observation point. For in-phase coherent sources, it determines whether the waves arrive in phase or out of phase.

Apply the conditions

Constructive interference occurs when path difference $=n\lambda$. Destructive interference occurs when path difference $=(n+\tfrac12)\lambda$, where $n$ is a whole number.

Count fringes carefully

Start from the central bright fringe when the path difference is zero. Each additional bright fringe changes the path difference by $\lambda$; dark fringes lie halfway between adjacent bright conditions.

Common trap

Do not assign $\lambda/2$ to every dark point. The first dark condition is $\lambda/2$, then $3\lambda/2$, $5\lambda/2$, and so on.

Set up two-source interference

Two coherent sources emit waves with the same frequency and a constant phase difference. At each observation point, compare the two source-to-point distances to find the path difference.

Map bright and dark regions

For in-phase sources, path difference $n\lambda$ gives constructive interference and a bright or high-amplitude region. Path difference $(n+\tfrac12)\lambda$ gives destructive interference and a dark or low-amplitude region.

Read the pattern

Points equidistant from the two sources have zero path difference and form a central constructive line. Further maxima and minima occur where the path difference changes by half-wavelength steps.

Common trap

Do not use source-to-source separation as the path difference. It is the difference between the two travel distances to the same observation point.

Relate fringe spacing to the apparatus

For Young’s double-slit interference, fringe separation is $s=\lambda D/d$, where $\lambda$ is wavelength, $D$ is slit-to-screen distance and $d$ is slit separation.

Rearrange before substituting

Use $\lambda=sd/D$, $d=\lambda D/s$, or $D=sd/\lambda$ as needed. Measure the separation between adjacent bright or dark fringe centres; if several fringes are measured, divide the total width by the number of intervals.

Check the trends

Fringes spread farther apart when wavelength or screen distance increases, and become closer when slit separation increases. The small-angle model assumes $D\gg d$ and approximately plane wavefronts normal to the slits.

Common trap

Do not use the total width across several fringes as s without dividing by the number of fringe spacings, and do not confuse slit separation d with screen distance D.

A boundary can split wave energy

When a travelling wave reaches a boundary, part of its energy may be reflected back into the first medium and part may be transmitted into the second. The transmitted wave is refracted when its speed changes and it crosses the boundary at a non-zero angle.

Read rays and wavefronts together

Angles are measured from the normal. Rays show energy-transfer direction and remain perpendicular to wavefronts. Frequency stays fixed at a stationary boundary; a change in speed therefore changes wavelength and wavefront spacing.

Boundary cases

At normal incidence the transmitted ray does not bend, although its speed and wavelength may change. Unless absorption is stated, reflected and transmitted energy together account for the incident energy; their amplitudes do not generally add directly.

Diffraction is wave spreading

A wave diffracts when its wavefront bends into the region behind an obstacle or spreads after passing through an aperture. The wave remains in the same medium, so diffraction itself does not require a change of speed or frequency.

Compare wavelength with the opening or body

Spreading is most noticeable when the aperture width or obstacle size is comparable to the wavelength. An aperture much wider than the wavelength gives a broad central region that travels nearly straight with limited edge spreading; narrowing the aperture increases the angular spread.

Wavefront–ray representation

Before a straight aperture, incident wavefronts are parallel. Beyond a narrow aperture, draw curved outgoing wavefronts; rays stay perpendicular to them and fan outward. Around a body, wavefronts curve into the geometrical shadow.

Do not confuse mechanisms

Refraction is a direction change caused by a speed change at a boundary. Diffraction is spreading caused by an edge or aperture, even when the medium is unchanged.

Recognize the pattern

A monochromatic wave passing through a narrow rectangular slit spreads and forms a broad central maximum with weaker side maxima separated by minima. The pattern results from interference between contributions across the slit.

Use the first-minimum condition

For slit width $b$, the first minimum satisfies $\theta\approx\lambda/b$ for small angles. On a screen a distance $x$ away, the central maximum width is approximately $2x\lambda/b$.

Check the trends

A narrower slit or longer wavelength produces greater angular spreading and a wider central maximum. A wider slit or shorter wavelength produces a narrower pattern. The syllabus treatment is monochromatic light and rectangular slits at normal incidence.

Common trap

Do not confuse the distance from the central maximum to the first minimum with the full central-maximum width; the latter is twice the first-minimum distance on the screen.

Separate the two patterns

In a multiple-slit intensity pattern, the fine interference maxima are contained within a broad single-slit diffraction envelope. The envelope sets the overall intensity scale; the double-slit or multiple-slit interference determines the rapid fringe structure.

Relate the widths

The single-slit envelope depends on slit width $b$, while the separation of fine interference maxima depends on source or slit separation $d$. A smaller $b$ makes the envelope wider; a larger $d$ makes interference fringes closer together.

Explain missing or unequal maxima

Interference maxima at different angles can have different intensities because the envelope changes across the screen. Some interference maxima may fall at an envelope minimum and disappear.

Common trap

Do not treat every interference maximum as having the same height, and do not confuse the fine fringe spacing with the width of the single-slit envelope.

Core idea

A diffraction grating has many equally spaced parallel slits. Bright principal maxima occur when the path difference between adjacent slits is an integer number of wavelengths:

$$n\lambda=d\sin\theta$$

Here, $n$ is the order number $0,1,2,\ldots$, $\lambda$ is the wavelength, $d$ is the spacing between adjacent slits, and $\theta$ is measured from the central maximum to the chosen maximum.

Build the model

If a grating has $N$ lines per metre, the slit spacing is $d=1/N$. For a selected maximum, identify its order $n$, convert $d$ and $\lambda$ to consistent units, and solve for the unknown angle, wavelength or spacing. The central maximum is $n=0$; the first maxima on either side are $n=1$.

Check the allowed orders

Because $\lvert\sin\theta\rvert\le 1$, a wavelength can only produce orders satisfying $n\lambda\le d$. The largest possible order is therefore the greatest integer not exceeding $d/\lambda$. For overlapping wavelengths, equate their path-difference conditions: if the second-order maximum of $\lambda_1$ coincides with the third-order maximum of $\lambda_2$, then $2\lambda_1=3\lambda_2$.

Common trap

Do not use the number of lines per metre as $d$; invert it first. Do not count the central maximum as first order, and do not replace the grating equation with the small-angle approximation unless the question explicitly permits that approximation.

Core idea

A standing wave forms when two waves with the same frequency, wavelength and amplitude travel in opposite directions and superpose. In practice, one wave is often the incident wave and the other is its reflection. The pattern oscillates in place rather than travelling along the medium.

Build the physical model

At each point, add the displacements of the two waves. Where they always cancel, the amplitude is zero: these fixed positions are nodes. Where they reinforce most strongly, the amplitude is greatest: these fixed positions are antinodes. The wave pattern repeats every half-wavelength, so adjacent nodes and adjacent antinodes are separated by $\lambda/2$, while a node and its nearest antinode are $\lambda/4$ apart.

Interpret what is and is not moving

The particles of the medium still oscillate between nodes and antinodes, but the locations of the nodes and antinodes do not move. A standing wave does not transfer energy progressively from one end to the other in the way a travelling wave does; energy is stored and exchanged locally within each segment between adjacent nodes.

Check the boundary

Do not describe a standing wave as a single wave travelling forward. First identify the two counter-propagating waves and then use superposition to explain the fixed pattern. The model here is limited to two identical opposite-travelling waves; the syllabus does not require superposition of more than two waves.

Identify the positions

A node is a fixed position where the displacement is always zero. An antinode is a fixed position where the amplitude is greatest. Adjacent nodes or adjacent antinodes are separated by $\lambda/2$; a node and its nearest antinode are separated by $\lambda/4$.

Read relative amplitude

Every point between two adjacent nodes oscillates at the same frequency, but its amplitude depends on position: zero at a node, maximum at an antinode, and intermediate elsewhere. The standing-wave envelope therefore describes amplitude, not a travelling displacement profile at one instant.

Read phase

Points in the same segment between adjacent nodes oscillate in phase. Points in neighbouring segments oscillate in antiphase, with phase difference $\pi$ (180°). At a node the phase is not useful to assign because the displacement amplitude is zero.

Common trap

Do not infer phase only from distance. First locate the nodes: crossing one node changes the phase by $\pi$; staying within the same node-to-node segment leaves the phase difference zero.

Start with the boundary conditions

A fixed end of a string is a displacement node; a free end is a displacement antinode. For air displacement in a pipe, a closed end is a displacement node and an open end is a displacement antinode. These end conditions determine which standing-wave patterns are allowed.

Use the string patterns

For a string fixed at both ends, or with two free ends, the nth harmonic has $n$ half-wavelengths in length $L$: $\lambda_n=2L/n$ and $f_n=nv/(2L)$. For one fixed and one free end, the allowed patterns contain an odd number of quarter-wavelengths: $\lambda_n=4L/(2n-1)$ and $f_n=(2n-1)v/(4L)$, with $n=1,2,3,\ldots$.

Apply the same geometry to pipes

An open pipe has displacement antinodes at both ends and follows the two-open-end pattern. A closed pipe has a displacement node at the closed end and an antinode at the open end, so only the odd sequence of harmonics is allowed. Use $v=f\lambda$ after finding the wavelength from the boundary pattern. End corrections for open pipes are not required.

Common trap

Do not use the closed-pipe formula for an open pipe, and do not count pressure nodes or pressure antinodes here: the syllabus asks for air-displacement nodes and antinodes. Also use “first harmonic” for the lowest-frequency mode; the syllabus does not require the terms fundamental or overtone.

Separate the frequencies

The natural frequency is the frequency at which a system oscillates after a disturbance when it is left alone. The driving frequency is imposed by an external periodic force. Resonance occurs when the driving frequency is equal or very close to the system’s natural frequency, producing a large amplitude response.

Explain the large amplitude

At resonance, the driving force supplies energy efficiently to the oscillator each cycle because its timing is well matched to the motion. The amplitude rises until the energy supplied per cycle is balanced by energy dissipated. Greater energy dissipation means a smaller maximum amplitude.

Read a frequency-response graph

Plot amplitude against driving frequency. The peak identifies the resonant frequency; the peak height is the maximum amplitude. A practical system may have its peak slightly displaced from its undamped natural frequency when damping is significant, but the syllabus requires only a qualitative frequency-response analysis.

Recognize useful and destructive resonance

Resonance is useful when a large, frequency-selective response is wanted, such as tuning a receiver or producing a strong musical sound. It can be destructive when repeated driving builds damaging oscillations in a bridge, building or machine. Designs then change the natural frequency, avoid the matching driving frequency, or add damping.

Common trap

Do not call the driving frequency the natural frequency. A large amplitude alone is not enough to establish resonance: connect it to the driving frequency being close to the natural frequency and to efficient energy transfer.

Read the response peak

Damping removes mechanical energy from an oscillator. On an amplitude-versus-driving-frequency graph, increasing damping lowers the maximum amplitude and makes the peak less sharp. The resonant frequency also shifts slightly to a lower value. These are qualitative changes; the syllabus does not require a detailed damped-oscillator derivation.

Connect damping to energy

More damping means more energy is dissipated during each cycle. The driver must supply that lost energy, but the oscillator cannot build up as large an amplitude before input and loss balance. With little damping, energy accumulates more efficiently and the resonance peak is taller and narrower.

Apply the model

If a suspension or bridge is damped, the oscillation amplitude is reduced and the resonant response occurs at a slightly lower driving frequency. This can be useful for controlling vibration, although damping also reduces the sharpness of frequency selection.

Common trap

Do not draw a damped response with a taller peak. More damping lowers the peak and shifts it left on a frequency axis whose driving frequency increases to the right.

Classify the response

Light damping lets the system oscillate about equilibrium while its amplitude decreases gradually. Critical damping returns the system to equilibrium in the shortest time without oscillating. Heavy damping also avoids oscillation, but returns to equilibrium more slowly than critical damping.

Choose the response from the design goal

A system that must settle quickly without repeated oscillation is adjusted close to critical damping. Too little damping allows repeated crossings of equilibrium; too much damping resists the motion so strongly that the return takes longer.

Common trap

Critical and heavy damping are both non-oscillatory, but they are not equally fast. Critical damping is the fastest return without overshoot; heavy damping is slower.

Core idea

The Doppler effect is the observed change in frequency, and therefore usually wavelength, caused by relative motion between a wave source and an observer. When the source and observer approach, wavefronts arrive more frequently and the observed frequency is higher. When they separate, the observed frequency is lower.

Apply it to sound

For sound, the wave travels through a medium. A moving source changes the spacing of emitted wavefronts in the medium; a moving observer changes how quickly the observer meets the wavefronts. In either case, motion toward one another gives a higher observed frequency and motion apart gives a lower one. The source frequency itself has not changed merely because the observer hears a different frequency.

Apply it to light

The Doppler effect also occurs for electromagnetic waves. A source moving toward an observer produces a shorter observed wavelength and a higher frequency (blueshift); moving away produces a longer wavelength and lower frequency (redshift). Unlike sound, the measured speed of light remains $c$; the observed change is in frequency and wavelength.

Use the shift as a measurement

Medical Doppler ultrasound uses a frequency shift in reflected sound to infer blood-flow speed. Radar uses a shift in reflected microwaves to infer the radial speed of a vehicle, aircraft or storm. In both cases the detected shift is tied to motion toward or away from the receiver, not to a change in the emitted frequency at the source.

Check the boundary

Do not explain light Doppler shift by adding the source speed to $c$. In this course the low-relative-speed approximation is used for the change in frequency or wavelength; the light speed remains $c$.

Start with the reference case

A stationary point source emits circular wavefronts whose centres remain at the source and whose spacing is the emitted wavelength. Use this as the comparison before adding motion. The wavefronts travel through the medium at the wave speed.

Move the source

If the source moves toward a stationary observer, successive wavefronts are emitted from progressively advanced positions, so the wavefronts are compressed in front of the source and spread behind it. The observer in front receives a shorter wavelength and higher frequency; behind it, the wavelength is longer and the frequency lower.

Move the observer

If the source is stationary and the observer moves toward it, the wavefront spacing in the medium is unchanged. The observer meets wavefronts more often, so the observed frequency increases. Moving away gives a lower observed frequency; it does not change the wavelength in the medium.

Common trap

For a moving source, shift the centres of successive wavefronts; for a moving observer, keep the wavefronts equally spaced and change only the rate at which the observer encounters them. Do not combine moving-source and moving-observer cases unless the question explicitly asks for both.

Use the low-speed model

This approximation applies when the relative source–observer speed $v$ is much smaller than the speed of light $c$. Here $f$ and $\lambda$ are emitted values, while $\Delta f$ and $\Delta\lambda$ are the magnitudes of the observed shifts.

Keep the direction

Motion away produces a redshift: $\Delta\lambda>0$ and the observed frequency decreases. Motion toward produces a blueshift: wavelength decreases and frequency increases. If a question asks for speed, use the absolute shift; if it asks for direction, use the sign of the wavelength or frequency change.

Worked example — spectral line

A line emitted at $4.86\times10^{-7}\,\text{m}$ is observed from a receding galaxy at $5.21\times10^{-7}\,\text{m}$. Then $\Delta\lambda=3.5\times10^{-8}\,\text{m}$. Substitution gives $v=(3.00\times10^8)(3.5\times10^{-8}/4.86\times10^{-7})=2.16\times10^7\,\text{m s}^{-1}$. The longer observed wavelength identifies recession.

Common trap

Do not put $\Delta\lambda$ in the denominator, and do not add $v$ to $c$. The approximation changes the observed frequency or wavelength, not the invariant speed of light.

Use the line pattern as a fingerprint

Elements produce characteristic sets of spectral lines. Compare the same line pattern measured in a laboratory with the pattern observed from a star or galaxy. If every characteristic line is displaced by the same fractional amount while the pattern remains recognisable, the displacement is evidence of relative motion along the line of sight.

Interpret redshift and blueshift

A shift toward longer wavelengths and lower frequencies is a redshift, indicating recession along the line of sight. A shift toward shorter wavelengths and higher frequencies is a blueshift, indicating approach. The line pattern itself identifies the element; the displacement carries the motion information.

State exactly what the shift reveals

A Doppler spectral shift gives the component of relative motion along the observer’s line of sight. Redshift indicates that separation is increasing; blueshift indicates that it is decreasing. The shift alone does not measure motion across the line of sight, and it does not imply that the emitting element or the speed of light has changed.

Common trap

Do not say that a redshift means the element has changed. The spectral identity remains in the line pattern; the wavelengths have shifted because of relative motion.

Set the scope

These equations apply to sound or other mechanical waves travelling at speed $v$ through a medium. Assume the source and observer move along the straight line joining them, and treat one moving object at a time as required by the syllabus. They do not apply to electromagnetic waves.

Choose the sign from the motion

For a moving source, subtract $u_s$ when it approaches and add it when it recedes. For a moving observer, add $u_o$ when it approaches and subtract it when it recedes. This makes approach give $f'>f$ and separation give $f'<f$.

Worked example — moving source

A $480\,\text{Hz}$ sound source approaches a stationary observer at $50.0\,\text{m s}^{-1}$ while sound travels at $350\,\text{m s}^{-1}$. Use the source-toward sign: $f'=480[350/(350-50.0)]=560\,\text{Hz}$. The result is above $480\,\text{Hz}$, matching the approach check.

Keep the syllabus boundary

Use one moving object at a time and motion along the line joining source and observer. Keep every speed in the same units. These mechanical-wave equations do not apply to electromagnetic waves; reflected medical or radar signals may involve a shift on both outward and return paths.