C. Wave behaviour

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.

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.

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.

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 $f=196\,\mathrm{Hz}$.

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

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

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 $k=84\,\mathrm{N\,m^{-1}}$,

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

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.