Unit 2: Waves and Electricity

On a displacement graph, amplitude is measured from the equilibrium line to a crest or trough, not from crest to trough. A displacement-time graph at one position gives period; a displacement-distance snapshot at one time gives wavelength.

The wave equation is $v=f\lambda$. During one period $T$, a wavefront moves one wavelength, so $v=\lambda/T=f\lambda$ because $f=1/T$. Use speed in $\mathrm{m\,s^{-1}}$, frequency in hertz and wavelength in metres.

Identify the wave speed in the relevant medium, convert prefixes such as MHz or nm, and rearrange before substituting. If speed stays constant, increasing frequency shortens wavelength in inverse proportion.

A sound wave of frequency $680\,\mathrm{Hz}$ travels at $340\,\mathrm{m\,s^{-1}}$. Its wavelength is $\lambda=v/f=340/680=0.50\,\mathrm{m}$.

In a longitudinal wave, molecules or particles oscillate parallel to the direction in which the wave travels. Their alternating crowding and separation produce compressions and rarefactions, so pressure varies as the disturbance passes.

The wavelength is the distance between neighbouring compressions, neighbouring rarefactions, or any two nearest points in the same phase. A molecule oscillates about its equilibrium position rather than travelling with the wave from source to receiver.

In a transverse wave, the oscillating quantity is perpendicular to the direction of wave propagation. For a wave travelling horizontally along a string, each element of string may move vertically while the disturbance travels horizontally.

A displacement-distance snapshot may show crests and troughs. These mark positive and negative displacement from equilibrium, while wavelength is measured between neighbouring points in the same phase, such as crest to crest.

Transverse waves can be plane polarised because their oscillations have directions perpendicular to travel. This distinguishes them from longitudinal waves, whose oscillations lie along the travel direction.

Label axes with quantity and unit, use the equilibrium line consistently, and apply any scale factor. Frequency comes from $f=1/T$ after reading a full cycle. For a standing wave, neighbouring nodes or neighbouring antinodes are $\lambda/2$ apart.

A longitudinal wave may still be represented by a sinusoidal pressure-distance or displacement-distance graph. The vertical graph coordinate is a measured quantity, not a direction in which the wave itself travels.

Connect a signal generator to a loudspeaker and also to one channel of a 2-beam oscilloscope. Connect a microphone to the second channel. Display stable traces with the same timebase and place the microphone in line with the speaker.

Record the microphone position when the two traces have a clearly defined phase relation, such as in phase. Move the microphone away until that same phase relation next occurs; the displacement is one wavelength. For lower percentage uncertainty, move through several repeats and divide the total displacement by the number of wavelengths.

Read the period from the oscilloscope and calculate $f=1/T$, or use the calibrated generator frequency. Then calculate $v=f\lambda$. Repeat positions and use a best-fit relation where possible; record the air temperature because sound speed depends on conditions.

Waves arriving in phase reinforce to give constructive interference and a larger resultant amplitude. Waves arriving in antiphase oppose and give destructive interference; equal amplitudes can cancel completely.

A path difference of one wavelength corresponds to one complete phase cycle. Therefore $\Delta\phi=2\pi(\Delta x/\lambda)$ radians, or $\Delta\phi=360^\circ(\Delta x/\lambda)$.

If $\Delta x=3\lambda/8$, then $\Delta\phi=2\pi(3/8)=3\pi/4$ radians, or $135^\circ$. Equivalent phases may differ by any whole multiple of $2\pi$.

A standing or stationary wave forms when two coherent waves of the same frequency travel in opposite directions and superpose, commonly because an incident wave reflects. Their interference fixes a pattern of nodes and antinodes in space.

Neighbouring nodes are $\lambda/2$ apart, as are neighbouring antinodes; a node and its nearest antinode are $\lambda/4$ apart. There is no net energy transfer along an ideal stationary wave, although energy moves locally between stores.

For a transverse wave on a stretched string, $v=\sqrt{T/\mu}$, where $T$ is tension in newtons and $\mu$ is mass per unit length in $\mathrm{kg\,m^{-1}}$. Linear density can be measured from $\mu=m/L$.

At fixed $\mu$, multiplying tension by four doubles speed. At fixed tension, multiplying $\mu$ by four halves speed. The square-root dependence means speed is not directly proportional to either quantity.

For $T=90\,\mathrm{N}$ and $\mu=2.5\times10^{-3}\,\mathrm{kg\,m^{-1}}$, $v=\sqrt{90/(2.5\times10^{-3})}=190\,\mathrm{m\,s^{-1}}$ to two significant figures.

Drive a stretched string with a vibration generator and signal generator. Adjust frequency until a clear stationary-wave pattern with a fixed number of loops appears. Measure vibrating length $L$, obtain tension from a hanging load where $T=mg$, and determine $\mu$ from the mass and length of a sample.

For the fundamental, $\lambda=2L$ and $f=(1/2L)\sqrt{T/\mu}$. Change one variable at a time, retune to the same mode, repeat readings and use a best-fit line to judge the predicted relationship.

Radiation intensity is power incident per unit area perpendicular to propagation: $I=P/A$. Its SI unit is $\mathrm{W\,m^{-2}}$. Rearranging gives $P=IA$ and, over time $t$, transferred energy $E=IAt$.

Radiation of intensity $250\,\mathrm{W\,m^{-2}}$ falls normally on a $0.40\,\mathrm{m^2}$ surface. The incident power is $P=(250)(0.40)=100\,\mathrm{W}$, so $300\,\mathrm{J}$ arrives in $3.0\,\mathrm{s}$.

If a source radiates power uniformly in all directions, the relevant area at radius $r$ is $4\pi r^2$, so $I=P/(4\pi r^2)$. This inverse-square result follows from the growing spherical area.

Refractive index is $n=c/v$, where $c$ is light speed in vacuum and $v$ is light speed in the medium. A larger $n$ means lower wave speed.

At an interface, $n_1\sin\theta_1=n_2\sin\theta_2$. Measure both angles from the normal. Light entering a higher-index medium bends toward the normal; entering a lower-index medium bends away.

Light travels from air $(n_1=1.00)$ into glass $(n_2=1.50)$ at $30^\circ$. Then $\sin\theta_2=(1.00/1.50)\sin30^\circ=0.333$, so $\theta_2=19.5^\circ$.

The critical angle $C$ is the incidence angle in the higher-index medium for which the refraction angle in the lower-index medium is $90^\circ$. For a material of refractive index $n$ meeting air, $\sin C=1/n$.

Confirm that the ray travels from higher $n$ to lower $n$, calculate $1/n$, then use the inverse sine in degree mode. For $n=1.52$, $C=\sin^{-1}(1/1.52)=41.1^\circ$.

For two non-air media, Snell's law gives $\sin C=n_{lower}/n_{higher}$. The ratio must not exceed 1, consistent with the higher-to-lower condition.

If both conditions hold, no refracted ray propagates into the second medium and the wave reflects internally. At $i=C$, the refracted ray travels along the boundary. At $i<C$, some wave is transmitted by refraction and some may be reflected.

Identify the two media, locate the normal, calculate or read $C$, and compare the incidence angle measured from the normal. State both the comparison and the resulting path.

Place a transparent block on paper and trace its outline. Direct a narrow monochromatic ray at one face. Mark the incident and transmitted paths, remove the block, join the marks, and draw the normal at the entry point. Measure incidence $i$ and refraction $r$ from the normal with a protractor.

Repeat for several incidence angles. For air entering the solid, calculate $n=\sin i/\sin r$, or plot $\sin i$ vertically against $\sin r$ horizontally; the best-fit gradient is $n$ when $n_{air}\approx1$.

Use a thin ray, widely separated path marks and angles neither extremely small nor near grazing incidence. Repeat measurements and use monochromatic light so different wavelengths do not refract by different amounts.

Unpolarised transverse radiation has oscillations in many directions perpendicular to travel. Plane-polarised radiation has oscillations restricted to one plane containing the direction of propagation.

A polarising filter transmits the component aligned with its transmission axis. A second filter acts as an analyser: aligned axes give strong transmission, while perpendicular axes ideally give no transmission. An intermediate angle transmits an intermediate intensity.

Polarisation is possible only when oscillations have a direction perpendicular to propagation. Its observation therefore supports the transverse nature of electromagnetic waves.

Diffraction is the spreading of waves after passing through a gap or around an obstacle. It is most noticeable when the gap or obstacle size is comparable with the wavelength.

In Huygens' construction, every point on an existing wavefront acts as a source of secondary wavelets. After a short time, the new wavefront is the envelope tangent to those wavelets.

Most wavelets are blocked at a barrier, but wavelets from points across a slit spread into the region beyond it. Their envelope is curved, predicting diffracted wavefronts. At an obstacle edge, unblocked wavelets extend into the geometric shadow.

For normal incidence on a diffraction grating, principal maxima satisfy $n\lambda=d\sin\theta$. Here $n=0,1,2,\ldots$ is order, $d$ is slit spacing and $\theta$ is measured from the central normal.

Convert line density to spacing: if a grating has $N$ lines per metre, $d=1/N$. For lines per millimetre, first multiply by $1000$ to obtain lines per metre.

Identify the order, convert $\lambda$ and $d$ to metres, solve for $\sin\theta=n\lambda/d$, then use inverse sine. A possible order must satisfy $n\lambda\le d$ because $\sin\theta\le1$.

Mount a diffraction grating of known line density perpendicular to a narrow light beam. Place a screen a measured perpendicular distance $D$ away. For a laser, keep the beam below eye level, secure it, never view the beam directly and remove reflective objects.

Mark the central maximum and matching order $+n$ and $-n$ maxima. Measure their separation and halve it to obtain $x$, reducing centre-location error. Calculate $\theta$ from $\tan\theta=x/D$.

Convert line density to $d$, then calculate $\lambda=d\sin\theta/n$. Repeat with several orders or distances and average consistent values. A graph of $\sin\theta$ against $n$ has gradient $\lambda/d$, providing a stronger multi-point result.

A beam of electrons passing through a thin crystalline target produces a diffraction pattern, such as concentric rings. The regular atomic spacing acts like a diffraction grating.

Diffraction is a wave phenomenon, so the pattern is evidence that electrons have wave behaviour. It occurs when the electron de Broglie wavelength is comparable with the spacing of atomic planes, allowing waves associated with different paths to interfere.

Changing electron momentum changes the de Broglie wavelength and therefore changes the diffraction geometry. The systematic pattern change links the effect to wavelength rather than to random particle scattering.

The de Broglie relationship is $\lambda=h/p$, where $h=6.63\times10^{-34}\,\mathrm{J\,s}$ and $p$ is momentum in $\mathrm{kg\,m\,s^{-1}}$. For a non-relativistic particle, $p=mv$.

Calculate momentum first, then divide $h$ by it. A larger momentum gives a shorter wavelength, so diffraction becomes harder to observe for everyday macroscopic objects.

An electron with momentum $2.0\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}$ has $\lambda=(6.63\times10^{-34})/(2.0\times10^{-23})=3.3\times10^{-11}\,\mathrm{m}$.

When a wave reaches an interface between media, part of its energy may return as a reflected wave and part may enter the second medium as a transmitted wave. The proportions depend on the media and boundary conditions.

For a straight boundary, reflection obeys equal incidence and reflection angles measured from the normal. Transmission may involve refraction when the wave speed changes.

A transducer sends a short pulse and detects its echo from an interface or object. If wave speed is $v$ and the delay from emission to echo is $t$, the one-way distance is $d=vt/2$ because the pulse travels out and back.

An ultrasound echo returns after $40\,\mu\mathrm{s}$ in material where $v=4000\,\mathrm{m\,s^{-1}}$. Then $d=(4000)(40\times10^{-6})/2=0.080\,\mathrm{m}$.

The wave model grew from evidence for interference and diffraction, but a wave-only account could not explain all observations of energy exchange with matter. The photon model was developed to represent those exchanges as discrete interactions. Later evidence therefore extended the available modelling rather than simply erasing the successful wave description.

Choose the model that exposes the behaviour under study. Electromagnetic radiation can display wave behaviour during propagation and interference while exchanging energy with matter in photon-sized amounts.

The energy of one photon is $E=hf$, where $h=6.63\times10^{-34}\,\mathrm{J\,s}$ and $f$ is frequency in hertz. Since $f=c/\lambda$ in vacuum, shorter-wavelength radiation has greater photon energy.

For $f=6.0\times10^{14}\,\mathrm{Hz}$, $E=(6.63\times10^{-34})(6.0\times10^{14})=4.0\times10^{-19}\,\mathrm{J}$ per photon to two significant figures.

Photon energy and beam intensity are different. At fixed frequency, greater intensity means more photon energy arriving per unit area per unit time, usually through a greater photon rate, not more energy per photon.

In the photoelectric effect, one photon is absorbed in a one-to-one interaction with one electron. The photon transfers its energy $hf$ as a single amount.

Part of that energy may be required to release the electron from the metal surface. If the photon energy is sufficient, the remaining energy becomes electron kinetic energy. If it is insufficient, increasing the number of identical low-energy photons does not release an electron in this model.

At a frequency above threshold, increasing intensity increases the rate at which photons arrive, so more electrons can be emitted per second. It does not increase the energy of each unchanged-frequency photon.

The work function $\phi$ is the minimum energy needed to release an electron from a metal surface. The threshold frequency $f_0$ is the minimum radiation frequency that can cause emission, so $\phi=hf_0$.

Energy conservation for the most energetic emitted electrons is $hf=\phi+\tfrac12mv_{max}^2$. Photon energy pays the work function first; any remainder becomes maximum kinetic energy.

If $hf=5.0\times10^{-19}\,\mathrm{J}$ and $\phi=3.2\times10^{-19}\,\mathrm{J}$, then $E_{k,max}=1.8\times10^{-19}\,\mathrm{J}$.

One electronvolt is the energy transferred when a particle with elementary charge moves through a potential difference of one volt: $1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}$.

$3.2\,\mathrm{eV}=(3.2)(1.60\times10^{-19})=5.1\times10^{-19}\,\mathrm{J}$. Conversely, $8.0\times10^{-19}\,\mathrm{J}=5.0\,\mathrm{eV}$.

These observations support a particle description in which electromagnetic energy arrives in discrete photons. In particular, intense radiation below threshold still fails to emit electrons, whereas a continuous-wave-only energy accumulation picture would not predict this frequency cutoff and immediate response together.

Electrons in an atom can occupy only discrete energy levels. An electron moving from a higher level $E_h$ to a lower level $E_l$ emits one photon with $hf=E_h-E_l$. Absorption occurs when a photon supplies the matching energy difference for an upward transition.

Because only particular level differences exist, only particular photon frequencies and wavelengths are emitted or absorbed. These appear as separate spectral lines rather than a continuous range.

For an energy difference $\Delta E=3.0\times10^{-19}\,\mathrm{J}$, $f=\Delta E/h=(3.0\times10^{-19})/(6.63\times10^{-34})=4.5\times10^{14}\,\mathrm{Hz}$. If required, wavelength follows from $\lambda=c/f$.

Electric current is the rate at which charge passes a point: $I=\Delta Q/\Delta t$. One ampere is one coulomb per second. Rearranging gives $\Delta Q=I\Delta t$.

Current can be carried by different charged particles. In a metal the mobile carriers are electrons, but conventional current is defined in the direction positive charge would move, opposite to electron drift.

A steady current of $0.40\,\mathrm{A}$ flows for $3.0\,\mathrm{min}=180\,\mathrm{s}$. The charge transferred is $Q=(0.40)(180)=72\,\mathrm{C}$. The number of electrons is $Q/e=4.5\times10^{20}$.

The potential difference between two points is the energy transferred per unit charge moving between them: $V=W/Q$. One volt is one joule per coulomb. Rearrangements are $W=VQ$ and $Q=W/V$.

Across a component, electrical energy is transferred to other stores. A larger p.d. means more energy is transferred for each coulomb, not necessarily that more charge flows; current also depends on the circuit resistance.

Moving $2.5\,\mathrm{C}$ through a p.d. of $12\,\mathrm{V}$ transfers $W=(12)(2.5)=30\,\mathrm{J}$. An electron accelerated through $108\,\mathrm{V}$ gains energy of magnitude $eV$.

Resistance at an operating point is defined by $R=V/I$ and measured in ohms. It compares the p.d. across a component with the current through it.

Ohm's law states that $I\propto V$ for a conductor at constant temperature. The ratio $V/I$ is then constant, so an $I$-against-$V$ graph is a straight line through the origin.

If $V=6.0\,\mathrm{V}$ produces $I=0.25\,\mathrm{A}$, then $R=6.0/0.25=24\,\Omega$. Doubling $V$ doubles $I$ only if conditions, especially temperature, keep $R$ constant.

Current division is a charge-flow statement; p.d. division is an energy-per-charge statement. A larger series resistance receives a larger share of p.d. because the same current flows and $V=IR$.

In series, current $I$ is common and p.d.s add. Using $V_T=V_1+V_2+\cdots$ and $V=IR$ gives $IR_T=IR_1+IR_2+\cdots$, so $R_T=R_1+R_2+\cdots$.

In parallel, p.d. $V$ is common and branch currents add. Using $I_T=I_1+I_2+\cdots$ and $I=V/R$ gives $V/R_T=V/R_1+V/R_2+\cdots$, so $1/R_T=1/R_1+1/R_2+\cdots$.

Two resistors $6.0\,\Omega$ and $3.0\,\Omega$ give $9.0\,\Omega$ in series but $2.0\,\Omega$ in parallel. A parallel equivalent must be smaller than the smallest branch resistance because it provides more paths for charge.

Substituting $V=IR$ into $P=VI$ gives $P=I^2R$. Substituting $I=V/R$ gives $P=V^2/R$. The choice matters when comparing circuits: at fixed current, power rises with $R$; at fixed p.d., power falls with $R$.

A $12\,\Omega$ resistor carries $0.50\,\mathrm{A}$. It dissipates $P=(0.50)^2(12)=3.0\,\mathrm{W}$ and transfers $W=(3.0)(40)=120\,\mathrm{J}$ in $40\,\mathrm{s}$.

At any point, resistance is $V/I$, not simply the gradient of an $I$-against-$V$ curve. A steeper ray from the origin corresponds to larger $I/V$ and therefore smaller resistance.

To obtain a graph, place an ammeter in series and a voltmeter across the component, vary p.d. safely, and reverse polarity when negative values are required. Include a current-limiting resistor for a diode.

For a uniform conductor, $R=\rho l/A$, where $l$ is length, $A$ is cross-sectional area and $\rho$ is resistivity in $\Omega\,\mathrm{m}$. Rearranging gives $\rho=RA/l$.

For the same material and temperature, resistance is directly proportional to length and inversely proportional to area. Doubling length doubles $R$; doubling diameter makes area four times larger and reduces $R$ to one quarter.

A wire has $R=3.0\,\Omega$, $l=2.0\,\mathrm{m}$ and $A=1.5\times10^{-6}\,\mathrm{m^2}$. Then $\rho=(3.0)(1.5\times10^{-6})/2.0=2.3\times10^{-6}\,\Omega\,\mathrm{m}$.

Measure wire diameter with a micrometer at several positions and orientations, correct any zero error, average it and calculate $A=\pi d^2/4$. Measure the test length $l$ between electrical contacts with a metre rule.

Connect the test wire and ammeter in series with a d.c. supply, switch and variable resistor; connect a voltmeter across the measured length. Use a low current and close the switch only for readings so heating does not change resistivity.

For several lengths, record $V$ and $I$ and calculate $R=V/I$. Plot $R$ vertically against $l$ horizontally. From $R=(\rho/A)l$, best-fit gradient $m=\rho/A$, so $\rho=mA$. A single-length result may instead use $\rho=RA/l$, but repeated lengths give a stronger test.

The current carried through cross-sectional area $A$ is $I=nqvA$. Here $n$ is mobile charge-carrier number density, $q$ is charge per carrier and $v$ is mean drift speed.

For a given area, current is larger when more mobile carriers are available or when their drift response is greater. Metals have a large density of conduction electrons and therefore conduct readily. Insulators have extremely few mobile carriers, producing tiny current for an applied field; semiconductors can change carrier number strongly with temperature or illumination.

If two wires carry the same current with the same $n$ and $q$, but one has four times the area, its drift speed is one quarter as large. Conversely, simultaneous fourfold increases in $I$ and $A$ leave $v$ unchanged.

A uniform wire has constant resistivity and cross-sectional area, so resistance from one end to distance $x$ is $R_x=\rho x/A$. In steady current $I$, the p.d. across that length is $V_x=IR_x$, hence $V_x\propto x$.

A graph of potential against distance is a straight line for a uniform wire carrying constant current. Its sign of gradient depends on the chosen direction and reference potential; its magnitude is the potential gradient.

If a wire drops $6.0\,\mathrm{V}$ uniformly over $1.5\,\mathrm{m}$, the drop is $4.0\,\mathrm{V\,m^{-1}}$. A point $0.40\,\mathrm{m}$ from the high-potential end is $1.6\,\mathrm{V}$ below that end.

Two resistors $R_1$ and $R_2$ in series carry the same current $I=V_{in}/(R_1+R_2)$. The output across $R_2$ is therefore $V_{out}=IR_2=V_{in}R_2/(R_1+R_2)$.

With $V_{in}=12\,\mathrm{V}$, $R_1=2.0\,\mathrm{k\Omega}$ and $R_2=4.0\,\mathrm{k\Omega}$, the output across $R_2$ is $12(4.0/6.0)=8.0\,\mathrm{V}$.

For an unloaded divider, the output across a component is $V_{out}=V_{in}R_{out}/R_{total}$. First mark exactly which resistor the output spans; then change the sensor resistance and recompute the ratio.

A control circuit switches when $V_{out}$ crosses a threshold. To predict its response, find sensor resistance at the stated condition, calculate output, then determine how the output moves on either side of that condition.

When a cell supplies current, energy conservation gives $\varepsilon=V+Ir$, so $V=\varepsilon-Ir$. Increasing current increases the lost volts and lowers terminal p.d. if $\varepsilon$ and $r$ remain constant.

With negligible current, $Ir\approx0$ and terminal p.d. is approximately the e.m.f. Under load, the source behaves like an ideal e.m.f. in series with its internal resistance.

Connect the cell, ammeter, switch and variable resistor in series. Connect a voltmeter directly across the cell terminals. Use high-resistance voltmeter and vary the external resistance to obtain a safe range of current values.

For each setting, close the switch briefly, record current $I$ and terminal p.d. $V$, then open it to reduce heating and cell discharge. Repeat readings and do not include a zero-resistance short circuit.

From $V=\varepsilon-Ir$, plot $V$ vertically against $I$ horizontally. Draw a best-fit line: the vertical intercept is $\varepsilon$, and the negative gradient is $-r$, so internal resistance is the positive magnitude of the gradient.

Use a large gradient triangle, include $\mathrm{V\,A^{-1}}=\Omega$, and inspect scatter or curvature before accepting a constant-$r$ model. The intercept is extrapolated, so a well-spread data range improves it.

For the same applied p.d., a metal's rising resistance reduces current. In an NTC thermistor, higher temperature increases conduction-electron number, lowering resistance and increasing current.

In a potential divider, translate this resistance change only after identifying the output position. As an NTC thermistor warms, p.d. across it falls if it is the output resistor, while p.d. across the fixed series resistor rises.

When illumination increases, more light energy arrives at an LDR and more electrons become available as conduction carriers. The increased carrier number allows a larger current for the same p.d., so the LDR's resistance decreases.

In lower illumination, fewer conduction electrons are available and resistance is higher. This carrier-number model connects the environmental input to the electrical response rather than treating the LDR as a switch with only two states.

In a fixed-voltage series circuit, lower LDR resistance lowers total resistance and raises current. In a potential divider, output across the LDR falls with illumination, while output across the fixed resistor rises.