Past exam of the physics B course of the University of Cambridge 2017 Paper 2 Section B B7 Solution 2026-10-18
Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length . Each has series resistance and inductance , and shunt electrical conductance and capacitance . The primes denote quantities per unit length, with units , , and , respectively.
Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, , differentiation gives and the same wave equation for . The propagation speed and characteristic impedance areA general lossless signal has and . Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.
For sinusoidal quantities with convention , the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately , using the lossless real in this estimate.
Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have and hence . The distributed circuit model follows from the field energy and charges in each slice.
For a coaxial cable with inner radius and inner surface of the outer conductor at , Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.
For wide parallel plates of width separated by , neglect fringing. The electric field crosses the gap, , while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, . ThereforeFor a forward wave, points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power . For complex peak amplitudes, average power is , or for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.
Terminations and reflection. At a load , require . If and are the incident and reflected voltage amplitudes there, the total current is . ThusA matched load, , absorbs the incident power with no reflection. An open circuit has , a current node and a doubled voltage at the end. A short circuit has , a voltage node and a doubled current there. A purely reactive load on a lossless line has and accepts no average power. For real , the reflected power fraction is and the accepted fraction is .
Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio . The input impedance of a lossless length isThis gives for a short and for an open. A quarter-wave section gives , enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is . Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.
An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux . Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance appears at the primary as , enabling impedance matching and electrical isolation.
For sinusoidal flux, gives . At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.
Past exam of the physics B course of the University of Cambridge 2018 Paper 2 Section B B9 Solution 2026-10-18
Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector describes the mode shape, while is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.
Let be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix . This guarantees real and a complete orthogonal basis in mass-weighted coordinates.
For distinct eigenvalues, the fact that and are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing decouples the motion into , with total energy . A positive-definite matrix gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.
Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.
For two equal masses , each attached to a fixed support by a spring of stiffness and coupled to the other by a spring of stiffness ,The symmetric mode leaves the coupling spring unstretched and has . The antisymmetric mode stretches it and has . Superposing the two gives beats when their frequencies are close.
Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length , tension and mass per unit length , the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.
A uniform chain of equal masses joined by springs , with spacing , has travelling normal modes withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion . These examples show how restoring forces and geometry determine the frequency spectrum.
Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to or can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.
Past exam of the physics B course of the University of Cambridge 2018 Paper 2 Section C C11 Solution 2026-10-18
Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.
In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers , , . Combining the vacuum dispersion relation and givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires , leaving two independent polarizations of an electromagnetic wave.
Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with , , only above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.
Three-dimensional mode count. A lattice cell in positive space has volume , where . In a continuum approximation, a shell of radius and thickness contains its positive octant, multiplied by two polarizations:Using givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.
Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies , . The zero-point energy does not enter its thermal occupation. Put . Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.
Thermal energy. Multiply the density of states by the mean excitation energy and integrate:The substitution isolates all the dependence on cavity dimensions and temperature:Expanding gives an integral . HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires large compared with the geometric level spacings in all three directions.
Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as , while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to , and increases the integrated area in proportion to . A plot per unit wavelength has a different peak position because the integration variable changes.
Thin-cavity radiation. Let . The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the scaling. For a cavity with fixed finite , cooling still further eventually freezes its lowest in-plane mode as well: the strict thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.
Past exam of the physics B course of the University of Cambridge 2019 Paper 2 Section A A5 Solution 2026-10-18
For an ideal electrical transformer, let . The reflected secondary load is . Matching a load to an source therefore requiresWith the generator replaced by its internal resistance, its resistance referred to the secondary isThis is the source output resistance relevant to matching and maximum power transfer.
To find the windings, use the inductance of a toroidal winding. The mean magnetic path is , the core area is and . ThusThe ideal secondary turn count is , giving the practical nominal designWith integer turns the irrational ratio cannot be exact: reflects the load as , an approximately resistance mismatch. These turn counts use the stated self-inductance; no operating frequency was specified from which to set a separate minimum inductance.
The condition must refer to the unloaded magnetizing reactance, not to the matched loaded input resistance. With ,The secondary self-inductance is , so its corresponding unloaded coil impedance satisfiesIf output impedance means this transformed coil reactance, is its minimum magnitude. It is distinct from the transformed source resistance. This is the magnetizing reactance and reflected resistance in a transformer distinction. A finite magnetizing inductance adds a parallel branch, , so exact purely resistive matching is an ideal large-reactance approximation; interpreting the loaded input itself as both and at least would be impossible.
Past exam of the physics B course of the University of Cambridge 2019 Paper 2 Section B B6 b Solution Created 2026-10-18 Updated 2026-10-20
A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, , soThe electric susceptibility describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.
Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude , weak-field orientational response gives when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.
At a static interface, and the tangential electric field is continuous. Bound surface charge is already included through . For a homogeneous filled capacitor, , and a reversible linear dielectric stores energy density . At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.
At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.
In a homogeneous Ohmic conductor, . Combining this with Maxwell's equations givesFor a good conductor, , so conduction dominates displacement current. With time dependence and depth , take . ThenThe field is consequently proportional to : it both attenuates and changes phase with depth. The skin depth is the distance for amplitude to fall by ; power density falls as , so its attenuation length is .
Physically, changing fields induce currents whose magnetic field opposes penetration. Equivalently, after neglecting displacement current, magnetic diffusion obeys . Faster changes have less time to diffuse into the metal. Increasing frequency, conductivity or magnetic permeability reduces , provided the relevant material parameters remain applicable.
For a thick conductor the tangential surface electrical impedance is with the stated time convention and consistent field orientation. Its real part represents dissipative resistance per square. Current crowding near the surface raises the AC resistance of a wire when is small compared with its radius; thin strands and laminated magnetic cores reduce suitable eddy-current losses. When , the full complex propagation constant must replace the good-conductor approximation. In a magnetic metal the magnetic permeability can itself be nonlinear and frequency dependent.
Past exam of the physics B course of the University of Cambridge 2019 Paper 2 Section B B7 Solution Created 2026-10-18 Updated 2026-10-20
Magnetization and field response. A magnetic material contains microscopic magnetic moments associated with electron spin and orbital motion. The magnetization is magnetic dipole moment per unit volume. In SI the macroscopic fields satisfyHere separates free-current excitation from the material response. A uniform isotropic linear material has , givingThis constant scalar magnetic permeability model is useful only within its linear range. Crystals can have a susceptibility tensor; strongly magnetic materials depend on field and magnetic history. For a finite specimen the internal also includes its demagnetizing field. In a uniformly magnetized ellipsoid, along a principal axis, so sample shape affects the apparent susceptibility.
Diamagnetic and paramagnetic response. Diamagnetism is a weak negative susceptibility caused by induced orbital response opposing the applied field. It occurs in all matter but can be masked by larger positive contributions. Closed-shell materials such as bismuth or many organic compounds show it clearly. Ordinary diamagnetism is often only weakly temperature dependent; superconducting flux exclusion is a separate collective phenomenon and is not described by an arbitrarily small negative susceptibility.
Paramagnetism arises when available magnetic moments partially align with a field. Thermal disorder prevents complete alignment at weak fields. For independent localized moments of magnitude , the classical weak-field result iswhere the second expression uses in a weakly magnetic sample. Thus localized moments obey a Curie law in this regime. Quantum angular momentum changes the coefficient to the appropriate expression and gives saturation at strong field. Conduction electrons instead exhibit Pauli paramagnetism: only states near the Fermi surface can change spin populations, so a degenerate gas has a nearly temperature-independent response. Its orbital diamagnetism must also be included for a full metallic susceptibility.
Collective magnetic order. Exchange interactions can favor parallel alignment, producing ferromagnetism below the Curie temperature. Iron, cobalt and nickel are familiar examples. A material may contain regions of spontaneous magnetization, or magnetic domains, while having little net external moment because the domains point in different directions. Applied fields move domain walls and rotate moments; at large fields the magnetization approaches saturation. Pinning and energy barriers make this process hysteretic. Remanence is the magnetization left after removing the driving field, and coercivity measures the reverse field needed to bring the chosen response back to zero.
The area of a closed – loop is the energy dissipated per unit volume per quasistatic cycle, , with positive dissipated magnitude. Consequently the magnetic permeability is neither one universal constant nor enough information to describe a ferromagnet: differential magnetic permeability , saturation and hysteresis all matter. Above the ordering temperature the spontaneous moment disappears, and a Curie–Weiss law is often an appropriate approximate paramagnetic description away from the critical region.
In antiferromagnetism, ordered opposing sublattices have equal moments and cancel macroscopically in the ideal compensated case. The order disappears above the Néel temperature. In ferrimagnetism, opposing sublattices have unequal moments and leave a net magnetization. Ferrites combine ferrimagnetic response with relatively high electrical resistivity, making them valuable when eddy-current losses would be substantial in a metallic core.
Electromagnets and engineering choice. A coil supplies free-current magnetomotive force . For a simple uniform magnetic path, Ampère's law gives and flux . A high-magnetic permeability soft magnetic core concentrates flux and reduces the current required for a given field. With an air gap ,Even a short gap can dominate the magnetic reluctance. The approximate attractive force across one uniform gap is , obtained from the field energy, with fringing neglected.
Soft magnetic materials have small coercivity and narrow hysteresis loops; they suit relays, electrical transformer cores and controllable electromagnets. Hard magnetic materials retain their magnetization and suit permanent magnets. Laminating metallic cores breaks large eddy-current loops; ferrites are useful at higher frequencies because of their resistivity. Increasing current cannot increase indefinitely through the linear formula: core saturation, coil heating and leakage limit performance. A good design selects material for the field amplitude, switching frequency and acceptable loss, rather than simply choosing the largest quoted magnetic permeability.
Past exam of the physics B course of the University of Cambridge 2019 Paper 2 Section B B9 Solution 2026-10-18
Inertia is a tensor. For a rigid body, distances between material points are fixed. Write each position relative to the centre of mass as , so its rotational velocity is . Integrating gives angular momentumThe rotational kinetic energy isThe inertia tensor is real and symmetric, so orthogonal principal axes diagonalize it. Angular momentum is parallel to angular velocity only along a principal axis, or within a subspace of equal principal moments. For translation plus rotation about the centre of mass, . Moving the reference to a parallel axis adds to the tensor, the tensor form of the parallel-axis theorem.
Euler's equations. The external torque about the centre of mass obeys . In body axes, the transport rule givesThe principal moments are constant in that frame, henceThese are Euler's equations. With zero torque, spatial angular momentum and kinetic energy are conserved, although the body components generally vary. For distinct moments, small perturbations about rotation around the largest or smallest principal moment oscillate, whereas those about the intermediate moment grow. This is the intermediate-axis instability.
A lamina. Choose the lamina's plane as . Then , so the normal is a principal axis, andThis perpendicular-axis theorem applies to a planar mass distribution; it is not a general identity for a three-dimensional body. Rotate the in-plane axes to diagonalize the remaining symmetric two-by-two block. For a uniform rectangular lamina with sides and mass , its central principal moments are , and . If , then : torque-free rotation about the long in-plane axis or the normal is stable, while that about the short in-plane axis is unstable. Coincident moments in a square require the degenerate, rather than strict intermediate-axis, case. Under a constraint to rotate only about the normal, the motion reduces to the scalar equation .
A symmetric top without torque. Let and . Euler's equations reduce toThus rotates at signed body-frame frequency , while the axial component stays fixed. The symmetry axis has constant angle to fixed spatial angular momentum. Indeed , soIt precesses in space about with angular speed . This spatial precession and the body-frame frequency describe different motions and should not be identified.
A heavy top. If a symmetric top is pivoted a distance from its centre of mass, gravity gives torque . Here the relevant moments are about the fixed pivot. The axial angular momentum and the vertical angular-momentum component are conserved, because gravity supplies neither an axial torque nor a torque about the vertical. Using tilt , precession angle and spin angle , its Lagrangian isEliminating the two conserved cyclic momenta leaves one-dimensional motion in an effective tilt potential; bounded oscillation in tilt is nutation. Steady precession at and fixed tilt satisfiesFor a rapidly spinning top on the slow branch, the first term is small andGravity then changes the direction of angular momentum rather than simply tipping the top straight down. The approximation fails near conditions where nutation or the fast-precession branch matters; the full equations retain those possibilities.
Past exam of the physics B course of the University of Cambridge 2022 Paper 1 Section A A5 Solution 2026-10-18
For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency and carrier frequency , . ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check follows directly from the dispersion relation. A different uniform filling medium would require replacing by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.
For the edge-hinged cube with a central spherical cavity, use the parallel-axis theorem: for a parallel axis at perpendicular distance from the centre-of-mass axis, . Let . The centre lies from the edge hinge, soAt equilibrium the centre is vertically below the hinge. Its gravitational torque at angular displacement is . Therefore the small-angle equation is , withThe removed sphere has mass and the same centre. Its hinge inertia isSuperposition leaves the centre at the same location but changes both inertia and weight:The frequency decreases: the removed central material contributes less hinge inertia per unit mass than the original cube average.
For the series-fed parallel LC voltage response, use phasors with time dependence . The parallel branch has impedance , where . ThereforeAt the ideal parallel resonance its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.
The capacitor voltage equals the parallel-branch voltage. Define . ThenAt low frequency ; at high frequency ; at resonance . Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from through zero to . Half-power points satisfy and
For the supplied component values and , and . The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.
The displacement is along , transverse to propagation in the membrane's plane. Write ; then . Here is the displacement amplitude, the initial phase, the angular frequency and the wavevector normal to constant-phase lines. Its magnitude is , and its direction specifies phase propagation for the chosen sign convention.
For equal-phase two-wave membrane modes, two equal complex amplitudes with opposite sum toThe edge condition gives , hence , , for this even-in- family. Opposite relative phase gives the sine family; together they yield all Dirichlet transverse modes , . For the equal-phase family,The lowest mode has . ThereforeIt approaches for and diverges as ; there is no real travelling-mode beyond the cutoff. This is an axial phase velocity, not the group velocity: . Here denotes the membrane wave speed, not the speed of light.
Membrane cutoff and evanescent response
. Original lowest-mode phase-velocity curve, with group velocity distinguished, and a schematic below-cutoff displacement envelope of exponential form. All lengths and amplitudes are dimensionless choices; no source apparatus drawing is copied. Details.For the evanescent membrane response below cutoff, below , put for decay along positive , with . A localized central drive creates an evanescent, sinusoidally oscillating deformation decaying away in both axial directions. Far enough from the drive the leading profile is proportional to . Higher evanescent modes contribute near the point source. The pattern has no steadily advancing axial phase front or far-field propagating wave; decreasing frequency increases its spatial localization.
For a mass change in a viscously damped spring oscillator, use linear viscous damping , holding the spring constant and damping coefficient fixed. Measuring from removes the constant force. The envelope is , whereDoubling halves both and . ThereforeNeglecting damping in the frequency gives , indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.
In sinusoidal steady state, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage lead. With the convention, , so .
The PDF labels across , not across the complete driven network. If is the current through , then and , whereThe resistor-referenced parallel-resonator response isWriting ,The maximum occurs at , where and the phase is zero. Below resonance the phase is positive, tending to ; above it the phase is negative, tending to . The amplitude tends to zero at both extremes: the inductor shorts the parallel network at low frequency, and the capacitor does so at high frequency. At resonance their reactive currents cancel, leaving only the resistor current.
Fixed endpoints require , with . Writing ,For equal amplitudes , the two requested profiles areThe first is asymmetric, with an interior zero at ; the second has its interior node at . For two-mode string recurrence, orthogonality means the whole initial profile recurs only when both cosine coefficients equal one. Since ,
Two-mode string profiles and the small change produced by weak positive dispersion at the original quarter-period
. Details. The weak-stiffness string frequency shift follows by expanding the supplied dispersion relation:Higher normal modes acquire larger fractional frequency shifts. Let and retain for the original ideal frequency. At the originally defined time ,Thus the sketch is unchanged, whereas the original quarter-period sketch gains a small negative fundamental component. If is redefined using the new fundamental frequency, its fundamental component is exactly zero and the second-mode change there is only second order. Both conventions agree that generic later shapes change because of relative phase drift.
In series the same current passes through both resistors, while their voltage drops add: . In parallel the same voltage drives currents and , which add. Thus and .
Use phasors proportional to . Then and . The printed diagram places C in parallel with the series RL branch, so the input admittance isIts nonzero zero-phase frequency exists for :At zero frequency, and . At , , givingFor , , so current lags voltage. At very high frequency, , so current leads by . Zero phase need not coincide exactly with the minimum current magnitude in this lossy network.
Substitution into the wave equation gives and . Therefore . The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.
The element's mass is , and its speed is . Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are and , soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy and use the dispersion relation.
For , the two lowest modes are and :The fundamental has one antinode at and no interior nodal line. The next mode has a node along and opposite-sign antinodes at and . Every boundary is a node.
For sinusoidal steady state with convention , complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since ,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are at and at , a ratio . The current resonance is , so the circuit favors the first transmitter.
The resistor voltage is in phase with the current; the inductor voltage leads it by and the capacitor voltage lags by . Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.
Reducing the capacitance raises to , between the two transmitters. Direct evaluation gives and , respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of also depends on frequency.
The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith , this is approximatelyThus a rapid oscillation has a beat envelope repeating in magnitude at . The signed envelope cosine has frequency . Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.
Original calculated inductor-voltage gains and illustrative two-tone output, including the circuit's frequency-dependent phase shifts
. Details. For the supplied linear meter model let and . If the applied step is , the settled deflection is . An underdamped response overshoots and rings about that value; a critically damped response reaches it monotonically without ringing; an overdamped response is also monotone from these initial conditions but has a slower final approach. The original plot uses the common final value and natural-frequency time scale.
The critical step response of a galvanometer has and repeated homogeneous root . Apply to the general constant-force solution to obtainIts derivative is , with the sign of the step. It is the fastest nonoscillatory response within this second-order family at fixed , avoiding a needle that repeatedly crosses the reading or an unnecessarily slow overdamped settling.
For sinusoidal drive choose phasors. Writing , the response amplitude isFor positive , choose the continuous lag branch from zero to as frequency increases; an ordinary single-argument arctangent without the quadrant would give the wrong high-frequency phase.
For the critical galvanometer transfer function, the denominator factors as . ThusFor , the meter follows the quasi-static deflection with a small lag . For , its amplitude is and its lag approaches .
The stated components give s⁻¹ and, if critical, rad/s, or Hz. At 50 Hz, rad/s andThe needle barely responds compared with its DC sensitivity and its mean deflection for a pure sinusoid is zero in this linear model. It is unsuitable for directly measuring 50-Hz amplitude or RMS current without an additional rectifier or another readout method. The calculation treats as the stipulated torque/back-emf coefficient of the given equation; it does not infer an unprovided winding geometry.








