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Use $e^{i\omega t}$ phasors. The relation $V=L\dot{I}$ gives $Z_L=i\omega L$. The three branches form a parallel RLC circuit and share the source voltage. The source-current phasor and its real time dependence areConsequently its amplitude is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L){)}^2}$ and its phase relative to the voltage is ${\tan }^{-1}[R(\omega C-1/(\omega L))]$. At low frequency it is inductive and lags toward $-\pi /2$; at high frequency it is capacitive and leads toward $+\pi /2$. The ideal amplitude diverges at either extreme and has a finite minimum between them; real component losses and source electrical impedance limit those asymptotes.

Zero phase requires cancellation of electrical susceptances:The inductor and capacitor currents are nonzero, equal in magnitude and opposite in phase, while the resistor supplies the in-phase source current. energy oscillates between the reactive elements; dissipation occurs in the shunt resistor. A source-current maximum would describe a different, series-resonant circuit.

The specified ratio isIt exceeds $1/\sqrt{2}$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and ${\omega }_0=1/\sqrt{LC}$. The two positive endpoints are ${\omega }_{\pm }=[\sqrt{b^2+4{\omega }_0^2}\pm b]/2$, soThe endpoints have geometric mean ${\omega }_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a series RLC circuit. Equivalently, $Q={\omega }_0{U}_{\max }/P_{\mathrm{loss}}$, with $U_{\max }=C{V}_0^2/2$ and $P_{\mathrm{loss}}=V_0^2/(2R)$. The parallel RLC bandwidth here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.

Use an RC low-pass filter: put the resistor in series with the input and the capacitor between the output node and the return conductor. Measure the output voltage across the capacitor, with negligible output loading. The voltage divider then has transfer functionAt the high frequency, the attenuation requirement isAt the low frequency, approximately unchanged amplitude means $2\pi f_{l}RC\ll 1$, soThe upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-frequency amplitude would sharpen the upper bound to approximately $68.7\mu \mathrm{F}$.

A practical illustrative choice is $C=22\mu \mathrm{F}$. Its voltage gain is approximately $0.99896$ at $1\mathrm{Hz}$ and $0.001096$ at $20\mathrm{kHz}$: the latter amplitude is reduced by a factor of about $912$. Its cutoff frequency is $1/(2\pi RC)\simeq 21.9\mathrm{Hz}$.

After switching, Kirchhoff's voltage law gives the free series RLC circuit equationIts characteristic roots are $-\gamma \pm i{\omega }_d$, whereA real, nonzero oscillation frequency requires $R^2<4L/C$. For a passive resistor,The endpoint $R=0$ is the undamped limit; strictly decaying oscillations have $0<R<894\Omega$. The critical endpoint itself does not have a nonzero ${\omega }_d$.

Let $Q=C{V}_0$ denote the initial electric charge, so $Q=30\mu \mathrm{C}$ and $\dot{q}(0)=0$. Distinguish $Q$ from the amplitude coefficient $q_0$ in an exponential-times-cosine representation: imposing zero initial electric current generally requires a nonzero phase.

Measure the mass's displacement $x$ from its static equilibrium height. This removes the constant spring extension supporting its weight. If the spring's top moves with the box, its incremental extension is $y_2-x$, so its incremental upward force is $k(y_2-x)$. The specified damping force is against the mass's velocity in the static frame. Newton's second law therefore givesIn absolute height coordinates an additional constant on the right expresses the spring's rest length, its fixed attachment height and gravity; translating to static equilibrium removes it. Replacing the given damping by $-b(\dot{x}-{\dot{y}}_2)$ would describe a different base-excited oscillator.

After the decaying transient, write the steady-state response as $x=A(\omega )\cos (\omega t-\Phi )$. Using a complex sinusoid givessoFor positive frequency, the phase lag lies between zero and $\pi$, passing through $\pi /2$ at ${\omega }_0=\sqrt{k/m}$. For small $b$, the frequency response begins at $a_s$, develops a narrow resonance near ${\omega }_0$, and falls asThe precise amplitude-peak frequency is $\sqrt{k/m-b^2/(2m^2)}$ when that expression is positive.

The pen records the relative displacement $z=x-y_2$. Its complex transfer function isAt high frequency, $x$ is small while $z\simeq -y_2$. The inertial mass then stays nearly stationary and the recorded relative motion reproduces the ground displacement, with reversed sign. At low frequency the mass follows the box and their relative motion is small; near resonance the response is strongly frequency-dependent. Here “high” means compared with the suspension's natural frequency, rather than an absolute seismic-frequency threshold.

Comparing with the standard wave equation gives the shallow-water wave speedTwice differentiating the proposed standing wave multiplies its spatial factors by $-k_x^2$ and $-k_y^2$, and its temporal factor by $-{\omega }^2$. It solves the wave equation precisely whenTo obtain the local gravitational potential energy, integrate through a water column of height $d+h$ and unit horizontal area:The time averages are $⟨h⟩=0$ and $⟨h^2⟩=A^2{\cos }^2(k_{x}x){\cos }^2(k_{y}y)/2$. ThusThe constant is retained because the specified zero of gravitational potential is the tank bottom. This is local time-averaged potential energy, not the sum of potential energy and kinetic energy.

The no-normal-flow wall conditions are Neumann boundary conditions. Differentiating the cosine factors at the opposite walls givesNegative integers give the same cosine modes. The formal $m=n=0$ case has $\omega =0$: it is a uniform change of the mean level, not a finite-frequency oscillation. In a closed tank with fixed water volume even that mean-level change is excluded from the oscillatory disturbance.

Since $a<b$, the lowest finite frequency is the mode $(m,n)=(0,1)$:It is uniform across the shorter direction, has opposite-sign displacement antinodes at the two ends of the longer direction, and a displacement node midway along it.

For the open-ended Bay of Fundy model, approximate the ocean mouth as a displacement node and the closed head as an antinode. The fundamental is a quarter-wave basin resonance, so its wavelength is $4L$ andTaking a roughly $12$-hour semidiurnal tide gives $d\simeq 79\mathrm{m}$. Using the principal lunar period of approximately $12\mathrm{h}25\min$ gives $d\simeq 73\mathrm{m}$. The idealized estimate is therefore about $7\text{–}8\times 10^1\mathrm{m}$. The lunar period is half of a roughly $24$-hour $50$-minute lunar day.^[ref]

This estimate treats the mouth as an ideal open boundary and replaces the real basin by a uniform depth. The supplied large tidal height motivates resonance; it is not enough by itself to determine damping or an amplification factor in this model.

Use distinct symbols for the dimensions of mass, length, time, temperature and amount of substance. The molar gas constant has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}{\mathsf{\Theta }}^{-1}{\mathsf{N}}^{-1}$, while the given $M$ is a molar mass with dimensions $\mathsf{M}{\mathsf{N}}^{-1}$. Matching dimensions in the proposed power law giveswith the amount-of-substance equation repeating $\alpha +\gamma =0$. ThereforeDimensional analysis fixes these exponents but not a dimensionless multiplier. For example, an open-tube mode number and the gas's heat-capacity ratio enter the actual resonance frequency without changing the dimensions.

Use the PDF's notation $f_0$ for the source-frame pulse frequency, and set $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. Successive emissions occur at the same train-frame position and are separated by proper time $1/f_0$. The Lorentz transformation gives a station-frame emission interval $\Delta t_e=\gamma /f_0$ and displacement $v\Delta t_e$. For a receding beacon, the second pulse travels an extra distance $v\Delta t_e$, soThis derivation includes both time dilation and the change in light-travel time. Expanding for $\beta \ll 1$ gives $f/f_0=1-\beta +O({\beta }^2)$, recovering the stated nonrelativistic approximation.

Let ${\lambda }_0$ be the common wavelength of the identical lights at rest. The approaching and receding relativistic Doppler effect factors giveTheir ratio and product therefore determine the speed and rest wavelength separately:No assumption about the rest color was needed.

The last observation requires a transverse-geometry convention. The general observer-frame ray formula iswhere $\theta$ is the angle between source velocity and photon propagation in the platform frame at emission. Under the usual transverse-ray interpretation, $\theta =\pi /2$, so for $\beta =0.6$ and $\gamma =1.25$,This is the standard transverse redshift from time dilation.

There is a genuine distinction between emission and reception at the instant of passing. A side-mounted source emitting perpendicularly in its own frame instead has $\cos \theta =\beta$ by aberration of light, giving $\lambda ={\lambda }_3^{\prime }/\gamma =320\mathrm{nm}$. The same blue result occurs if “at the exact moment” means reception when the source's current position is at closest approach, with a small finite transverse offset. For offset $d$, reception and emission satisfy $t_r=t_e+\sqrt{v^2{t}_e^2+d^2}/c$. Setting $t_r=0$ gives $t_e=-\gamma d/c$ and $\cos \theta =\beta$; setting $t_e=0$ instead gives the transverse ray received at $t_r=d/c$. These limits remain different as $d\to 0$.

Thus 500 nm requires the transverse ray in the platform frame; 320 nm applies to reception at closest approach or transverse emission in the train frame. The wording does not distinguish these event/frame choices, so merely neglecting the source-observer separation is insufficient to select between them. The reusable derivation is transverse Doppler shift and observation time.

Expand the cloud's potential energy about a stable equilibrium: $U(z)=U(0)+U^{\prime }(0)z+U^{\prime \prime }(0)z^2/2+\cdots$. Equilibrium gives $U^{\prime }(0)=0$, stability gives $U^{\prime \prime }(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq k{z}^2/2$ and $F=-dU/dz\simeq -kz$.

Including the electric and damping forces, Newton's second law is $m\ddot{z}=qE(t)-b\dot{z}-kz$. Dividing by $m$ gives the Lorentz oscillator modelThe specified convention uses $2\gamma$ as the damping coefficient per unit mass.

For the steady response, represent the field by $E_0{e}^{i\omega t}$ and the displacement by $\tilde{z}{e}^{i\omega t}$, taking real parts for physical quantities. Substitution givesDefine the nonnegative amplitude $z_0=|\tilde{z}|$ and $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. ThenFor positive $q$, $\varphi =-\delta$. For electrons $q<0$, the phase of a negatively charged driven oscillator is $\varphi =\pi -\delta$: the low-frequency displacement is opposite to the field, rather than in phase with it. A signed amplitude could absorb this extra $\pi$, but the phase convention must then be stated. The induced dipole $p=qz$ has phase $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.

For the electron cloud, $q/m=-e/m_e$ andThe visible range corresponds to $\omega =2\pi c/\lambda$ between $2.35\times 10^{15}$ and $4.71\times 10^{15}\mathrm{rad}{\mathrm{s}}^{-1}$, below this resonance. The resonant wavelength is about $180\mathrm{nm}$, in the ultraviolet.

The acceleration amplitude is ${\omega }^2{z}_0$. At fixed incident field amplitude, the scattered intensity is therefore proportional toWell below resonance and outside strong damping corrections, $z_0$ is nearly frequency-independent and this becomes the Rayleigh scattering wavelength dependence $I_{\mathrm{sc}}\propto {\omega }^4\propto {\lambda }^{-4}$. The shortest visible wavelengths are consequently scattered much more strongly into directions away from the direct solar beam than the longest ones, producing a blue sky. The simple fourth-power approximation gives a factor 16 between 400 and 800 nm at equal incident intensity. Retaining the undamped denominator with the given resonance raises that factor to about 22.7, since the upper end is not extremely far below resonance. The exact ratio needs the unspecified damping and incident solar spectrum; the stronger scattering of the blue end is the robust conclusion here.

The model describes frequency-dependent scattering, not the detailed perceived color after solar-spectrum weighting and human visual response. It also explains why treating the electrons as free at visible frequencies would miss the bound-charge fourth-power limit.

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.

The Huygens-Fresnel principle constructs a new wavefront as the envelope of secondary wavelets emitted by points on an earlier wavefront. Let the phase velocities in the two media be $v_1=c/n_1$ and $v_2=c/n_2$, with refractive indices evaluated at the light's fixed frequency.

For refraction from secondary wavelets, let point $A$ of the incident wavefront reach the interface first. A second point $B$ reaches the interface at $C$ after time $\Delta t$, so $BC=v_1\Delta t$. During the same interval, the secondary wavelet centered on $A$ advances radius $AD=v_2\Delta t$ into the second medium. The refracted wavefront $CD$ is tangent to this wavelet, hence $AD\perp CD$; also $BC\perp AB$ because a ray is normal to its wavefront in these isotropic media.

Let ${\theta }_1,{\theta }_2$ be the incident and refracted ray angles from the interface normal. They also equal the corresponding acute angles between the wavefronts and the interface. The two right triangles have common hypotenuse $AC$, givingTheir ratio eliminates both $AC$ and $\Delta t$:This is Snell's law. In particular, entering a medium of larger refractive index reduces the ray's angle to the normal. The construction assumes a stationary interface and homogeneous isotropic media with a propagating refracted wave.

The force is conservative, with $\mathbf{F}=-\nabla U$ and potential energy zero at the origin:Released from rest at $(c,c)$, the trajectory has $x=y=c\cos \omega t$, givingThe right panel of the figure shows this nonnegative oscillation: it starts at a maximum $k{c}^2$, reaches zero at $t=\pi /(2\omega )$, and repeats after $\pi /\omega$. Its frequency is twice the coordinate frequency. Conservation of mechanical energy gives $K(t)=k{c}^2{\sin }^2\omega t$, so the missing potential energy is kinetic energy, not an energy loss.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

The original diagram has a series circuit containing $R$ and $L$, in parallel with $C$; it is not three separate parallel components. The electrical admittance is thereforeFor $R>0$, voltage and current are in phase when the imaginary part of $Y$ vanishes. Excluding the zero-frequency solution, division by $\omega$ givesA nonzero lossy parallel coil resonance exists only if $L>R^{2}C$. With $R=10\Omega$, $L=0.150\mathrm{H}$ and $C=1.00\times 10^{-5}\mathrm{F}$,For zero electrical resistance, the ideal parallel inductor-capacitor cancellation occurs atThe fractional reduction relative to the zero-resistance coil isAt exactly $R=0$ and $\omega ={\omega }_0$, $Y=0$, so finite applied voltage draws zero net current in the ideal model. The quoted frequency is the limiting parallel-resonance value; the phase of an exactly zero current is undefined.

Let $C_T=C+C_1$ be the total parallel capacitance, since capacitors in a parallel circuit have additive capacitances. Put ${\omega }_{\ast }=2\pi (50\mathrm{Hz})$ into the lossy parallel coil resonance condition:The additional capacitance is thereforeThis is the added capacitance, rather than the new total, and the conversion from frequency to angular frequency is essential.

Using the convention $e^{i\omega t}$, the series RLC circuit has complex impedanceAt the stated frequency, $\omega L=180.956\Omega$ and $1/(\omega C)=88.4194\Omega$. ThusThe vector is $(525,92.5)$ in the resistance–reactance plane. Positive reactance means that the net circuit is inductive and current lags voltage. With the alternative $e^{-i\omega t}$ convention the imaginary signs reverse, while the magnitude and physical phase relationship agree.

In the small-amplitude approximation, $\dot{x}$ is the speed along the track, and $y=B{x}^2$ supplies the gravitational potential. The conserved energy isDifferentiating and extending through turning points by continuity givesThis is simple harmonic motion with ${\omega }_0^2=4gB/\{(r_2/r_1{)}^2+2\}$. HenceThe frequency is lower than for a sliding point mass because energy is also stored in rotation.

Let $X_1,X_2$ be longitudinal displacements. Hooke's law givesThe symmetric and antisymmetric eigenvectors are respectively $(1,1)$ and $(1,-1)$, soOrdinary frequencies are these values divided by $2\pi$.

A normal mode is a small motion in which all coordinates oscillate at one common angular frequency with fixed relative amplitudes. For conservative linearized motion, these are eigenvectors of the mass-weighted stiffness matrix.

All three springs are initially stretched from $l$ to $2l$, giving equal tension $T_0=kl$ and zero resultant force on each mass. This prestress is essential for the transverse modes found below.

For the final mixed-axis trajectory of a prestressed spring pair, the initial displacements excite only the symmetric mode in each direction. Therefore, to first order in $ϵ/l$,The second mass has the same displacements about $4l$. Over the first transverse period $T_y=2\pi \sqrt{2m/k}$, write $s=t/T_y$:The trajectory is not closed after $T_y$, because the frequency ratio is $\sqrt{2}$. It ends at approximately $(-0.858,1)$ in these scaled coordinates rather than $(1,1)$.

The first-order grating equation is $d\sin \theta =\lambda$, so $d=2.879\mu \mathrm{m}$, about $347$ lines per millimetre. For $N$ equally illuminated lines,Principal maxima stay at the same angles, their peaks scale as $N^2$ for fixed illumination per line, and their angular widths shrink as $1/N$. The resolving power of a diffraction grating is $\lambda /\Delta \lambda =mN$. Thus first order needs $N=1000$ and illuminated width $Nd\simeq 2.88\mathrm{mm}$.

The grating dispersion follows by differentiation: $d\theta /d\lambda =1/(d\cos \theta )$. With a detector centred on the chief ray, one pixel corresponds approximately toThis exceeds the nominal $0.5\mathrm{nm}$ grating width. The practical resolution scale is therefore a few nanometres; two-pixel sampling gives about $4.5\mathrm{nm}$. Exact resolvability also depends on slit width and image sampling. Taking the detector coordinate as $f\tan \theta$ about the original axis instead changes the single-pixel estimate only to $2.20\mathrm{nm}$.

For ideal photon counting, Poisson distribution statistics give SNR $=\sqrt{N_{\gamma }}$. Hence $N_{\gamma }=10^4$ and $t_{\min }=10^4/(2\times 10^5)=0.050\mathrm{s}$.

Use a lock-in amplifier for phase-sensitive detection: chop or otherwise modulate the source before the entrance slit, retaining a synchronous reference. At each grating setting amplify the photodiode output, multiply it by a reference and low-pass filter. For a signal $s\cos ({\omega }_{r}t+\varphi )$ and reference $2\cos {\omega }_{r}t$, the low-pass output is $s\cos \varphi$; quadrature demodulation also recovers $s\sin \varphi$. Choose the modulation frequency above slow drift while within the source/detector bandwidth and away from strong electrical interference. Average for many cycles before changing the grating setting.

This rejects steady backgrounds and reduces low-frequency noise through narrow-band detection. It does not remove random dark-current/stray-light shot noise within that bandwidth, nor a background modulated coherently with the source. The chopping geometry must therefore distinguish wanted light from stray paths.^[ref]