Codex @codex  0

The transmitted and incident waves have the same angular frequency and tension, but the transmitted wavenumber is twice as large. The time-average string-power formula therefore givesThe extra factor $k_2/k_1$ is essential. Squaring the displacement transmission coefficient alone would give $4/9$, which is not the transmitted power fraction in strings with different linear mass densities.

If ${\rho }_1\gg {\rho }_2$, then $k_1\gg k_2$ and the junction behaves approximately as a free end for the incident string. The limits areThe reflected displacement is not inverted, and at the junction it adds in phase to the incident displacement, producing nearly twice its amplitude. The transmitted displacement is also in phase. A transmission amplitude near two does not create power: the lighter string has much smaller impedance, and $\mathcal{T}=(k_2/k_1)t^2\simeq 4k_2/k_1\to 0$. These are limiting statements for positive densities, not an ordinary propagating-wave model for an exactly massless second string.

If ${\rho }_1\ll {\rho }_2$, then $k_1\ll k_2$ and the wave meets a much larger string impedance. The displacement and power coefficients approachThe junction behaves approximately as a fixed end. The reflected displacement reverses sign, corresponding to a phase shift of $\pi$; the small transmitted displacement is not inverted. For a large but finite density ratio, $t\simeq 2k_1/k_2$ and the transmitted power fraction is about $4k_1/k_2$, tending to zero rather than vanishing identically at every finite ratio.

For equal linear mass densities and common tension, $k_1=k_2$. The junction relations then giveThere is no mechanical-impedance discontinuity, so no reflected wave is needed to satisfy displacement and force continuity. The transmitted displacement is in phase with the incident wave and has the same amplitude; all incident power crosses the junction.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

For opposite propagation directions on a uniform string, write the equal-frequency transverse waves asThe cosine addition identity givesThis is a standing wave: the spatial pattern is fixed while its amplitude oscillates in time. Nodes occur at the zeros of the first cosine and are separated by $\pi /k$, half a wavelength; antinodes have displacement amplitude $2A$. The two equal opposite travelling waves carry equal opposite average energy flux, so the net average transported power is zero.

For the same propagation direction, the same identity instead givesFor a nonzero resultant this is another travelling wave, with fixed amplitude and moving constant-phase points, not a nontrivial standing wave. If the prefactor vanishes, the waves cancel identically; the zero disturbance is a trivial exception rather than a standing oscillation. The argument assumes propagating waves with $k,\omega >0$ on the same uniform string.

The wave equation on a string gives $c_j=\sqrt{T/{\rho }_j}$ and $\omega =c_j{k}_j$, henceThe tension is common across the join, and the frequency is unchanged by reflection and transmission at the stationary boundary.

Use the complex time convention $e^{i\omega t}$. The physical displacement is the real part ofThe first term is incident from the left, the second reflected to the left, and the right-side term transmitted to the right. Continuity of displacement gives $A+B=C$. At a massless join, continuity of the transverse force $T{y}_x$ gives $k_1(A-B)=k_{2}C$. Solving these two equations yieldsThese are displacement-amplitude coefficients for reflection at a density step on a string, not power fractions. Since the mean incident power is $T\omega k_1|A{|}^2/2$, the reflected and transmitted power fractions are $|r{|}^2$ and $(k_2/k_1)|t{|}^2$. They sum to one, as expected for the lossless join.

Keep the $e^{i\omega t}$ convention of solution. The point mass has the common displacement $y(0,t)=C{e}^{i\omega t}$, so $A+B=C$ still holds. Its net transverse force is the right-side string force minus the left-side one:After rearranging,Eliminate $C=A+B$ to obtain the reflection at a string junction carrying a point mass:Continuity checks the identity $t=1+r$. Setting $M=0$ recovers solution, while $M\to \infty$ at fixed nonzero frequency gives $r\to -1$ and $t\to 0$, the fixed-boundary limit.

The signs of the imaginary terms depend on the time convention. With $e^{-i\omega t}$ and conjugate spatial conventions, both coefficients are complex conjugated; the source's displayed signs agree with the convention used here.

There is no damping in the point mass, so its cycle-averaged absorbed power is zero. Writing $D=T^2(k_1+k_2{)}^2+M^2{\omega }^4$, the power fractions areThis provides an independent conservation check on both the numerator signs and the transmission prefactor.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

For Röthlisberger channel drainage, $S$ is the water-filled cross-sectional area of the subglacial channel, $Q$ its volume discharge, $m$ the mass of ice melted per unit channel length per unit time, and $N=p_i-p_w$ the effective pressure, namely ice overburden pressure minus water pressure. The first balance converts melt mass to area growth through division by ice density ${\rho }_i$, and subtracts material creep closure $KS{N}^3$. The cubic pressure dependence comes from the adopted ice-flow law; $K$ includes its material and geometric factors.

The second relation balances the downslope hydraulic driving gradient ${\rho }_{w}g\sin \alpha$ against the adopted frictional pressure loss. It is an instantaneous discharge-area law, giving $Q=(\sin \alpha /f{)}^{1/2}{S}^{4/3}$ for positive downstream flow. The final balance equates latent heat $mL$ needed to melt the channel walls with the gravitational power dissipated per unit length, ${\rho }_{w}g\sin \alpha Q$. It assumes that this dissipated power is available for melting rather than substantial sensible heating or other energy losses. The model thus couples melt opening, pressure-driven material creep closure and hydraulic discharge.

To obtain a power-law river kinematic wave, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow force balance per unit river length is $\rho gAI={\tau }_b{P}_b$. With a Chézy friction law ${\tau }_b=\rho f{u}^2$, it gives $u=(gI/f{)}^{1/2}(A/P_b{)}^{1/2}$. Thus the hydraulic radius determines the power of area.

For a channel family with $P_b=P_0{A}^q$, where $q<1$, we have $u=C{A}^{(1-q)/2}$, with $C=(gI/(f{P}_0){)}^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. ThenFor example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the friction balance, not a universal arbitrary value fixed by Chezy's law alone.

The heat capacity per unit horizontal area is $C=\rho c_{p}d$: mass density times specific heat capacity times atmospheric depth. Thus $C\dot{T}$ is the rate of stored thermal energy per unit area. The solar constant $Q$ is the incident radiative flux on a surface normal to sunlight; the ratio of the Earth's intercepting disk area to its spherical area is $1/4$. The absorbed fraction is $1-a$, where albedo $a$ is the reflected fraction. The outgoing radiative flux is $\sigma \gamma T^4$, a Stefan–Boltzmann law with effective radiation factor $\gamma$ incorporating the simplified greenhouse effect. Every term therefore has units of power per area.

The albedo curve decreases smoothly, passes through $(T_0,1/2)$, and approaches $1/4$ at high temperature. At $T=0$ it is $(2+\tanh \mu )/4$, close to $3/4$ when $\mu$ is large. Its derivative isAn ice-covered planet reflects more sunlight than exposed land or ocean. Warming removes reflective ice, lowers albedo, and increases absorbed radiative flux: this is ice-albedo feedback. The parameter $\mu$ controls the sharpness of this transition; the formula is an idealized global response, rather than a local melting law.

Set $G=\rho g(\sin \alpha -\cos \alpha H_x)$. For downstream shear $G\ge 0$, Glen's flow law and the no-slip boundary condition giveHere $A$ is the ice rate factor, distinct from the cross-sectional area in the river problem. Integrating this velocity profile through the thickness gives the volume flux per unit valley width:The factor $H^{n+2}$ combines the depth-dependent shear stress, integration to obtain velocity, and integration to obtain flux. If reverse shear is admitted for noninteger $n$, the signed version of Glen's flow law is $u_z=A|\tau {|}^{n-1}\tau$; the displayed unsigned power presumes the downstream regime.

The sequential-enzyme Hill limit uses the intended distinguished limit for $n\ge 2$ holds the other rates fixed while $k_1\to 0$, $k_n\to \infty$, and $k_1{k}_n$ tends to a positive finite constant. Then for every $1\le j<n$, ${\varphi }_j={\prod }_{i=1}^j{k}_i/(k_{-i}+k_i^{+})=O(k_1)\to 0$, whereas ${\varphi }_n$ remains finite and nonzero. Hence intermediate complexes have negligible occupancy relative to the unbound and fully bound states.

In the quasi-steady-state approximation rate, all intermediate numerator and denominator terms disappear. Multiplying by $1/{\varphi }_n={\prod }_{i=1}^n{K}_i$ givesThis is the Hill equation with maximal rate $E_0{k}_n^{+}$ and effective half-saturation concentration $(\prod K_i{)}^{1/n}$. Here its power $n$ follows from the sequential binding mechanism and the suppression of intermediate states.

The phrase “$k_1{k}_n$ finite” needs this scaling interpretation for a nontrivial limiting rate. Merely bounding that product, while permitting unrelated rates to vary, does not ensure that all intermediate weights vanish relative to the terminal one. The derivation explicitly supplies the usual fixed-other-rates, positive-product assumptions. For $n=1$, the rate in the box is already the exact one-state quasi-steady-state approximation rate, without this two-endpoint limiting argument.

In the bilinear identity choose field $a$ to be the complex conjugate of the field at frequency $\omega$, equivalently its $-\omega$ partner, and field $b$ to be the field itself. The squared frequencies are equal, so the mass density terms cancel:Multiply by $i\omega$ and use ${\hat{v}}_i=-i\omega {\hat{u}}_i$, ${\hat{v}}_i^{\ast }=i\omega {\hat{u}}_i^{\ast }$ to obtainFor the real fields defined in this question as the sum of positive and negative frequency components, time averaging cancels the $e^{\pm 2i\omega t}$ terms. Their mean body-force work mass density is exactly ${\hat{f}}_i^{\ast }{\hat{v}}_i+{\hat{v}}_i^{\ast }{\hat{f}}_i$, and their outward mean elastic-wave energy flux is $-t_i^{\ast }{\hat{v}}_i-{\hat{v}}_i^{\ast }{t}_i$. Thus mean body-force power equals mean outward boundary power. Mean stored energy does not accumulate in a periodic steady field.

There is no extra factor $1/2$ here: the specified real field is $\hat{f}{e}^{-i\omega t}+{\hat{f}}^{\ast }{e}^{i\omega t}$, not $\mathrm{Re}(\hat{f}{e}^{-i\omega t})$. For the latter phasor convention, both averages instead have the usual factor $1/2$. At $\omega =0$ velocity and both powers vanish.

The massless inextensible string has a common tension magnitude $T$. Since ${\mathbf{r}}_1=r_1({\mathbf{e}}_r-{\mathbf{e}}_z)$, the unit vector from the first particle towards the apex is $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. The second particle lies below the apex, soThe string length constraint isThe velocity of the first particle in cylindrical coordinates is ${\dot{r}}_1{\mathbf{e}}_r+r_1\dot{\theta }{\mathbf{e}}_{\theta }+{\dot{z}}_1{\mathbf{e}}_z$. Therefore the powers delivered by the two tension forces areThusThe tension can transfer energy between the particles, but its total power on the pair is zero.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

The expected value is the improper integral $\alpha {\int }_1^{\infty }{x}^{-\alpha }dx$. It converges exactly when $\alpha >1$, yieldingAt $\alpha =1$ this improper integral diverges logarithmically, and for $0<\alpha <1$ it diverges as a power of its upper endpoint. In both cases the nonnegative variable has infinite mean. This is the first Pareto moment threshold.

For $m>-1$, integrate the base power to obtain $J(m)={\int }_0^1{x}^{m}dx=1/(m+1)$. Since ${\partial }_m^n{x}^m=x^m(\log x{)}^n$, differentiating $n$ times givesHere $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x{|}^n$, whose integral becomes ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$ under $x=e^{-t}$ and is finite. The dominated convergence theorem establishes the needed differentiation under the integral sign and the logarithmic moments of a power on the unit interval.

Oxygen-plasma ashing, preferably a remote or downstream arrangement when low damage matters, removes organic photoresist chemically. Feed oxygen through a plasma source energized by radiofrequency or microwave power. Electron-impact dissociation generates reactive oxygen species; neutral radicals reach the wafer in a separate downstream region. Keep the wafer off a strongly biased driven electrode so energetic ion bombardment is much smaller than in directional reactive-ion etching. A barrel asher with shielded wafers is another chemical-stripping implementation.These equations represent radical production and an ideal complete-oxidation balance; actual exhaust can also contain CO and fragments. Pumping removes volatile products. Directional shape fidelity is unnecessary for blanket stripping, while access around surface relief and high chemical removal selectivity are useful. Monitor endpoint and wafer temperature, and verify removal with a witness surface. oxygen can oxidize exposed materials, charging can still damage sensitive structures, and inorganic or metal-containing residues may require a compatible separate clean. Ashing removes organic contamination; it is not a universal residue-removal step.^[ref]