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Taking the gradient of the quadratic potential energy found in part (ii) resolves the linearized force into identical $x$ and $y$ components:Each component therefore performs simple harmonic motion withThus the angular frequency is

The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

The physical rapid phase isDefine the local wavevector, angular frequency, and group velocity byThe equality of mixed partial derivatives gives the phase-compatibility equationsA ray is defined to move with the local group velocity:

Differentiate the local dispersion relationwith respect to $x_i$. Along a ray,The last two terms cancel because $\mathbf{k}$ is a gradient. Similarly,Finally,Thus the Hamiltonian ray-tracing equations are

A stationary state has the formFor a time-independent quantum observable $Q$, the global phases cancel in the expectation value:It is therefore independent of time.

For the infinite square well,The possible energy measurement results arewith respective Born rule probabilities $|c_1{|}^2$ and $|c_2{|}^2$. Normalization gives $|c_1{|}^2+|c_2{|}^2=1$.

At time $t$,The required matrix elements areConsequentlywhere the angular frequency isFinally, $2|c_1{c}_2|\le |c_1{|}^2+|c_2{|}^2=1$, so

At time $t$ the state isDirect evaluation of the radial integral gives the matrix elementsandIt follows thatThus $R(t)$ oscillates sinusoidally about $39a/8$. Its angular frequency is

In the Toomre stability criterionthe disk's gravitational potential supplies the destabilizing attraction, gas pressure represented by $c_s$ suppresses short wavelengths, and the Keplerian shear represented by the angular frequency $\Omega$ prevents coherent collapse on long scales through epicyclic motion. Axisymmetric perturbations are stable for $Q>1$ and gravitationally unstable for $Q<1$.

The state $Q\sim 1$ is a self-regulating fixed point. If cooling lowers $c_s$ until $Q<1$, gravitational instability produces shocks and turbulence that heat the disk and raise its effective velocity dispersion. If $Q$ becomes appreciably larger than one, the instability and its heating switch off, while turbulent dissipation and radiative cooling reduce $c_s$. A sustained gravito-turbulent state therefore remains close to marginal stability.

Equating the two linearized expressions for $p(a,t)-p_{\infty }$ from parts (iii) and (iv) givesThis is the equation of simple harmonic motion. Thus the bubble performs small oscillations with angular frequencyThis is the small oscillation of a planar gas bubble in an annular liquid.

For this metric, $\sqrt{-g}=a^2$, while $g^{\eta \eta }=-a^{-2}$, $g^{zz}=a^{-2}$ and $g^{ij}={\delta }^{ij}$. The massless Klein-Gordon equation is consequentlyInsert the spatial Fourier transform given in the question. Each wavenumber mode then satisfiesThus every field mode is a simple harmonic oscillator with a time-dependent angular frequency.

In small-amplitude oscillatory shear, expand the measured shear stress asComparison with the definitionof the storage modulus and loss modulus givesand henceThus the measured phase directly gives the loss tangent.

At low angular frequency, a standard viscoelastic fluid has time to relax and responds mainly as a viscous liquid: stress is approximately in phase with strain rate, so $\delta \to \pi /2$ as $\omega \to 0$. At high frequency it cannot relax during one cycle and responds mainly as an elastic solid, so stress is approximately in phase with strain and $\delta \to 0$ as $\omega \to \infty$.

For a material dominated by one viscoelastic relaxation time $\lambda$, the crossover between these regimes occurs when $\omega \lambda$ is of order one. Measure the frequency ${\omega }_c$ at which $G^{\prime }=G^{\prime \prime }$, equivalently $\delta =\pi /4$, and estimateA broad or multi-peaked crossover instead indicates a spectrum of relaxation times.

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

Measure $x_1,x_2$ as horizontal displacements from the respective downward equilibrium positions; the absolute horizontal coordinates are $x_1$ and $L+x_2$. Each vertical displacement is of second order in its small angle. Thus the spring is horizontal to first order, its extension is $x_2-x_1$, and Hooke's law gives horizontal forces $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized simple pendulum restoring force yieldsFor the normal modes of two spring-coupled equal pendulums, add and subtract these ordinary differential equations. The normal mode coordinates $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$ satisfyHence the two natural angular frequencies and displacement patterns areOrdinary frequencies are ${\omega }_{+}/(2\pi )$ and ${\omega }_{-}/(2\pi )$. In the first normal mode, both particles swing together with equal amplitudes and the spring is unextended to first order. In the second normal mode, the particles swing in opposite directions with equal amplitudes, so the spring adds a restoring force and increases the angular frequency.

The small-angle approximation $\sin \theta =\theta +O({\theta }^3)$ reduces ii to $\ddot{\theta }+{\omega }^2\theta =0$, whereThe coefficient is positive, so the solution is simple harmonic motion, $\theta (t)=C\cos \omega t+D\sin \omega t$. Here $\omega$ is the angular frequency; the ordinary frequency and period are

Use separation of variables to seek a normal mode $y(x,t)=X(x)\cos (\omega t+\delta )$, with angular frequency $\omega >0$ and wavenumber $k=\omega /c$. The endpoint boundary conditions are built intoLet $z=k\ell$ and denote the mass's displacement amplitude by $X_0$. Continuity and the mass equation becomeThe two possibilities must be separated before dividing by $\sin z$.

Antisymmetric modes. If $\sin z=0$, then $z=n\pi$, $n=1,2,\dots$, and $X_0=0$. The force equation gives $A+B=0$, since $\cos z\ne 0$. Thus $X(-x)=-X(x)$ and the mass stays at rest. These allowed angular frequencies satisfyThe restoring forces on the mass cancel even though the string vibrates.

Symmetric modes. If $\sin z\ne 0$, continuity forces $A=B$. Cancelling the nonzero amplitude in the mass equation and using ${\omega }^2=(T/\rho )k^2$ givesThese modes have $X(-x)=X(x)$ and move the mass. In dimensionless form their equation is $\cot z=\alpha z$, where $\alpha =m/(2\rho \ell )>0$. This is the spectral splitting for a string with a central point mass.

For the antisymmetric family, the graph of $\sin z$ crosses zero at every positive multiple of $\pi$. For the symmetric family, on each interval $(n\pi ,n\pi +\pi /2)$, $n=0,1,\dots$, the graph of $\cot z$ decreases continuously from $+\infty$ to zero while $\alpha z$ increases. By the intermediate value theorem and strict monotonicity, there is exactly one intersection in each interval. On the remaining half-intervals $\cot z<0<\alpha z$, so there are no further positive roots. This proves graphically and analytically that both families contain infinitely many normal modes. The figure uses $\alpha =1$ for illustration.

The algebraic case $k=0$ is not a mode: a zero-frequency profile would be linear on each half, continuous and zero at the endpoints; the zero-acceleration mass equation also makes its slopes agree, forcing the entire profile to be zero. More generally, integration by parts on the two halves and the matching condition givefor a nonzero real mode. Thus no nonzero static or exponentially growing mode has been omitted.

The lowest positive root belongs to the symmetric family and lies in $(0,\pi /2)$; all antisymmetric roots are at least $\pi$. Let $\alpha =m/(2\rho \ell )\gg 1$. The equation $\cot z=\alpha z$ forces its first root to tend to zero: for any fixed $\epsilon >0$, the right side eventually exceeds $\cot \epsilon$ at $z=\epsilon$, so the decreasing-minus-increasing difference has its zero below $\epsilon$.

The Taylor series $\cot z=z^{-1}-z/3+O(z^3)$ then givesSince $\omega =cz/\ell$ and $c=\sqrt{T/\rho }$, the lowest angular frequency isPhysically, at such a low frequency each half-string is nearly straight. If the mass moves up by $Y$, the two slopes are approximately $Y/\ell$ and $-Y/\ell$, so the restoring force is $-2TY/\ell$. The resulting harmonic oscillator equation, $m\ddot{Y}+(2T/\ell )Y=0$, has exactly the leading angular frequency above. The next correction from the same expansion is $\omega /{\omega }_0=1-\rho \ell /(3m)+O((\rho \ell /m{)}^2)$.

The damped angular frequency is $\Omega =\sqrt{k-{\lambda }^2/4}$. The oscillation envelope is proportional to $e^{-\lambda t/2}$. A decrease by a quarter leaves three quarters of the original amplitude, so $e^{-\lambda t/2}=3/4$. ThereforeThis concerns the envelope; it does not ask when the instantaneous displacement first equals three quarters of a chosen value.

One period advances the fundamental phase by $2\pi$, so the angular frequency is

The initial displacement is already the first spatial normal mode, so sine orthogonality gives $A_1=1$ and every other $A_n=0$. Zero initial velocity gives every $B_n=0$. ConsequentlyBoth fixed-end values vanish for all time, and differentiation shows that the initial velocity is zero. This is the fundamental standing wave, with angular frequency $\pi c/L$.

Use the method of undetermined coefficients, choosing a particular solution $y_p=A+B\cos 2x$. Substitution givesMatching the constant and cosine coefficients to $\frac{1}{2}-\frac{1}{2}\cos 2x$ gives $A=1/2$, $B=1/6$. ThusNeither forcing term is resonant with the homogeneous angular frequency one.

A normal mode has all coordinates following one fixed displacement pattern with a common time factor. Its normal frequency is that factor's oscillation frequency; normal mode coordinates are amplitudes in a basis of such patterns, making the linear equations independent. A zero-frequency mode can instead have constant velocity along a symmetry direction.

For the unit circle, each angular displacement equals its corresponding arc displacement. The kinetic energy and change in potential energy areAny common spring prestress changes only the constant $V_0$, because the sum of the three arc lengths is fixed. The Euler-Lagrange equations giveThese are the normal modes of three beads connected by springs on a circular hoop, with third mass $M=\mu m$. For $\mu >0$, their three mode vectors and squared angular frequencies areThe zero mode rotates the entire configuration without stretching a spring. The antisymmetric mode moves the equal masses oppositely while leaving the third fixed. In the symmetric mode the equal masses move together against the third, with zero total angular momentum. These vectors are orthogonal in the mass inner product with matrix $m\mathrm{diag}(1,1,\mu )$.

Writing $\mathbf{\theta }=Q_0{v}_0+Q_A{v}_A+Q_S{v}_S$ gives independent equations ${\ddot{Q}}_0=0$, ${\ddot{Q}}_A+{\omega }_A^2{Q}_A=0$, ${\ddot{Q}}_S+{\omega }_S^2{Q}_S=0$. In particular $Q_0=A+Bt$, retaining possible slow common rotation. For $\mu =1$, the two positive frequencies coincide at $\sqrt{3k/m}$; their two-dimensional space is all patterns with ${\theta }_1+{\theta }_2+{\theta }_3=0$.

With downward position as the vertical coordinate, the full potential energy isThe equilibrium tensions are $2k(a-\ell )=6mg$, $k(b-\ell )=3mg$ and $k(c-\ell )=mg$. On expanding, these relations cancel every term linear in the displacements, leaving $V=k{x}^2+k(x-y{)}^2/2+k(y-z{)}^2/2+V_0$. Newton's second law therefore givesFor a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, set $s=m{\omega }^2/k$. The generalized eigenvalue problem isIts determinant is $2(1-s)[3(1-s{)}^2-2]$, so the angular frequencies and mode directions areThe square roots of these positive values give the requested frequencies. Only the lowest mode has all components of the same sign, so all three particles oscillate in phase at the lowest frequency. The mode shapes are shown below.

For release from rest with $Y$ remaining at equilibrium, chooseThis excites only the middle-frequency mode, so $y(t)=0$ for all time. It is a stationary middle mass in a three-mass spring chain: the equal and opposite adjacent spring forces cancel on $Y$.