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In neutral meson mixing, the mass eigenstates and flavor eigenstates differ. A state prepared as $D^0=c\bar{u}$ develops a ${\bar{D}}^0=u\bar{c}$ component because the effective Hamiltonian operator has off-diagonal mass and decay width entries. In the CP symmetry limit implicit in the source, the propagating states are equal superpositions, with average mass $M=(M_1+M_2)/2$ and average decay width $\Gamma =({\Gamma }_1+{\Gamma }_2)/2$.

To retain the source's positive $i\sin$ convention for $g_{-}$, take $M_2-M_1=\Delta M$ and ${\Gamma }_2-{\Gamma }_1=\Delta \Gamma$, with $g_{\pm }=(e_1\pm e_2)/2$, where $e_j=\exp (-i{M}_{j}t-{\Gamma }_{j}t/2)$. Then $x=\Delta M/\Gamma$ and $y=\Delta \Gamma /(2\Gamma )$ reproduce the given functions. Reversing an eigenstate label changes signs in intermediate formulae; it must also change the phase conventions consistently.

For equal widths, the mixed probability is proportional to $e^{-\Gamma t}{\sin }^2(\Delta Mt/2)$. Thus the angular angular frequency is $|\Delta M|$ in natural units, or $|\Delta M|/\hslash$ with units restored; the period is $2\pi \hslash /|\Delta M|$. A nonzero difference of decay widths adds hyperbolic time dependence.

The ordinary neutral flavour-carrying mesons that mix are $K^0$, $D^0$, $B_d^0$ and $B_s^0$ with their antiparticles. A self-conjugate ${\pi }^0$ has no distinct antiparticle flavour to oscillate into. There are no long-lived top-flavoured mesons, since the top quark decays before hadronization.

The Euler-Lagrange equation from a is $2M{R}^2\ddot{\theta }+MgR\sin \theta =0$. Linearizing about $\theta =0$ givesBoth angular frequency and ordinary frequency are given to distinguish the two conventions. The stable position places the extra mass below the pivot.

The uniform cylindrical torso contributes $I_c=M{r}^2/2=50(0.20{)}^2/2=1.00\mathrm{kg}{\mathrm{m}}^2$. Adding both arm point masses about the same vertical axis givesWith initial angular frequency ${\omega }_i=2\pi \mathrm{rad}{\mathrm{s}}^{-1}$, conservation of angular momentum gives $I_i{\omega }_i=I_f{\omega }_f$. ThereforeThis corresponds to $3.47788$ rotations per second; it is important not to confuse that frequency with the requested angular frequency.

For the attractive central force $\mathbf{F}=-k\mathbf{r}$, with $k>0$, the potential energy is $U(r)=k{r}^2/2$ up to an additive constant. Conservation of angular momentum confines the motion to a plane; writing its magnitude as $J=m{r}^2\dot{\phi }$, the kinetic energy isAt fixed $J$, the radial energy equation becomes $E=m{\dot{r}}^2/2+U_{\mathrm{eff}}(r)$, whereThe effective potential combines the true potential energy with the angular-motion contribution. Its negative derivative gives the radial equation $m\ddot{r}=-U_{\mathrm{eff}}^{\prime }=J^2/(m{r}^3)-kr$; the first term is the centrifugal contribution arising from elimination of the angular coordinate.

A circular orbit is a fixed radial position at a stationary point:For $J\ne 0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=4k>0$, so this is a stable minimum. Small radial disturbances have angular frequency $2\sqrt{k/m}$. For $J=0$, there is no nonzero-radius circular orbit.

Measure the vertical displacement $x$ downward from the spring's unstretched position. The kinetic energy is $m{\dot{x}}^2/2$, while the spring and gravitational potential energies are $k{x}^2/2$ and $-mgx$. Thus the Lagrangian isApplying the Euler-Lagrange equation, $d(\partial \mathcal{L}/\partial \dot{x})/dt-\partial \mathcal{L}/\partial x=0$, givesGravity shifts the equilibrium to $x_{\ast }=mg/k$. For displacement $\xi =x-x_{\ast }$ from that equilibrium, the equation of motion becomes $\ddot{\xi }+(k/m)\xi =0$, giving simple harmonic motion with angular frequency $\omega =\sqrt{k/m}$. Choosing an upward coordinate instead reverses the gravitational term, but leaves the equilibrium-relative angular frequency unchanged.

Write $r=r_0+\eta$ and keep $L$ fixed. The small radial oscillations in a power-law potential have squared angular frequencyThe circular orbital angular frequency has magnitude $\Omega =|L|/(m{r}_0^2)$, so ${\Omega }^2=(\beta k/m)r_0^{k-2}$. HenceA restoring radial oscillation exists for $k>-2$, subject to the attractive-force condition in part (b). For $k<-2$ the radial stationary point is unstable; $k=-2$ is the degenerate boundary. This agrees with circular-orbit stability for a power-law central force, since the actual force is proportional to $-r^{k-1}$.

Let $x_0$ be the equilibrium, so $V^{\prime }(x_0)=0$. For $\xi =x-x_0$, the Taylor expansion of the potential givesAn ordinary small-amplitude simple harmonic motion therefore requires a nondegenerate local minimum, $V^{\prime \prime }(x_0)>0$, with the quadratic term dominating the higher terms over the excursion. ThenA local minimum with $V^{\prime \prime }(x_0)=0$, such as a quartic minimum, need not give small-amplitude harmonic motion with a nonzero limiting angular frequency. If $V^{\prime \prime }(x_0)<0$, the equilibrium is unstable rather than oscillatory.

Constant azimuth. If $\varphi$ is constant, $\dot{\varphi }=0$ and the motion is a planar simple pendulum:For the small-angle approximation, $\sin \theta \simeq \theta$, giving simple harmonic motion withThe harmonic angular frequency requires small amplitude. At finite libration amplitude ${\theta }_{\max }<\pi$, the exact oscillation period is $4\sqrt{l/g}K(\sin ({\theta }_{\max }/2))$, where $K$ is the complete elliptic integral of the first kind; the nonlinear pendulum does not have the same amplitude-independent period.

Constant polar angle. If $\theta ={\theta }_0$ is nonzero and constant, the bob traces a horizontal circle: this is a conical pendulum. Setting $\ddot{\theta }=0$ and dividing by $\sin {\theta }_0$ givesThe sign distinguishes the two senses of rotation. The exactly downward configuration ${\theta }_0=0$ is stationary; its azimuth is undefined and it must not be obtained by dividing by $\sin {\theta }_0$. A nontrivial constant-angle orbit of this gravity-only system needs $\cos {\theta }_0>0$ even though the supporting rod can also push.

For a sphere with uniform mass density, $M=(4\pi /3)\rho a^3$. Its moment of inertia of a uniform solid sphere is $I=2M{a}^2/5$, and rotation about a diameter gives angular momentum magnitude $L=I\omega$. The spheres have the same angular frequency $\omega$, soThe mass grows by a factor of two, and the extra squared lever arm supplies a further factor of four. Equal angular frequencies alone do not mean equal angular momenta.

For the elastic pendulum, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin \theta$ and $Z=-r\cos \theta$, with $Z$ upward. Differentiating gives ${\dot{X}}^2+{\dot{Z}}^2={\dot{x}}^2+r^2{\dot{\theta }}^2$. Thus the Lagrangian isFor $x$, the Euler-Lagrange equation givesFor $\theta$, the canonical momentum is $p_{\theta }=m(l+x{)}^2\dot{\theta }$. Its Euler-Lagrange equation isor, for positive length,At an equilibrium, all time derivatives vanish, so $kx=mg\cos \theta$ and $\sin \theta =0$. Modulo full rotations, the two formal possibilities areThe lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.

The potential energy is $V=k{x}^2/2-mg(l+x)\cos \theta$. At either formal stationary point its mixed second derivative vanishes, whileBoth curvatures are positive at the lower state, making it a stable equilibrium; the upper state, when it has positive length, has negative angular curvature and is unstable.

To linearize about the stable state, define the small displacement $\xi =x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as ${\dot{\theta }}^2$, $\dot{\xi }\dot{\theta }$ and $\xi \ddot{\theta }$ are second order, while $\cos \theta =1+O({\theta }^2)$ and $\sin \theta =\theta +O({\theta }^3)$. The first-order equations of motion therefore decouple:There are two independent normal modes. The radial one is vertical simple harmonic motion with angular frequency ${\omega }_r=\sqrt{k/m}$. The transverse one is a simple pendulum oscillation at fixed, statically stretched length $r_0$, with angular frequency ${\omega }_{\theta }=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the elastic pendulum.

Choose the orientation $J>0$. To leading order in the radial disturbance, the azimuthal angular frequency is ${\omega }_{\theta }=J/(m{r}_0^2)$, soFor either orientation the positive frequency ratio is ${\omega }_r/|{\omega }_{\theta }|$. At $k=-1$, the ratio is $1$: there is one radial oscillation per revolution. The bound Kepler orbit is an ellipse with the force centre at a focus, with one closest and one furthest approach per revolution. Its exact polar coordinates equation isAt $k=2$, the ratio is $2$: there are two radial oscillations per revolution. The isotropic harmonic oscillator trajectory is an ellipse centred on the force centre. An exact parametrization isHere $r^2=a^2{\cos }^2\Omega t+b^2{\sin }^2\Omega t$ repeats twice during one full circuit. The two different placements of the force centre distinguish these ellipses. These are the two all-bound-orbit cases of Bertrand's theorem.

Put $t_1=0.005\mathrm{s}$, $\tau =t-t_1$, and $I_1=2(1-e^{-1.875})$. When the switch reopens, the circuit again contains the battery and all three series elements. Inductor current continuity and capacitor voltage continuity giveThus $I^{\prime }(0^{+})=(V_0-R{I}_1)/L\simeq 115.0\mathrm{A}{\mathrm{s}}^{-1}$, not zero. Differentiating the loop equation gives the homogeneous current equationThe battery is still connected; its constant voltage disappears on differentiating, and the eventual capacitor voltage is $V_0$.

With $\alpha =R/(2L)=187.5{\mathrm{s}}^{-1}$ and ${\omega }_0=1/\sqrt{LC}$, the inequality $\alpha <{\omega }_0$ makes the circuit underdamped. The damped angular frequency and ordinary frequency areThe full current satisfying both initial conditions isIts amplitude envelope decays as $e^{-\alpha \tau }$. ThereforeThis is measured after reopening; on the clock starting with the first closure it is $t\simeq 0.02098\mathrm{s}$. An instantaneous current zero is not an amplitude decay measurement.

For positive angular frequency, the squared amplitude response of the series-parallel RC band-pass filter isIts denominator is minimized exactly when $w=w^{-1}$, soThis is a band-pass filter maximum produced entirely by the two RC arms; the drawn electrical circuit contains no inductor. The low- and high-frequency limits areThus the output amplitude tends to zero linearly with $\omega$ at low frequency and as $1/\omega$ at high frequency. In real time,The upper series capacitor suppresses low frequencies, while the lower shunt capacitor suppresses high frequencies.

The half-power amplitude condition for the RC low-pass filter gives $1+(\omega RC{)}^2=2$. Since angular frequency is positive,At this cutoff angular frequency, the output amplitude is smaller by $1/\sqrt{2}$ and the phase shift is $-\pi /4$.

This is a plane electromagnetic wave with linear polarization. $\mathbf{E}$ is the electric field; $E_0$ its peak amplitude; $x$ the position along propagation; $t$ the time; $k=2\pi /\lambda$ the wavenumber; and $\omega =2\pi f$ the angular frequency. Here $\lambda$ is the wavelength, $f$ the frequency, and $\hat{\mathbf{y}}$ the unit vector in the direction of the electric field. Constant phase $kx-\omega t$ moves along $+x$ with phase velocity $\omega /k$.

Put $\theta =kx-\omega t$. The differential Faraday's law of induction givesThe accompanying oscillating magnetic field, excluding any independent static background, is therefore $\mathbf{B}=(k/\omega )E_0\cos \theta \hat{\mathbf{z}}$. To determine the propagation speed, also use the source-free Ampere-Maxwell law:ThusThe electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation; $\mathbf{E}\times \mathbf{B}$ points along $+x$.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

At the balance angular frequency, $Z=W$ and the two voltage dividers have the same ratio. ThusThis zero detector electric current is the operational condition used to find $R_v$ and ${\omega }_0$.

At zero angular frequency, the lower right capacitor is an open circuit, while the upper right electrical impedance tends to $R_x$. With the detector removed, no electric current flows down that branch, so $V_R=V_0$ and $V_L=V_0/2$. ThereforeFor the connected ideal ammeter, the resulting electric current from right to left is $V_0/(R+2R_x)$, despite its zero terminal voltage. This follows from $I_A=V_{\mathrm{oc}}/Z_{\mathrm{th}}$ with $Z_{\mathrm{th}}=R/2+R_x$.