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For $y^{\prime \prime }+ϵ(y^{\prime }{)}^n+y=0$, the method of multiple scales gives no first-order phase drift. If $n$ is odd, the amplitude obeys the displayed damping law with $C_n=2^{-(n+1)}\left(\genfrac{}{}{0ex}{}{n+1}{(n+1)/2}\right)$. If $n$ is even, the average is zero and the amplitude is unchanged on the first slow time $T=ϵt$. The odd case dissipates mechanical energy, whereas the even case is reversible and its force is not damping of both velocity signs.

A smooth wire has instantaneous shape $z=f(r)$ in cylindrical coordinates and rotates at constant angular velocity $\dot{\theta }=\omega$. The tangent vector at fixed time is ${\mathbf{e}}_r+f^{\prime }(r){\mathbf{e}}_z$. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by $\dot{r}$ proves conservation ofThis is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at $a>0$ satisfies $U^{\prime }(a)=g{f}^{\prime }(a)-{\omega }^{2}a=0$. Its linearized equation is $(1+f^{\prime }(a{)}^2)\ddot{ϵ}=-U^{\prime \prime }(a)ϵ$. A strict local minimum with $U^{\prime \prime }(a)>0$ gives Lyapunov stability, while $U^{\prime \prime }(a)<0$ gives an unstable equilibrium. If $U^{\prime \prime }(a)=0$, higher orders are needed.

An orbit that remains within a bounded spatial region for all times for which the motion is defined. In an attractive inverse-distance central potential with zero potential energy at infinity, a noncolliding orbit is bound when its total mechanical energy is negative. That sign test depends on the potential and its energy origin; it is not a universal definition of bounded motion. A closed orbit repeats its position and velocity, whereas a general bound orbit can precess without closing.

For constant masses, a time-independent potential energy $V$ and forces ${\mathbf{F}}_i=-{\nabla }_{i}V+{\mathbf{R}}_i$ with zero total extra power ${\sum }_i{\mathbf{R}}_i\cdot {\mathbf{v}}_i=0$, Newton's second law gives $dT/dt=-dV/dt$. Thus mechanical energy is conserved. Smooth stationary constraint reactions and a purely magnetic Lorentz force supply zero power. This is more restrictive than conservation of energy in general, which may include thermal or other internal forms.

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

For unit-mass motion in $V(x)=3x^2-2x^3$, the mechanical energy is $E=v^2/2+V(x)$. The origin is a center equilibrium, and $(1,0)$ is a saddle equilibrium. Energies $0<E<1$ have trapped periodic orbits around the origin as well as a separate escaping component to the right. The barrier energy has a homoclinic orbit returning to the saddle after a turn at $x=-1/2$, plus two saddle branches on $x>1$. Above the barrier, particles traverse the well and eventually escape to the right.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

Let $r$ be the instantaneous bond length, $r_e>0$ its equilibrium value, and $q=r-r_e$ the stretching displacement. Assume one fixed electronic quantum state, a smooth potential energy $V(r)$ with a nondegenerate local minimum at $r_e$, and omit molecular rotation initially. The Taylor series has no linear term because $V^{\prime }(r_e)=0$:Keeping only the quadratic term is the harmonic approximation near a potential minimum. The bond force constant $k$ has units $\mathrm{N}{\mathrm{m}}^{-1}$. It measures local stiffness; it is not the bond-dissociation energy, which concerns the depth and distant shape of the entire well.

The two-particle relative coordinate explains the reduced mass. For stretching along a fixed axis in a stationary centre of mass frame, let $u_1,u_2$ be the nuclear displacements from equilibrium. Then $u_2-u_1=q$ and $m_1{u}_1+m_2{u}_2=0$, soHere $T$ is the internal kinetic energy, and $m_1,m_2>0$ are the two nuclear masses. The lighter partner moves farther; the reduced mass is smaller than either individual mass. The restoring force $-dV/dq\simeq -kq$ gives simple harmonic motion:The angular frequency is $\omega$, the ordinary frequency is $\nu$, and ${\tilde{\nu }}_{\mathrm{harm}}$ is the corresponding spectroscopic wavenumber; $c$ is the speed of light. Use $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$ to obtain ${\tilde{\nu }}_{\mathrm{harm}}$ in $\mathrm{c}{\mathrm{m}}^{-1}$. The symbol $k$ here is a bond force constant, not an angular wavenumber.

For an original dimensionless example, choose $m_1=m_2=2$ and $k=4$ in consistent model units. Then $\mu =1$ and $\omega =2$. Releasing at $q(0)=0.1$ with $\dot{q}(0)=0$ gives $q(t)=0.1\cos (2t)$ and $u_1=-q/2$, $u_2=q/2$. Each atomic nucleus moves only half the stretching amplitude, while the conserved internal mechanical energy above $V_0$ is $\frac{1}{2}k(0.1{)}^2=0.02$. Fixing one partner instead would give $\omega =\sqrt{2}$, demonstrating why a one-body calculation changes the model.

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.

For constant symmetric mass and stiffness matrices, $E=\frac{1}{2}{\dot{q}}^{\mathsf{T}}M\dot{q}+\frac{1}{2}{q}^{\mathsf{T}}Kq$ satisfies $\dot{E}={\dot{q}}^{\mathsf{T}}(M\ddot{q}+Kq)$. Every equation $M\ddot{q}+Kq=G\dot{q}$ with antisymmetric $G$ conserves this same mechanical energy, because ${\dot{q}}^{\mathsf{T}}G\dot{q}=0$. Thus a scalar conservation law alone cannot select the dynamics of several degrees of freedom. The specified conservative-force or Lagrangian structure supplies the missing information.

For a damped harmonic oscillator $\ddot{y}+\gamma \dot{y}+{\omega }_0^{2}y=0$, differentiation of the mechanical energy $E=m({\dot{y}}^2+{\omega }_0^2{y}^2)/2$ gives $\dot{E}=-m\gamma {\dot{y}}^2$. This identity is exact. In the weak-damping regime $\gamma \ll {\omega }_0$, the leading period average has equal kinetic energy and potential energy, so $\overline{{\dot{y}}^2}\simeq \overline{E}/m$ and $\dot{\overline{E}}\simeq -\gamma \overline{E}$. Thus the averaged energy has exponential decay, while the instantaneous energy has a small oscillatory correction. Its derivative vanishes at every turning point, ruling out an exact pointwise nonconstant exponential. The quality factor is $Q\simeq {\omega }_0/\gamma$ in this approximation.

For the normalized damped pendulum, the sum of kinetic energy and gravitational potential energy, divided by $mg\ell$, isThe chain rule and the full nonlinear equation give this identity exactly, without a small-angle approximation. Here $\dot{\theta }=d\theta /d\tau$ as in damped pendulum. In physical units the mechanical energy and the power of the resisting torque areThe pendulum's lost mechanical energy is transferred to the damper and surroundings; it does not violate conservation of energy. In an ideal viscous dashpot this loss is frictional heating. The exact cumulative balance is particularly useful for checking a numerical solution:

The figure uses $c=0.20$ and release from rest. Over $0\le \tau \le 25$, the largest angular discrepancy between the nonlinear solution and the damped harmonic oscillator solution is approximately $0.000702\mathrm{rad}$ ($0.0402^{\circ }$) for a $10^{\circ }$ release, but $0.14995\mathrm{rad}$ ($8.592^{\circ }$) for a $60^{\circ }$ release. Their nonlinear initial energies are $1-\cos 10^{\circ }\approx 0.015192$ and $1-\cos 60^{\circ }=0.5$, respectively. At $\tau =25$ the remaining fractions are approximately $0.0064080$ and $0.0063757$. These are illustrative calculations from specified initial conditions, not observations. The generator uses a Runge-Kutta method and integrates $Q$ alongside the motion; halving its step and independently integrating $c{\dot{\theta }}^2$ checks the result.

For $\ell =1\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the first three nonlinear same-side swing intervals after a $60^{\circ }$ release are approximately $2.1008$, $2.0378$, and $2.0221\mathrm{s}$. They approach the linear-model oscillation period $2.0162\mathrm{s}$ as the amplitude shrinks. After a $10^{\circ }$ release the corresponding intervals are already about $2.0185$, $2.0168$, and $2.0164\mathrm{s}$. This makes the experiment's period drift a quantitative test of the small-angle approximation, rather than an automatic reason to change the damping coefficient.

At a turning point, $\dot{\theta }=0$ and $\dot{E}=0$ even though the swing is losing mechanical energy over the cycle. Its potential energy there is $E_n=1-\cos A_n$, so measured turning angles give successive energies without estimating angular velocity. In the proposed decay experiment, multiply these values by $mg\ell$ to obtain energies in joules, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small amplitudes is $E_n\simeq A_n^2/2$. In the linear underdamped model, same-side peak energies consequently satisfy $E_{n+1}/E_n=e^{-\gamma T_d}$, the square of the angular amplitude ratio. For the illustrative parameters above this energy ratio is approximately $0.2828$. Between turning points, kinetic energy and potential energy continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.

It is therefore inaccurate to replace the exact identity by $\dot{E}=-cE$. Even for the linear model, $E_{\mathrm{lin}}=({\dot{\theta }}^2+{\theta }^2)/2$ has ${\dot{E}}_{\mathrm{lin}}=-c{\dot{\theta }}^2$. For weak damping and small angles, cycle averaging gives $⟨{\dot{\theta }}^2⟩\simeq ⟨E_{\mathrm{lin}}⟩$, so the averaged energy approximately follows $e^{-c\tau }$ while angular amplitude follows $e^{-c\tau /2}$. This exponential decay is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.

The energy identity also gives a global conclusion without solving the nonlinear equation. If $E$ were constant on a positive-length interval, then $\dot{\theta }$ would vanish throughout it; the equation would force $\sin \theta =0$, making the entire trajectory an equilibrium. Every nonstationary trajectory therefore loses mechanical energy over every positive-length interval, even though its instantaneous derivative can vanish at a turning point. A nonconstant periodic motion would have to return to its initial mechanical energy, so it is impossible for this undriven damped pendulum. Oscillations near a stable focus are successive, shrinking swings, not a repeating cycle. Moreover, if $E(0)<2$, the angle cannot reach an upright equilibrium, whose potential energy is $2$; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.

A linearly damped oscillator $m\ddot{x}+b\dot{x}+kx=0$ has $x=A{e}^{-\gamma t}\cos ({\omega }_{d}t+\varphi )$, with $\gamma =b/(2m)$ and ${\omega }_d^2=k/m-{\gamma }^2>0$. After one damped period $T=2\pi /{\omega }_d$, both displacement and velocity are multiplied by $e^{-\gamma T}$. Their quadratic mechanical energy is therefore multiplied exactly by $e^{-2\gamma T}$, regardless of the starting phase. In particular successive same-direction equilibrium crossings have this ratio even though potential energy is zero at each. A cosine phase maximum is not exactly a displacement turning point when damping is nonzero.

For a uniform ideal fixed-end string, a real normal mode of displacement amplitude $A_n$ has total mechanical energy $T{\pi }^2{n}^2{A}_n^2/(4L)$. This follows by integrating the squared sine velocity and cosine slope over the length; each squared trigonometric function integrates to $L/2$. Thus finitely many excited modes with $|A_n|$ proportional to $1/n$ have equal total modal energies. Their instantaneous kinetic energies need not agree, because their time factors oscillate at different frequencies.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

For potential energy $\Phi (r)=-mk/r^2$, the inward central force has magnitude $2mk/r^3$. With conserved specific angular momentum $h=r^2\dot{\theta }$, the radial equation is $\ddot{r}=(h^2-2k)/r^3$. At the critical value $h^2=2k$, this reduces to $\ddot{r}=0$. An inward initial radial velocity $-v$, with $v>0$ and $r(0)=a$, therefore gives $r=a-vt$ until collision at $t_{\ast }=a/v$.

For $h>0$, integration of $\dot{\theta }=h/(a-vt{)}^2$ gives $\theta -\theta (0)=(h/v)(1/r-1/a)$. Thus the orbit is a hyperbolic spiral and the particle completes infinitely many turns before the finite collision time. The tangential speed $h/r$ diverges. The potential energy and kinetic energy individually diverge with opposite signs, while their sum remains the finite conserved mechanical energy $m{v}^2/2$.

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

For a mechanical system whose imposed motion is uniform rotation with angular velocity $\omega$ about a fixed axis, a conserved rotating-frame energy can exist even when the inertial mechanical energy $E$ changes. With angular momentum $L_z$ about that axis, it is $J=E-\omega L_z$. For a bead constrained to rotate with $\dot{\theta }=\omega$, this becomesFor $z=r^2/(2a)$ in uniform gravity, division by mass yieldsThis is the first integral that controls the bead's radial motion. Different normalizations and signs are used for the related Jacobi constant in celestial mechanics.

Total kinetic energy is unchanged between the incoming and outgoing states of an elastic collision. This does not require each body's kinetic energy to be unchanged, nor that kinetic energy remain constant during temporary elastic deformation. In general mechanical energy includes potential energy, so conservation of mechanical energy alone does not imply constant kinetic energy.