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Let the cylinder length be $\ell =0.020\mathrm{m}$, cross-sectional area $A$, and the fluid mass density ${\rho }_f$. At floating equilibrium, Archimedes' principle givesFor a small downward displacement $z$ from this equilibrium, the displaced fluid volume increases by $Az$. The additional upward buoyant force is ${\rho }_{f}gAz$, and Newton's second law becomesThis is simple harmonic motion, with angular frequency $\omega =\sqrt{2g/\ell }$ and oscillation periodThe result for vertical oscillations of a floating cylinder is independent of its cross-sectional area and fluid mass density because those quantities cancel using the known immersion fraction. It neglects drag, surface effects and the extra fluid inertia from the motion; the cylinder must remain partly immersed throughout the small motion.

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 momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

A primitive cell containing two atoms has two independent displacement coordinates for each chosen polarization. Their coupled normal modes split into an acoustic phonon branch, which translates the cell at long wavelength, and an optical phonon branch, which moves its constituents against one another. The nonzero restoring force for the relative displacement gives a nonzero optical frequency even at the zone centre.

For an explicit scalar model, take alternating masses $M,m$, equal nearest-neighbor spring constant $K$, and cell length $a$. If $u_n,v_n$ are the displacements of the two atoms in cell $n$, Newton's second law givesA mode proportional to $e^{i(kna-\omega t)}$ has a nonzero amplitude only when the two-by-two dynamical determinant vanishes. The resulting diatomic-chain dispersion relation with equal springs, in this cell-length convention, isThe minus branch is acoustic, with ${\omega }_{-}(0)=0$ and a linear small-$|k|$ dispersion. The plus branch is optical, withAt $k=0$, the acoustic eigenvector has $u=v$. The optical eigenvector instead satisfies $Mu+mv=0$: its cell centre of mass stays fixed while the atoms move oppositely. At the zone edge, for $M>m$, the frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, giving a gap between branches. Both slopes vanish there. The dispersion and displacement patterns are shown in the preceding original figure.

If the two atoms carry opposite effective charges, the relative motion creates an oscillating electric dipole moment, enabling infrared absorption by long-wavelength light. The photon momentum is tiny compared with the zone size, so ordinary optical absorption principally excites modes close to $k=0$. Raman activity instead depends on modulation of polarizability; the name optical does not imply that every such mode is infrared active. Three-dimensional crystals have separate longitudinal and transverse polarizations, and long-range Coulomb forces can split their optical frequencies. A monatomic primitive cell has no internal relative coordinate and therefore no optical branch of this type.

Let the nearest-neighbour spacing be $d$, so the primitive unit cell has length $\ell =2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K^{\prime }$ between cells. Hooke's law and Newton's second law giveFor normal modes $u_n=u{e}^{i(qn\ell -\omega t)}$ and $v_n=v{e}^{i(qn\ell -\omega t)}$, the two amplitudes satisfyIts vanishing determinant gives the alternating-spring chain dispersion relationThe minus sign is the acoustic phonon branch, with ${\omega }_{-}(0)=0$ and $u=v$ at the origin. The plus sign is the optical phonon branch, with $u=-v$ at the origin. For positive springs the acoustic phonon branch rises from zero to $\sqrt{2\min (K,K^{\prime })/m}$ at the Brillouin zone edge; the optical phonon branch falls from $\sqrt{2(K+K^{\prime })/m}$ to $\sqrt{2\max (K,K^{\prime })/m}$. Their long-wavelength limits areThus the leading frequency in the strong-bond limit of the alternating-spring chain is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.

Newton's first law says that a body retains uniform rectilinear motion, including rest as the zero-velocity case, unless impressed forces change that motion. Its physical content includes the existence of a common class of inertial frames: free bodies do not each require an unrelated frame chosen just to straighten their own trajectory. Uniformity refers to the inertial time parameter, and straightness to the spatial structure of those frames. Everyday slowing under friction is compatible with this claim because it involves an interaction, not an exception to inertia.

Newton's second law relates the impressed resultant force to change of momentum. In modern continuous-force notation,so for constant mass it becomes $m\mathbf{a}={\mathbf{F}}_{\mathrm{net}}$. An impulsive version is $\Delta \mathbf{p}=\int \mathbf{F}dt$. The change is directed along the impressed resultant, not necessarily along the body's current velocity. The original wording concerns change of quantity of motion; the differential equation is its modern continuous-time formulation.^[ref]

These are substantive physical claims only when mass, interactions and the inertial structure are constrained independently. If “force” merely abbreviates whatever equals $m\mathbf{a}$ in any arbitrarily chosen coordinates, the second law by itself predicts nothing. Actual applications combine the law with independently testable interaction laws, vector addition of forces, initial conditions and a specification of the inertial frames.

Within an already specified inertial frame, with constant mass, substituting zero resultant force into Newton's second law does give zero acceleration. In that limited mathematical sense, force-free uniform motion is a special case. The reason for retaining Newton's first law is that its physical role is broader than this substitution.

The second law does not, just by its algebra, identify the coordinate systems in which acceleration measures the response to physical interactions. A free body described in coordinates $x^{\prime }=x-\frac{1}{2}A{t}^2$ has ${\ddot{x}}^{\prime }=-A$, despite experiencing no impressed interaction. If the second law is restricted to inertial frames, the first law supplies a criterion for that restriction and asserts that a common inertial structure exists. If instead one introduces a fictitious force $-mA$ to preserve the equation in the accelerated coordinates, that coordinate-dependent bookkeeping should not be confused with a newly discovered physical interaction.

Nor can the first law simply be used as a circular definition in which “free” means “unaccelerated” and “inertial” means “whatever makes the chosen body free”. The framework requires empirical identification of interactions and a mutually consistent class of free-body trajectories. Removing an influence or improving isolation should make the same inertial description work for many bodies, not merely one fitted trajectory.

Thus the two laws have distinguishable jobs: the first articulates the force-free inertial background; the second quantifies departures caused by impressed forces. A modern axiomatization may build that background into its spacetime structure and then omit a separately numbered first law. That is a legitimate reorganization, but it incorporates the first law's content rather than obtaining all of it from the equation $\mathbf{F}=m\mathbf{a}$. Newton states the laws separately in the opening axioms.^[ref]

Oscillatory response. A material with viscoelasticity stores some deformation energy and dissipates some over a loading cycle. In small-amplitude oscillatory shear, represent the strain by $\gamma (t)=\mathrm{Re}({\gamma }_0{e}^{i\omega t})$. The shear-stress amplitude is related to the strain amplitude by the complex viscoelastic modulusHere $G_1=G^{\prime }$ is the storage modulus, the component in phase with strain, and $G_2=G^{\prime \prime }$ is the loss modulus, the dissipative quadrature component. Both have units of pressure. The stress leads the strain by an angle $\delta$, withFor this $e^{i\omega t}$ convention, a passive material has $G_2\ge 0$ at positive frequency. Integrating $\tau \dot{\gamma }$ over one period gives dissipated energy per unit volume $\pi {\gamma }_0^2{G}_2$, and the average input power per unit volume is $\omega {\gamma }_0^2{G}_2/2$. Thus the loss tangent compares the dissipative and elastic parts of the response. The frequency dependence reflects which internal relaxation processes can follow the imposed deformation.

Torsional measurement. Use $\rho$ for radial position inside the rod, reserving $r$ for its outer radius. A small angular displacement $\theta$ of one end relative to the fixed other end gives shear strainIn harmonic amplitude notation, the shear stress is $G^{\ast }\rho \Theta /\ell$. Its restoring torque is obtained by multiplying by the lever arm and integrating over the circular section:The factor $J={\int }_A{\rho }^{2}dA=\pi r^4/2$ is the polar second moment of area; it is not the disc's mass moment of inertia $I$. The torsional spring constant becomes the complex stiffness $K^{\ast }=J{G}^{\ast }/\ell$. Angular Newton's second law therefore gives the paper's compact representationIts precise interpretation is a torsional measurement of a complex shear modulus. At a specified frequency, with complex angular amplitude $\Theta$, the driven response obeysSetting the external torque to zero supplies the free-mode condition, whose damped frequencies are generally complex. A dissipative free oscillator does not execute a sustained nonzero oscillation at a real frequency without driving. In real time, a general frequency-dependent complex viscoelastic modulus represents a causal memory response, rather than a complex instantaneous force on a real displacement. The derivation assumes a uniform slender circular rod, small twist, negligible rod inertia relative to the disc, and no additional bending, slip or external damping.

Frequency regimes and sketches. A permanently crosslinked elastomer has a low-frequency rubbery plateau $G_R>0$. Slow chain rearrangements can follow the strain, so the network is relatively compliant but still elastic. At high frequency the relevant segmental motions are frozen, giving a much larger unrelaxed or glassy stiffness $G_U$. Between these regimes the motions lag behind the loading and dissipate strongly.

A standard linear solid model supplies a consistent illustrative sketch of this process. Put $x=\omega {\tau }_r$, where ${\tau }_r$ is the stress viscoelastic relaxation time, and $\Delta G=G_U-G_R>0$. ThensoAt very low frequency, $G_1\to G_R$, $G_2\sim \Delta Gx$ and $\tan \delta \to 0$. Through the relaxation region $G_1$ rises, $G_2$ has a maximum and $\tan \delta$ has a loss peak. At high frequency, $G_1\to G_U$, $G_2\sim \Delta G/x$ and $\tan \delta \to 0$ again. Real polymers usually have a spectrum of relaxation times, so these features are broader and can have further structure.

The two kinds of loss peak need not coincide. The peak frequencies in a standard linear solid areThe first follows by differentiating $x/(1+x^2)$; the second by differentiating $x/(G_R+G_U{x}^2)$. This is why the sketch marks the peaks separately. If the polymer is an uncrosslinked melt rather than a permanent network, there is an additional ultimate low-frequency terminal-flow regime, with $G_1\propto {\omega }^2$ and $G_2\propto \omega$; a permanent elastomer does not have that zero-frequency fluid limit.

Interstitial atoms in a metal. A suitable mechanism is the Snoek effect, a form of anelasticity. An interstitial defect distorts its local crystal structure anisotropically. Different site orientations become unequally favorable under an applied stress, so hopping atoms redistribute towards the preferred orientations and supply additional strain. The hopping time is thermally activated, approximately ${\tau }_a={\tau }_0{e}^{E_a/(k_{B}T)}$, with activation energy $E_a$.

At low frequency, $\omega {\tau }_a\ll 1$, the populations nearly equilibrate during each cycle. The defect contribution to strain is nearly in phase with stress, giving a relaxed compliance and little dissipation. At intermediate frequency, $\omega {\tau }_a\sim 1$, redistribution lags behind the stress and energy dissipation is largest. At high frequency, $\omega {\tau }_a\gg 1$, the populations are effectively frozen; the extra relaxation strain cannot develop, and the response approaches the stiffer unrelaxed elastic limit. For weak single-mode relaxation, the loss tangent is approximately proportional to $\omega {\tau }_a/[1+(\omega {\tau }_a{)}^2]$. The metal remains a solid in the low-frequency regime: the extra delayed strain is recoverable, not unrestricted viscous flow.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

Choose $x$ vertically upward. Let $G(x,t)$ be the gas-occupied part of a pipe cross-section, of area $\alpha A$. Define the phase mean mass density, mass-weighted velocity, second moment of velocity, and phase mean pressure byThe void fraction $\alpha$ is a volume fraction, not a mass fraction. The gas momentum in a slice of length $dx$ is $A\alpha {\bar{\rho }}_g\bar{v}dx$; its advected momentum flux is $A\alpha {\bar{\rho }}_g\overline{v^2}$. The two end pressures give the net force $-A{\partial }_x(\alpha {\bar{p}}_g)dx$. Include every other vertical force on the gas, including interfacial forces, in $M_{g}dx$. Newton's second law for this fixed slice, followed by division by $Adx$, therefore givesThis is a cross-section averaged phase momentum balance. In particular, mass weighting makes the momentum and its flux exact averages; one must not replace $\overline{v^2}$ by ${\bar{v}}^2$ without an additional closure. With liquid mean velocity $\bar{u}$, constant liquid mass density ${\rho }_{\ell }$, and liquid second moment $\overline{u^2}$, the same calculation givesWith the common pressure $p={\bar{p}}_g={\bar{p}}_{\ell }$, the specified pressure contributions to $M_g,M_{\ell }$ cancel the phase-fraction derivatives from the product rule. The remaining pressure forces per unit mixture volume are $-\alpha p_x$ and $-(1-\alpha )p_x$.

Assume that the bubbles at a given height have the stated radius $a$. Their number per unit mixture volume is $N=3\alpha /(4\pi a^3)$. The drag force on a rising bubble opposes its relative velocity, so multiplication of the single-bubble law by $N$ gives the signed magnitudeGravity acts downward; the drag force on the liquid is the opposite of that on the gas. With no wall force,A bubble retains its mass in the absence of transfer between phases. Consequently $a^3{\bar{\rho }}_g=a_0^3{\rho }_0$ andFor clarity suppress bars on $p,{\rho }_g,v$ in the steady calculation. On a macroscopic variation length $\ell$, gas inertia per unit volume scales like $\alpha {\rho }_g{v}^2/\ell$, whereas the quadratic drag force scales like $\alpha {\rho }_{\ell }{v}^2/a$. Their ratio is $O[({\rho }_g/{\rho }_{\ell })(a/\ell )]$. Thus ${\rho }_g=O({\rho }_0)\ll {\rho }_{\ell }$, together with the scale separation $a\ll \ell$ inherent in this averaging model, justifies the leading-order neglect of gas inertia. Small reference density alone does not justify it in an arbitrarily short or strongly compressed region. Neglect gas weight at the same order, with the liquid fraction not vanishing. For a stationary liquid, summing the two momentum equations givesThe leading gas balance is $0=-\alpha p_x-D_v$. Substitution of the bubble-radius law yieldsThe gas continuity equation, with constant pipe area and no mass exchange, is ${\partial }_t(\alpha {\rho }_g)+{\partial }_x(\alpha {\rho }_{g}v)=0$. Hence the steady gas mass flux $J=\alpha {\rho }_{g}v$ is constant. For $0<\alpha <1$ the pressure equation selects $v>0$, andEvaluate the latter constant at a reference point ${\rho }_g={\rho }_0$, $\alpha ={\alpha }_0$ and divide. The terminal-drift density law for compressible bubbles isThe exponent $3/5$ comes from combining mass conservation with the density dependence of bubble radius; no gas equation of state was needed.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

Specify the convention before quoting a stability parameter. Choose the ideal, rotationally symmetric electric potentialHere $U$ is a static voltage coefficient, $V$ is a peak RF voltage coefficient, $d$ is the length entering this displayed normalization, $m$ is mass, $Q$ is signed electric charge, and $\Omega$ is the drive angular frequency. The coefficient $V$ is neither a peak-to-peak voltage nor an RMS voltage. An apparatus's ring-to-cap voltage requires matching its electrode surfaces to this electric potential; electrode radius alone does not define our $d$. The minus sign fixes the RF phase at $t=0$.

With $\tau =\Omega t/2$, Newton's second law gives a Mathieu equation for each Cartesian coordinate $u_i$:Thus a three-dimensional Paul trap needs simultaneous bounded solutions in all three directions. Changing the sign of $V$ reverses $q_i$ and shifts the RF phase by half a cycle; it does not change the ideal infinite-time stability regions. Rescaling the voltage or length convention changes the numerical parameters. Floquet theory determines stability of the full Mathieu equation; the small-parameter approximation below does not establish stability for arbitrary $q_i$. Bounded mathematical motion can still hit an electrode if its amplitude exceeds the available space.

A physical law expresses a reproducible relation used to describe or predict physical phenomena within its domain of applicability. Applying it requires a model specifying the system and conditions: Newton's second law, together with a force model, predicts a body's acceleration; conservation of energy constrains exchanges within a specified system. Its predictions must be compared with physical observations.

In a functional model, a driven piston reduces the chamber's volume. A surrounding cylinder carries the wall stress; a deformable packing or gasket closes a potential leak path around a moving or removable closure. A sample and a separate pressure sensor occupy the chamber. A transmitting liquid can load the sample on all sides; the sensor's signal passes to an external readout. This is a division of experimental functions, not a scale reconstruction of one historical machine.

For a slowly moving frictionless drive piston of effective surface area $A$, external pressure $p_{\mathrm{ext}}$ and inward applied force $F$, Newton's second law at mechanical balance givesThus $20\mathrm{kN}$ on a stipulated $20\mathrm{m}{\mathrm{m}}^2$ effective surface area corresponds ideally to $1\mathrm{GPa}$ of differential pressure. These are illustrative numbers, not apparatus specifications. The effective surface area is the one associated with the pressure difference, not necessarily an external ram's visible cross section. With resisting friction $F_f$ during inward motion, the balance becomes $p-p_{\mathrm{ext}}=(F-F_f)/A$. During outward motion the sign of the friction contribution reverses. An applied-load reading can therefore show hysteresis even when the chamber's actual pressure is well defined.

A resistance pressure gauge instead supplies an internal signal. In a generic electrical readout, measure an electric potential difference $V_e$ with a known electric current $I$ and use Ohm's law to infer electrical resistance $R=V_e/I$. This describes a possible readout principle; the lecture passage cited below identifies the sensor material and calibration landmarks without specifying this wiring.

Two semi-infinite strings with common tension $T$ and a point mass $M$ at their join have continuous transverse displacement, $A+B=C$. With time factor $e^{i\omega t}$ and a right-moving incident factor $e^{-i{k}_{1}x}$, Newton's second law at the join gives $k_{1}T(A-B)=(k_{2}T+iM{\omega }^2)C$. Solving gives the displayed reflection coefficient and $t=C/A=2k_{1}T/[T(k_1+k_2)+iM{\omega }^2]$. Their power fractions obey $|r{|}^2+(k_2/k_1)|t{|}^2=1$. Reversing the time convention complex-conjugates the coefficients. The limits $M=0$ and $M\to \infty$ recover a massless join and a fixed boundary, respectively.

For a semi-infinite ideal string on $x\le 0$ with tension $T$ and linear mass density $\mu$, let a free transverse endpoint mass be $m$. Use a time-harmonic wave $y=\mathrm{Re}\{A(e^{ikx}+R{e}^{-ikx})e^{-i\omega t}\}$, where $k=\omega \sqrt{\mu /T}>0$. The endpoint boundary condition from Newton's second law is $m{y}_{tt}=-T{y}_x$. It gives the title displacement reflection coefficient, with characteristic impedance of a stretched string $Z=\sqrt{T\mu }$. Since numerator and denominator have equal moduli, all mean incident power is reflected. The endpoint kinetic energy varies within a cycle, but its mean rate of change is zero. A free mass changes the reflected phase, and does not by itself provide dissipation. The limits $m\to 0$ and $m\to \infty$ give $R=+1$ and $R=-1$ respectively at fixed positive angular frequency.

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.