Key examples — Newtonian mechanics

Where the chapter’s machinery shows up across the bookshelf.

This sub-page collects worked examples that apply the mechanics chapter to specific systems in Sound, Hearing, and Cavitation. Each example points back to the relevant section of the core chapter and forward to the consumer lesson where the result is used.

Example 1: the middle-ear lever

The malleus and incus of the middle ear function as a lever, with the malleus arm slightly longer than the incus arm (ratio 1.3\approx 1.3). Applying the torque-balance law FLLL=FRLRF_L L_L = F_R L_R from the chapter, the force amplification across the ossicular chain is LL/LR1.3L_L/L_R \approx 1.3. Combined with the eardrum-to-stapes area ratio of about 1717, the total pressure-amplification factor from air-side eardrum to oval-window stapes is roughly 2222 — close to the impedance ratio that needs to be matched between air and cochlear fluid.

This is the operative argument in Hearing Ch 3.3.

Example 2: pressure as the rate of momentum delivery

The kinetic-theory derivation of pressure (kinetic theory chapter) is exactly the impulse–momentum theorem applied to a molecular collision. A molecule of mass mm hitting the wall with xx-velocity vxv_x and bouncing back elastically delivers an impulse Δp=2mvx\Delta p = 2 m v_x to the wall. The number of such molecules striking per unit area per unit time is 12nvx\tfrac12 n |v_x|, so the pressure — the time-averaged rate of impulse delivery per unit area — is

p  =  2mvx12nvxf(vx)dvx  =  nmvx2  =  13nmv2.p \;=\; \int 2 m v_x \cdot \tfrac12 n |v_x|\, f(v_x)\, dv_x \;=\; n m \langle v_x^2\rangle \;=\; \tfrac13 n m \langle v^2\rangle.

The factor of two is from the elastic bounce (e=1e = 1); the factor of one-half is from the half of molecules moving toward the wall. This is the same elastic-collision idealisation visualised in the chapter’s collision visualizer — see Sound Ch 1.2.

Example 3: Euler’s equation as Newton on a slab

For a small slab of fluid with area AA and thickness dxdx, the net force from pressure on the two faces is (p(x+dx)p(x))A=Adxp/x-(p(x + dx) - p(x))\cdot A = -A\, dx \cdot \partial p/\partial x. Newton’s second law F=maF = m a with m=ρAdxm = \rho A\, dx gives

ρDuDt  =  px.\rho \frac{Du}{Dt} \;=\; -\frac{\partial p}{\partial x}.

The vector generalisation is Euler’s equation, ρDu/Dt=p\rho\,D\mathbf{u}/Dt = -\nabla p. This is the most direct application of Newton’s second law to a continuum — every step of the derivation is a free-body diagram on a small fluid element, exactly as the chapter advocates. See Sound Ch 4.3 and the fluid-mechanics chapter for the full treatment.

Example 4: the Rayleigh–Plesset equation

The radial dynamics of a spherical bubble are derived by integrating Newton’s second law (Euler’s equation in spherical coordinates) from the bubble wall out to infinity, accounting for the surface-tension pressure at the wall and the polytropic gas inside. The result is

RR¨+32R˙2  =  1ρ ⁣[pBp2σR4μR˙R].R\,\ddot{R} + \tfrac{3}{2}\dot{R}^2 \;=\; \frac{1}{\rho}\!\left[p_B - p_\infty - \frac{2\sigma}{R} - \frac{4\mu \dot{R}}{R}\right].

The left side is the inertia of the radially-flowing liquid (a continuum Newton’s second law in spherical symmetry); the right side is the net of internal gas pressure, ambient pressure, Young–Laplace term, and viscous-stress damping. See Cavitation Ch 3.1.

Example 5: acoustic energy density

In a plane sound wave, the kinetic energy density is 12ρ0v2\tfrac12 \rho_0 v'^2 and the potential energy density (the work done compressing the fluid against its bulk modulus) is p2/(2ρ0c2)p'^2/(2\rho_0 c^2). On time-average they are equal — acoustic equipartition — and their sum is the total energy density that the waves chapter uses to define intensity. The argument is the work–energy theorem applied to an oscillating fluid element: integrate the work done by pressure over the compression cycle and verify it stores in the elastic restoring force on the way in and releases it on the way out. See Sound Ch 5.2.