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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.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

A point mass constrained to a fixed-length massless arm under uniform gravitational acceleration is a simple pendulum. With no drag its angular equation is $\ddot{\theta }+(g/\ell )\sin \theta =0$. The small-angle approximation yields simple harmonic motion, whereas the full restoring force is nonlinear. Adding drag gives a damped pendulum.

For an ideal point mass $m$ swinging in a vertical plane, the torque about the pivot is $-mg\ell \sin \theta$. Dividing by its moment of inertia $m{\ell }^2$ gives the displayed equation. Both the inertia and the gravitational torque scale with $m$, explaining why the point mass cancels. This derivation and the small-amplitude period are given in Ling, Sanny and Moebs, University Physics, pendulums.

Released from rest at a small angular amplitude ${\theta }_0$, the linearized solution is $\theta (t)\simeq {\theta }_0\cos (\sqrt{g/\ell }t)$, with oscillation period $T_0=2\pi \sqrt{\ell /g}$. For $g=9.81m/{\mathrm{s}}^2$, lengths one and four metres give $T_0\approx 2.0061\mathrm{s}$ and $4.0121\mathrm{s}$. This same scaling follows from dimensional analysis; the dynamical calculation supplies the coefficient $2\pi$.

Finite angular amplitude changes the exact ideal period. Conservation of energy gives ${\dot{\theta }}^2=(2g/\ell )(\cos \theta -\cos {\theta }_0)$ for release from rest. Integrating a quarter oscillation and substituting $\sin (\theta /2)=\sin ({\theta }_0/2)\sin \varphi$ givesHere $K$ is the complete elliptic integral of the first kind. The small-angle approximation gives a ratio tending to one; at $10^{\circ },60^{\circ },90^{\circ }$, the ratios are approximately $1.00191,1.07318,1.18034$. Thus almost constant period at small amplitude has a quantitative limit. As ${\theta }_0\to \pi$, the period diverges, connecting to the pendulum separatrix approach takes infinite time.

These predictions assume the specified ideal system. Drag changes the mechanical energy; a massive support changes the moment of inertia. The distinction is useful in experiments: a change in period can test an approximation rather than merely indicate an arithmetic mistake.

For $a=a_0(1+ϵ)$, equilibrium outer radius $b_0$, and a bubble gas obeying $p_0\pi a^2=\text{constant}$, the linearization isThe bubble therefore undergoes simple harmonic motion with

The spring constant is the positive proportionality coefficient in the ideal spring restoring force $F=-kx$. Its SI unit is newtons per metre. With a constant attached mass $m$, it sets the angular frequency $\omega =\sqrt{k/m}$ of simple harmonic motion. Doubling $k$ doubles the restoring force at a fixed displacement, but increases that angular frequency only by a factor $\sqrt{2}$.

The torsion constant, or torsional spring constant, is the positive restoring torque per unit twist angle of a complete suspension in its linear elasticity range:Its SI unit is $\mathrm{N}m/rad$. The minus sign describes restoration toward the untwisted orientation. The derivative specifies a local stiffness if the response is not globally linear; using one constant over a finite range additionally requires a straight calibration curve. Return to the same zero after unloading tests recoverable elastic deformation; repeat loading in both directions to check whether the response depends on its previous loading. Staying below permanent deformation is necessary, but does not alone prove proportionality between torque and twist.

For an ideal uniform circular wire of length $\ell$, radius $r$, and shear modulus $G_s$, the filament torsional stiffness is $C=G_s{J}_p$, with polar second moment of area $J_p=\pi r^4/2$. Uniform twist gives $\theta /\ell$ as the twist per unit length, soHere $U$ is stored elastic energy. Doubling the wire's radius multiplies the ideal torsion constant by $16$; doubling its length halves it. This explains why a long thin fiber is useful for weak torque measurement. The formula assumes linear elasticity with uniform material and circular section; it does not calibrate arbitrary ribbons, clamps, or a compound suspension. Within this ideal model, the equipartition theorem predicts an equilibrium thermal mean-square angular displacement $⟨{\theta }^2⟩=k_B\mathcal{T}/\kappa$, where $k_B$ is the Boltzmann constant and $\mathcal{T}$ the uniform absolute temperature. A softer suspension magnifies both the signal and its thermal fluctuations.

Static calibration supplies known torques without assuming the unknown interaction being measured. Apply a known perpendicular force $F$ at lever arm $a$, record the change from the unloaded twist, and useSeveral positive and negative settings test proportionality; fit the slope of applied torque against twist and retain a possible intercept. A nonzero intercept must not be silently absorbed into the slope. Convert degrees to radians before reporting $\mathrm{N}m/rad$. PASCO's procedure turns the Coulomb balance on its side and adds small known masses to generate a gravitational weight at the sphere's lever arm, returning the rod to its index mark by turning the dial. Its tabulated coefficient is force per degree: multiply by the lever arm and $180/\pi$ to obtain our torsion constant. ^[ref]

The figure's ideal static calibration uses $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$: an applied torque of $2.0\times 10^{-10}\mathrm{N}\mathrm{m}$ gives $0.020\mathrm{rad}$. Both signs lie on the same straight line. The plotted points are calculated examples, not experimental observations; they cannot demonstrate a real fiber's linearity.

Dynamic calibration instead measures the oscillation period $T_0$ with known total suspended moment of inertia $I$. For negligible damping, small free oscillations about a fixed reference obey simple harmonic motion:Include the rod, mirrors, and any attachments in $I$ unless their contributions are demonstrably negligible. Keep the applied torque absent or independent of angle. If an external torque has derivative $g_{\tau }=d{\tau }_{\mathrm{ext}}/d\theta$ at equilibrium, linearization gives effective stiffness $\kappa -g_{\tau }$: timing alone then measures that combination, and stability requires $\kappa -g_{\tau }>0$. A constant applied torque merely shifts equilibrium. Non-negligible damping changes the measured period; damped torsional oscillation gives a correction under a specified damping model. Agreement between static and dynamic calibrations is a useful check on the assumed moment of inertia, reference stability, and restoring law, rather than a reason to ignore their assumptions.

Two fixed outer wires at $x=\pm d$ attract a middle wire carrying the same electric current $I$. If the middle wire is displaced in the perpendicular direction $y$, the net magnetic force per unit length is $-{\mu }_0{I}^{2}y/[\pi (d^2+y^2)]$. For $|y|\ll d$, the wire's linear mass density ${\mu }_m$ gives the displayed simple harmonic motion frequency. Motion towards either outer wire in the original plane is unstable. Gravity must be absent from or balanced in the oscillation direction.

Trigonometric functions turn an angle into a coordinate or a ratio of lengths. Start at $(1,0)$ on the unit circle and rotate counterclockwise through $\theta$ radians: the resulting point isThe cosine records horizontal position and the sine records vertical position. This definition continues around the whole circle, including negative angles, so the signs encode direction rather than requiring a separate rule for each right triangle.

The unit circle picture explains why these functions are useful beyond measuring triangles. The two coordinates of a point rotating at constant angular speed are periodic functions. Scaling the radius to $A$ and using $\theta =\omega t+\varphi$ gives $x=A\cos (\omega t+\varphi )$ and $y=A\sin (\omega t+\varphi )$. A circular motion becomes two simple repeating traces, connecting Euclidean geometry, simple harmonic motion, and Fourier series.

For an acute angle in a right triangle, rescaling its hypotenuse to one recovers the familiar ratios: sine is opposite side divided by hypotenuse, and cosine is adjacent side divided by hypotenuse. The tangent function is their ratio, $\tan \theta =\sin \theta /\cos \theta$, wherever the denominator is nonzero; the secant function, cosecant, and cotangent give the other reciprocal ratios.

The same names extend to complex numbers through the complex exponential:These definitions agree with the circle interpretation for real angles. For complex arguments, the values need not lie between $-1$ and $1$; the literal circle-coordinate picture is restricted to real arguments. ^[ref]

A vertical cylinder of constant cross-sectional area $A$, partly immersed to depth $h_0$ in a fluid of mass density ${\rho }_f$, satisfies $mg={\rho }_{f}gA{h}_0$ by Archimedes' principle. A small extra downward displacement $z$ increases the buoyant force by ${\rho }_{f}gAz$, so $m\ddot{z}=-{\rho }_{f}gAz$. This is simple harmonic motion with the displayed angular frequency and oscillation period $2\pi \sqrt{h_0/g}$. The ideal model neglects fluid added inertia, drag and surface effects; it requires partial immersion throughout the motion.

A test star inside an infinite planar slab of uniform mass density $\rho$ experiences acceleration $\ddot{z}=-4\pi G\rho z$. Reflection symmetry gives zero vertical gravitational field at the mid-plane, and integrating the gravitational Poisson equation gives the field linear in height. The interior orbit is therefore simple harmonic motion; release from rest at any height within the slab gives its first mid-plane crossing after $\pi /(2\sqrt{4\pi G\rho })$. Finite radial extent, a varying vertical mass density or leaving the slab changes this result.