Simple Harmonic Motion from a Linear Restoring Force

Derivation

Simple Harmonic Motion from a Linear Restoring Force

D-057 Home PU-103 Threads force · energy Depends on newtons-second-law, taylor-expansion-of-potential

Statement

For a particle of mass \(m\) confined near a stable equilibrium of a smooth one-dimensional potential \(U(x)\), small displacements \(x\) from the minimum obey \(m\ddot{x} = -kx\) with \(k \equiv U''(x_0) > 0\), and the motion is sinusoidal, \(x(t) = A\cos(\omega t + \varphi)\), with a single angular frequency \(\omega = \sqrt{k/m}\) that is independent of amplitude.

Why it matters

The simple harmonic oscillator is the universal low-energy behaviour of matter. Almost every stable system — a molecular bond, a pendulum, an atom in a crystal lattice, an \(LC\) circuit, a field mode in the vacuum — looks harmonic when disturbed gently, because any analytic potential with a genuine minimum is locally parabolic. Mastering this one result therefore buys you the leading-order dynamics of an enormous class of physical problems.

It is also the entry point to deeper theory: normal modes, the equipartition theorem, and the quantum harmonic oscillator (with its equally spaced ladder \(E_n = \hbar\omega\left(n+\tfrac12\right)\)) all rest on the linear restoring force established here. The amplitude-independence of \(\omega\) — isochronism — is precisely the property that makes the result so powerful and so widely exploited in clocks and resonators.

Assumptions

The potential has a stable equilibrium at \(x_0\), i.e. \(U'(x_0)=0\) and \(U''(x_0)>0\).If \(U''(x_0)=0\) the leading restoring term is cubic or higher and the motion is anharmonic with amplitude-dependent period; if \(U''(x_0)<0\) the equilibrium is unstable and displacements grow exponentially rather than oscillate.

The displacement is small enough that the potential is well approximated by its quadratic Taylor term.If dropped, higher-order terms \(\tfrac16 U'''x^3+\cdots\) contribute; the frequency acquires an amplitude dependence and the orbit is no longer a pure sinusoid.

The potential is smooth (at least twice differentiable) at the minimum.If \(U\) has a cusp or is only piecewise linear (e.g. a V-shaped well), \(U''\) does not exist at \(x_0\); the restoring force is not linear and the period depends on amplitude even for the smallest oscillations.

Mass is constant and the motion is conservative (no dissipation, no driving).If dropped, \(\ddot x\) gains a damping term \(-\gamma\dot x\) or a forcing term; the free undamped sinusoid is replaced by decaying or driven solutions and \(\omega\) shifts to \(\sqrt{\omega_0^2-\gamma^2/4}\).

Derivation

\[ U(x) = U(x_0) + U'(x_0)\,(x-x_0) + \tfrac12 U''(x_0)\,(x-x_0)^2 + O\!\left((x-x_0)^3\right) \]

Taylor-expand the smooth potential about the equilibrium point \(x_0\) (prior result: taylor-expansion-of-potential). A

\[ U'(x_0) = 0, \qquad k \equiv U''(x_0) > 0 \]

Equilibrium means the potential is stationary; stability means the curvature is positive. Rename the curvature \(k\) (the effective spring constant). A

\[ q \equiv x - x_0, \qquad U(q) \approx U(x_0) + \tfrac12 k\,q^2 \]

Shift the origin to the minimum and discard the constant \(U(x_0)\) (dynamically irrelevant) and the vanished linear term, keeping only the quadratic term valid for small \(q\). A

\[ F(q) = -\frac{dU}{dq} = -k\,q \]

The conservative force is minus the potential gradient; differentiating the quadratic gives a force linear in displacement and directed back toward equilibrium. A

\[ m\,\ddot{q} = -k\,q \]

Insert the linear restoring force into Newton's second law \(m\ddot q = F\) (prior result: newtons-second-law); note \(\ddot q=\ddot x\) since \(x_0\) is constant. A

\[ \ddot{q} + \frac{k}{m}\,q = 0 \qquad\Longrightarrow\qquad \ddot{q} + \omega^2 q = 0,\quad \omega^2 \equiv \frac{k}{m} \]

Divide by \(m>0\) and define \(\omega^2=k/m\); this is the canonical linear, homogeneous, second-order ODE with constant coefficients. B

\[ \lambda^2 + \omega^2 = 0 \;\Rightarrow\; \lambda = \pm i\omega, \qquad q(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t} \]

The trial solution \(q\propto e^{\lambda t}\) gives a characteristic equation with purely imaginary roots, so the general solution is a superposition of the two oscillatory modes. C

\[ x(t) = x_0 + A\cos(\omega t + \varphi) \]

Requiring \(q(t)\) real and combining the two modes gives a single real sinusoid; \(A\ge0\) (amplitude) and \(\varphi\) (phase) are the two integration constants fixed by initial conditions. B

Result

\[ m\ddot{x} = -k x, \qquad x(t)=x_0+A\cos(\omega t+\varphi), \qquad \boxed{\;\omega=\sqrt{\dfrac{k}{m}}\;},\quad T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}} \]

Reading. Near any stable minimum the force is a spring: it pulls back in proportion to displacement. The response is a pure cosine whose angular frequency is set entirely by the ratio of the potential's curvature \(k\) to the inertia \(m\). Crucially, \(\omega\) contains neither \(A\) nor \(\varphi\) — the oscillation is isochronous, ticking at the same rate whether the swing is large or vanishingly small (within the small-displacement regime). Stiffer wells and lighter masses oscillate faster.

Units check. \([k]=[U'']=\mathrm{J\,m^{-2}}=\mathrm{N\,m^{-1}}=\mathrm{kg\,s^{-2}}\). Then \(\sqrt{k/m}=\sqrt{\mathrm{kg\,s^{-2}}/\mathrm{kg}}=\sqrt{\mathrm{s^{-2}}}=\mathrm{s^{-1}}=\mathrm{rad\,s^{-1}}\), the correct dimension for an angular frequency; hence \(T=2\pi/\omega\) has units of seconds.

Limiting cases

\(k\to 0^+\) (flattening well): \(\omega\to 0\), \(T\to\infty\) — the restoring force vanishes and the particle drifts freely; the harmonic approximation degenerates.
\(m\to\infty\) (heavy mass): \(\omega\to 0\) — inertia dominates and the oscillation slows without bound.
\(m\to 0\): \(\omega\to\infty\) — an idealised massless spring responds instantaneously; physically capped by neglected effects.
\(A\to 0\) (infinitesimal swing): \(\omega\) is unchanged — isochronism holds exactly in the linear regime, which is why the pendulum keeps time.
Springs combined: parallel curvatures add, \(k_\mathrm{par}=k_1+k_2\) (stiffer, faster); in series \(k_\mathrm{ser}^{-1}=k_1^{-1}+k_2^{-1}\) (softer, slower), rescaling \(\omega\) accordingly.

Breaks when

Large amplitude (anharmonic regime). Once the cubic/quartic Taylor terms are non-negligible, the period becomes amplitude-dependent. For the pendulum, \(U=-mgL\cos\theta\) gives \(T=T_0\left(1+\tfrac{1}{16}\theta_0^2+\cdots\right)\) — the small-angle result fails at large swing.
Inflection or vanishing curvature, \(U''(x_0)=0\). The leading restoring term is then \(\propto x^3\) (or higher); \(\omega=\sqrt{k/m}\) is undefined and the true period scales with amplitude, e.g. \(T\propto A^{-1}\) for a pure quartic well.
Non-smooth potentials. A V-shaped (triangular) well has no second derivative at the vertex; motion is oscillatory but the period depends on energy and the linear formula does not apply.
Strong damping or driving. With dissipation \(\gamma\), free oscillation occurs only if \(\gamma<2\omega_0\); above critical damping the motion no longer oscillates at all, and a driven oscillator locks to the drive frequency, not \(\sqrt{k/m}\).

Failure modes

Confusing \(\omega\) with \(f\). Writing \(T=1/\omega\) instead of \(T=2\pi/\omega\); \(\omega=2\pi f\) is the angular frequency in rad/s, not the cyclic frequency in Hz.
Retaining the linear Taylor term. Forgetting that \(U'(x_0)=0\) at equilibrium and carrying a spurious constant force, which shifts the equilibrium rather than describing oscillation about it.
Using \(k\) from a formula instead of \(U''\). For a non-spring system (pendulum, molecular bond) students plug in a literal spring constant; the correct \(k\) is the curvature \(U''(x_0)\) evaluated at the minimum.
Sign error in \(F=-dU/dx\). Dropping the minus sign yields a repulsive force and predicts runaway growth instead of oscillation.
Assuming amplitude affects frequency. Expecting a bigger push to give a faster oscillation — true for real (anharmonic) systems but false in the linear regime, the whole point of isochronism.
Pendulum length/mass confusion. Writing \(\omega=\sqrt{g/m}\) or including the bob mass; for the simple pendulum \(\omega=\sqrt{g/L}\) and the mass cancels.

Discussion

The deep content of this derivation is not the spring but the Taylor expansion. Stability forces \(U'(x_0)=0\), and a genuine minimum forces \(U''(x_0)>0\); the first surviving term is therefore always quadratic. Since a quadratic potential produces a force linear in displacement, and a linear restoring force produces sinusoidal motion, every smooth stable equilibrium is a harmonic oscillator at low energy. The specific system — atoms in a lattice, a torsion balance, a superconducting circuit — enters only through the two numbers \(k\) and \(m\).

Isochronism is the physically startling consequence. Because \(\omega\) is independent of \(A\), the restoring force scales up in exact proportion to displacement, so a larger orbit covers proportionally more distance at proportionally higher speed and the two effects cancel in the timing. This is what Galileo noticed watching a swinging lamp, and it underlies every pendulum clock and quartz resonator. It is exact only in the linear regime; anharmonic corrections are what ultimately limit the precision of real oscillators.

The result threads directly into energy language: with \(U=\tfrac12 kq^2\) and \(K=\tfrac12 m\dot q^2\), the total energy \(E=\tfrac12 kA^2\) is conserved and continuously traded between kinetic and potential forms twice per cycle. In phase space \((q,\dot q)\) the trajectory is an ellipse of fixed area \(\propto E/\omega\) — the adiabatic invariant that survives into the quantum theory as the quantised action \(\oint p\,dq = \left(n+\tfrac12\right)h\).

At the deepest level the harmonic oscillator is the bridge to quantum field theory. Quantising \(H=\frac{p^2}{2m}+\tfrac12 m\omega^2 q^2\) gives the equally spaced ladder \(E_n=\hbar\omega\left(n+\tfrac12\right)\) and the operators \(a,a^\dagger\); reinterpreting each normal mode of a field as an independent oscillator turns those ladder rungs into particle number. The photon itself is a quantum of a harmonic field mode. Thus the same \(\omega=\sqrt{k/m}\) that times a pendulum, read as \(\omega=\sqrt{U''/m}\) for a field's stiffness, ultimately sets the energy of a light quantum — a striking reach for a result built from one Taylor expansion.

Common misconceptions. SHM is not "any back-and-forth motion" — a bouncing ball and a large-angle pendulum oscillate but are not simple harmonic. The defining feature is a restoring force strictly proportional to displacement (equivalently, a strictly parabolic potential), which alone guarantees a single amplitude-independent frequency. And \(\omega=\sqrt{k/m}\) is a property of the system, fixed by curvature and inertia, not of how hard you push it.

Worked examples

A block of mass \(m=0.50\ \mathrm{kg}\) on a frictionless surface is attached to a spring of constant \(k=200\ \mathrm{N\,m^{-1}}\), pulled \(A=0.04\ \mathrm{m}\) from equilibrium, and released from rest. Find \(\omega\), \(T\), the maximum speed, and the total energy.

Set up: linear restoring force, so \(\omega=\sqrt{k/m}\). A

\[ \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.50}}=\sqrt{400}=20\ \mathrm{rad\,s^{-1}} \]

Substitute numbers into the boxed result. A

\[ T=\frac{2\pi}{\omega}=\frac{2\pi}{20}\approx 0.314\ \mathrm{s} \]

Period from \(T=2\pi/\omega\). A

\[ v_\mathrm{max}=\omega A=20\times0.04=0.80\ \mathrm{m\,s^{-1}},\qquad E=\tfrac12 kA^2=\tfrac12(200)(0.04)^2=0.16\ \mathrm{J} \]

Released from rest at \(x=A\), so all energy is potential initially; \(v_\mathrm{max}\) occurs at equilibrium where \(\tfrac12 m v_\mathrm{max}^2=E\). B

\[ \omega=20\ \mathrm{rad\,s^{-1}},\quad T\approx0.31\ \mathrm{s},\quad v_\mathrm{max}=0.80\ \mathrm{m\,s^{-1}},\quad E=0.16\ \mathrm{J} \]

Reading. Consistency check: \(\tfrac12 m v_\mathrm{max}^2=\tfrac12(0.50)(0.80)^2=0.16\ \mathrm{J}=E\), confirming energy conservation between the turning point and equilibrium.

A diatomic bond is modelled near its minimum by \(U(r)=\tfrac12 k(r-r_0)^2\) with effective stiffness \(k=1.86\times10^{3}\ \mathrm{N\,m^{-1}}\) (typical of HCl). The reduced mass is \(\mu=\dfrac{m_\mathrm{H}m_\mathrm{Cl}}{m_\mathrm{H}+m_\mathrm{Cl}}\). Estimate the vibrational angular frequency and the wavenumber \(\tilde\nu=\omega/(2\pi c)\).

A two-body vibration reduces to a one-body oscillator of mass \(\mu\); apply \(\omega=\sqrt{k/\mu}\). B

\[ \mu=\frac{(1.008)(34.97)}{1.008+34.97}\ \mathrm{u}=0.980\ \mathrm{u}=0.980\times1.6605\times10^{-27}\ \mathrm{kg}=1.627\times10^{-27}\ \mathrm{kg} \]

Compute the reduced mass and convert atomic mass units to kg. A

\[ \omega=\sqrt{\frac{k}{\mu}}=\sqrt{\frac{1.86\times10^{3}}{1.627\times10^{-27}}}=\sqrt{1.143\times10^{30}}\approx3.38\times10^{15}\ \mathrm{rad\,s^{-1}} \]

Substitute into the boxed result with \(m\to\mu\). A

\[ \tilde\nu=\frac{\omega}{2\pi c}=\frac{3.38\times10^{15}}{2\pi\,(3.00\times10^{10}\ \mathrm{cm\,s^{-1}})}\approx 1.79\times10^{3}\ \mathrm{cm^{-1}} \]

Convert angular frequency to spectroscopic wavenumber using \(c\) in cm/s. B

\[ \omega\approx3.4\times10^{15}\ \mathrm{rad\,s^{-1}},\qquad \tilde\nu\approx1.8\times10^{3}\ \mathrm{cm^{-1}} \]

Reading. The measured HCl fundamental is \(\approx 2886\ \mathrm{cm^{-1}}\); our harmonic estimate is the right order of magnitude, and the shortfall reflects both the chosen \(k\) and real anharmonicity (the Morse potential) — exactly the "breaks when" regime. The same \(\omega=\sqrt{k/m}\) that timed the block sets an infrared absorption line.

Problems

A \(0.25\ \mathrm{kg}\) mass on a spring oscillates with period \(0.50\ \mathrm{s}\). Find the spring constant \(k\).
Solution
\(T=2\pi\sqrt{m/k}\Rightarrow k=\dfrac{4\pi^2 m}{T^2}=\dfrac{4\pi^2(0.25)}{(0.50)^2}=\dfrac{9.87}{0.25}\approx 39.5\ \mathrm{N\,m^{-1}}\).
A simple pendulum has length \(L=1.0\ \mathrm{m}\). Using the small-angle result \(\omega=\sqrt{g/L}\) with \(g=9.81\ \mathrm{m\,s^{-2}}\), find its period, and explain why the bob mass does not appear.
Solution
\(\omega=\sqrt{9.81/1.0}=3.13\ \mathrm{rad\,s^{-1}}\); \(T=2\pi/\omega\approx 2.01\ \mathrm{s}\). For \(U=mgL(1-\cos\theta)\) in the arc coordinate \(s=L\theta\), the curvature gives an effective restoring stiffness \(k_\mathrm{eff}=mg/L\) per unit arc length while the inertia is \(m\); the mass cancels between the gravitational restoring term and the inertia, so \(\omega=\sqrt{g/L}\) depends only on \(g\) and \(L\).
An atom of mass \(m=4.0\times10^{-26}\ \mathrm{kg}\) sits in a lattice well of curvature \(k=60\ \mathrm{N\,m^{-1}}\). Find \(\omega\) and \(f\).
Solution
\(\omega=\sqrt{k/m}=\sqrt{60/(4.0\times10^{-26})}=\sqrt{1.5\times10^{27}}\approx3.9\times10^{13}\ \mathrm{rad\,s^{-1}}\); \(f=\omega/2\pi\approx6.2\times10^{12}\ \mathrm{Hz}\) (a few THz, typical of lattice vibrations).
A particle moves in \(U(x)=U_0\left[(x/a)^2 - \tfrac12(x/a)^4\right]\). Show it is harmonic for small \(x\), find \(\omega\), and state where the harmonic approximation breaks down.
Solution
\(U'=U_0\left[\dfrac{2x}{a^2}-\dfrac{2x^3}{a^4}\right]\), so \(x=0\) is an equilibrium. \(U''(0)=2U_0/a^2\equiv k>0\), hence \(\omega=\sqrt{2U_0/(ma^2)}\). The quartic term becomes comparable to the quadratic when \((x/a)^2\sim 1\), i.e. \(|x|\sim a\); beyond this the restoring force weakens (the well turns over near its maxima) and the motion is strongly anharmonic.
A block of mass \(m=1.0\ \mathrm{kg}\) on a spring with \(k=100\ \mathrm{N\,m^{-1}}\) starts at \(x(0)=0.10\ \mathrm{m}\) with \(\dot x(0)=-1.0\ \mathrm{m\,s^{-1}}\). Write \(x(t)=A\cos(\omega t+\varphi)\) explicitly (find \(A\), \(\varphi\), \(\omega\)).
Solution
\(\omega=\sqrt{100/1.0}=10\ \mathrm{rad\,s^{-1}}\). Amplitude: \(A=\sqrt{x_0^2+(\dot x_0/\omega)^2}=\sqrt{(0.10)^2+(-1.0/10)^2}=\sqrt{0.01+0.01}=0.141\ \mathrm{m}\). Phase: \(x_0=A\cos\varphi\Rightarrow\cos\varphi=0.10/0.141=0.707\); \(\dot x_0=-A\omega\sin\varphi=-1.0\Rightarrow\sin\varphi=+0.707\), so \(\varphi=+\pi/4\ \mathrm{rad}\). Thus \(x(t)=0.141\cos\!\left(10t+\tfrac{\pi}{4}\right)\ \mathrm{m}\).