Differential equation of SHM

Push a swing, pluck a guitar string, watch a block bob on a spring — very different objects, yet the mathematics governing the small back-and-forth motion is identical in every case. The common ingredient is a restoring force that grows in direct proportion to displacement and always points back toward equilibrium: $F=-kx$ . Feed that single fact into Newton's second law and out falls one compact differential equation, $\ddot x+\omega_0^2x=0$ , whose solution is the cosine wave you already associate with oscillation. This article works through that derivation from scratch — not quoting the result but building it, solving the resulting second-order linear ODE by the auxiliary-equation method, and pinning down what the two constants of integration, amplitude $A$ and phase $\phi$ , actually mean physically. This is the mathematical seed of the whole oscillations block: the damped and driven equations you meet next simply bolt extra terms onto the same left-hand side.

The core insight

Newton's second law applied to a linear restoring force $F=-kx$ produces the differential equation $\ddot x+\omega_0^2x=0$ with $\omega_0=\sqrt{k/m}$ , whose general solution $x(t)=A\cos(\omega_0t+\phi)$ describes oscillation at a frequency fixed purely by the system, never by how far it was displaced.

The core idea

Setting up the equation. Consider a particle of mass $m$ moving along one axis under a Hooke's-law restoring force $F(x)=-kx$ , where $k>0$ and $x$ is displacement from equilibrium. Newton's second law, $F=m\ddot x$ , converts this force law directly into a statement about acceleration:

m\ddot{x} = -kx \quad\Longrightarrow\quad \ddot{x} + \omega_0^2 x = 0, \qquad \omega_0 \equiv \sqrt{\frac{k}{m}} .

where: $m$ is the mass (kg), $x(t)$ is displacement from equilibrium (m), $k$ is the force constant (N/m), and $\omega_0$ is the natural angular frequency (rad/s). This is a linear, second-order, homogeneous, constant-coefficient ordinary differential equation — the defining equation of SHM, and every system that obeys it is, by definition, undergoing simple harmonic motion.

Reading the solution's shape before deriving it. Because the equation is second-order, its general solution needs exactly two independent constants, fixed by two initial conditions — typically the displacement $x(0)$ and the velocity $\dot x(0)$ at the start of the clock. The full solve-out is carried in the derivation block below; the punchline is $x(t)=A\cos(\omega_0t+\phi)$ , a cosine wave of fixed amplitude $A$ and fixed angular frequency $\omega_0$ , merely shifted in time by the phase $\phi$ .

A second realization: the pendulum. The same equation reappears wherever a linear restoring "force" (or torque) exists. For a simple pendulum of length $L$ , Newton's second law along the arc gives $\ddot\theta=-(g/L)\sin\theta$ , which is not linear in $\theta$ and so is not exact SHM. But for small angular amplitude, $\sin\theta\approx\theta$ (radians), collapsing the equation to $\ddot\theta+(g/L)\theta=0$ — precisely the SHM form, with $\omega_0^2=g/L$ and $\theta$ playing the role of $x$ . This confirms the differential equation is a universal template, not a spring-specific accident.

Figure

Displacement versus time for a mass on a spring tracing a cosine curve of amplitude A and period T

Derivation ASolving the SHM differential equation

The goal is to start from Hooke's-law force $F=-kx$ and Newton's second law and end with the fully justified general solution $x(t)=A\cos(\omega_0t+\phi)$ , deriving both the differential equation and its solution through the auxiliary-equation method rather than quoting the cosine answer, so that the origin of the two free constants $A$ and $\phi$ is completely traceable to the two initial conditions of the physical system.

Step 1. Write Newton's second law for the restoring force. With $F(x)=-kx$ and $F=m\ddot x$ , substitution gives $m\ddot x=-kx$ . Dividing through by $m$ and defining $\omega_0^2\equiv k/m$ produces $\ddot x+\omega_0^2x=0$ , a linear, homogeneous, second-order ODE with constant coefficients.

Step 2. Propose a trial solution and find the auxiliary equation. Try $x(t)=e^{\lambda t}$ for some constant $\lambda$ ; then $\ddot x=\lambda^2e^{\lambda t}$ , and substituting into the ODE gives $\lambda^2e^{\lambda t}+\omega_0^2e^{\lambda t}=0$ . Since $e^{\lambda t}\neq0$ , this requires the auxiliary (characteristic) equation $\lambda^2+\omega_0^2=0$ .

Step 3. Solve the auxiliary equation and identify both independent solutions. $\lambda^2=-\omega_0^2$ gives $\lambda=\pm i\omega_0$ — purely imaginary roots, with no real part, because the undamped equation has no term proportional to $\dot x$ . The two independent solutions are therefore $e^{i\omega_0t}$ and $e^{-i\omega_0t}$ , and the general solution of a linear ODE is any combination of them, $x(t)=c_1e^{i\omega_0t}+c_2e^{-i\omega_0t}$ .

Step 4. Convert to real form and name the physical constants. Using Euler's formula and demanding $x(t)$ be real, this combination can always be rewritten as $x(t)=C_1\cos(\omega_0t)+C_2\sin(\omega_0t)$ , or equivalently in amplitude–phase form,

x(t) = A\cos(\omega_0 t + \phi),

where $A$ (amplitude) and $\phi$ (phase constant) are the two constants of integration. Being a second-order equation, exactly two constants are needed, and they are fixed uniquely by the two initial conditions $x(0)$ and $\dot x(0)$ — nothing about $A$ or $\phi$ is left free once the system's starting state is specified. Crucially, $\omega_0=\sqrt{k/m}$ never appears among the constants fixed by initial conditions: it is set entirely by the physical system ( $k$ and $m$ ), so the oscillation frequency is independent of how far the system was initially displaced — the defining signature of true SHM.

Worked example

Problem. A block of mass $m=0.50$ kg is attached to a spring of spring constant $k=200$ N/m and rests on a frictionless horizontal surface. The block is pulled $6.0$ cm from equilibrium and released from rest at $t=0$ . Find (a) the angular frequency, (b) the period, (c) the position $x(t)$ , and (d) the position at $t=0.10$ s.

Step 1. Find $\omega_0$ from the physical constants of the system, exactly as the derivation prescribes: $\omega_0=\sqrt{k/m}=\sqrt{200/0.50}=\sqrt{400}=20\ \text{rad/s}$ . Note this uses only $k$ and $m$ — the release distance of $6.0$ cm has not entered yet, confirming frequency is amplitude-independent.

Step 2. Convert to the period: $T=2\pi/\omega_0=2\pi/20\approx0.314\ \text{s}$ .

Step 3. Apply the two initial conditions to fix $A$ and $\phi$ in $x(t)=A\cos(\omega_0t+\phi)$ . Here $x(0)=0.06$ m and $\dot x(0)=0$ (released from rest). Differentiating, $\dot x(t)=-A\omega_0\sin(\omega_0t+\phi)$ ; setting $\dot x(0)=0$ forces $\sin\phi=0$ , so $\phi=0$ or $\pi$ . Since $x(0)=A\cos\phi=0.06>0$ , the consistent root is $\phi=0$ , giving $A=x(0)=0.06$ m. So

x(t) = 0.06\cos(20t)\ \text{m}.

Step 4. Evaluate at $t=0.10$ s: the argument is $20\times0.10=2.0$ rad, and $\cos(2.0\ \text{rad})\approx-0.416$ , so $x(0.10)\approx0.06\times(-0.416)\approx-0.025\ \text{m}$ .

The block sits at $x\approx-2.5$ cm at $t=0.10$ s — it has already swung past equilibrium onto the other side of its motion.

Competitive stretch

Swap the spring for a simple pendulum of length $L=1.0$ m near Earth's surface ( $g=9.8\ \text{m/s}^2$ ), released from a small angle at rest. By the small-angle result derived above, $\omega_0=\sqrt{g/L}=\sqrt{9.8/1.0}\approx3.13\ \text{rad/s}$ and $T=2\pi/\omega_0\approx2.01\ \text{s}$ — over six times slower than the stiff spring above, purely because $g/L$ is smaller than $k/m$ , with the underlying differential equation and solution method unchanged.

Common traps

Writing $F=-kx$ and jumping straight to $x(t)=A\cos\omega_0t$ without solving the ODE — a complete derivation must pass through the auxiliary equation and identify both independent solutions before combining them, or the origin of $A$ and $\phi$ stays mysterious.
Believing $\omega_0$ depends on amplitude $A$ — for genuine SHM, $\omega_0=\sqrt{k/m}$ is fixed by the system alone, completely independent of how far it is displaced.
Applying the pendulum's small-angle formula $\omega_0=\sqrt{g/L}$ to large swings, where $\sin\theta\approx\theta$ breaks down and the motion is no longer strictly simple harmonic.
Guessing the sign or value of $\phi$ instead of checking it against the actual initial conditions — both roots of $\sin\phi=0$ or $\cos\phi=0$ must be tested against $x(0)$ and the sign of $\dot x(0)$ .

Ask the tutor: if the auxiliary-equation method, the imaginary-roots step, or picking the correct root for $\phi$ from initial conditions feels shaky, ask the AI tutor to walk through another worked derivation or check your own algebra.

Why do we need the auxiliary equation at all — can't we just guess x equals A cos omega0 t and check it works?

Guessing and checking does confirm that $A\cos(\omega_0t)$ satisfies $\ddot x+\omega_0^2x=0$ , but it silently throws away information: it hides where the two constants of integration ( $A$ and $\phi$ , or equivalently a sine term) come from, and it gives no way to be sure you have found the general solution rather than one lucky special case. The auxiliary-equation method proves systematically that $e^{\pm i\omega_0t}$ are the only two independent solutions, so any real solution must be some combination of them — that completeness is exactly what a check-by-substitution cannot give you.

Why are the roots of the auxiliary equation imaginary here but real for something like exponential decay?

The sign of the coefficient in front of $x$ decides this. For SHM, $\ddot x+\omega_0^2x=0$ has $\lambda^2=-\omega_0^2$ , a negative number, whose square roots are imaginary — that is precisely what produces oscillation rather than growth or decay, since $e^{i\omega_0t}$ traces a circle in the complex plane, not a drifting exponential. A decay equation like $\dot x+kx=0$ instead has a first-order structure whose root is real and negative, giving $e^{-kt}$ . The imaginary roots here are the mathematical fingerprint of an undamped restoring force.

Since the pendulum equation is only approximately SHM, why does it still get called simple harmonic motion in this course?

Because for the amplitudes this course cares about — a few degrees at most — the approximation $\sin\theta\approx\theta$ is excellent, and the resulting equation $\ddot\theta+(g/L)\theta=0$ matches the exact SHM template $\ddot x+\omega_0^2x=0$ term for term. The pendulum is therefore SHM in the small-angle limit, not in general; at larger swings the true period grows with amplitude (unlike genuine SHM, where it cannot), which is exactly why this article flags amplitude-independence as a defining test you can use to check whether a system is really undergoing SHM.

Differential equation of SHM

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads