How Do You Create a Second-Order Equation from a Spring Constant?

A spring constant tells you how strongly a spring pulls or pushes when stretched or compressed. To turn that information into a second-order differential equation, combine Hooke’s law with Newton’s second law. The equation describes how the mass’s displacement changes over time.

Start with Hooke’s law

For a spring that obeys Hooke’s law, the force exerted by the spring is proportional to its displacement from equilibrium and points back toward equilibrium:

$$F_{\text{spring}}=-kx.$$

Here, $k$ is the spring constant, measured in newtons per meter (N/m), and $x$ is displacement in meters. The minus sign matters: if the mass is displaced in the positive direction, the spring force points in the negative direction.

Newton’s second law says that the net force on a mass $m$ is $m$ times its acceleration. If $x(t)$ is the displacement at time $t$, then acceleration is $x''(t)$, so for an ideal spring with no other forces,

$$m x''=-kx.$$

Rearranging gives the standard equation for undamped free motion:

$$m x''+kx=0,$$

or, after dividing by $m$,

$$x''+\frac{k}{m}x=0.$$

This is a second-order differential equation because it contains the second derivative of displacement.

Include damping or an external force

Real systems may have resistance from air, friction, or a shock absorber. A common model assumes that the damping force is proportional to velocity and opposite its direction: $-c x'$, where $c$ is the damping coefficient. If an external force $F(t)$ also acts on the mass, Newton’s law becomes

$$m x''+c x'+kx=F(t).$$

This is the general linear mass–spring–damper equation. Set $c=0$ if there is no damping, and set $F(t)=0$ if no external force is applied. For a spring with no damping or external input, it reduces to $m x''+kx=0$.

Worked example: find the equation and motion

Suppose a $0.50$ kg mass is attached to a spring with spring constant $8$ N/m. Let $x=0$ represent the equilibrium position, and choose the positive direction to be the initial direction of displacement. Ignore damping and external forces.

The spring’s force is $-8x$. Applying Newton’s law gives

$$0.50x''=-8x,$$

so

$$x''+16x=0.$$

The equation is second order. Its angular frequency is

$$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{8}{0.50}}=4\text{ rad/s}.$$

If the mass is released from rest $0.10$ m from equilibrium, the initial conditions are $x(0)=0.10$ and $x'(0)=0$. The general solution of $x''+16x=0$ is

$$x(t)=A\cos(4t)+B\sin(4t).$$

Using the initial position gives $A=0.10$. Differentiating gives $x'(t)=-4A\sin(4t)+4B\cos(4t)$, so the zero initial velocity requires $B=0$. Therefore,

$$x(t)=0.10\cos(4t).$$

The mass oscillates about equilibrium with amplitude $0.10$ m and angular frequency $4$ rad/s. The spring constant alone is not enough to determine the motion: you also need the mass, and solving for a particular motion requires initial position and velocity.

What if the spring hangs vertically?

A hanging mass also experiences gravity. If $y$ is measured downward from the spring’s unstretched position, then the force balance is $my''=mg-ky$, or $my''+ky=mg$. The mass’s static equilibrium extension is $y_{\mathrm{eq}}=mg/k$. Define displacement from that equilibrium by $x=y-y_{\mathrm{eq}}$. Substitution removes the constant gravity term and again gives $mx''+kx=0$. Thus, measuring displacement from equilibrium is usually the simplest way to write the equation.

Common mistakes

• Dropping the restoring-force sign: The spring force points toward equilibrium, so it is $-kx$ when positive displacement is defined away from equilibrium.
• Using the spring constant by itself as the frequency: The coefficient in the normalized equation is $k/m$, and the angular frequency is $\sqrt{k/m}$.
• Confusing position with displacement: For vertical motion, the unstretched length and equilibrium position differ. State which reference point defines $x=0$.
• Leaving out forces in the model: Add gravity if measuring from the unstretched position, damping if resistance is modeled, and any applied force on the right-hand side.
• Assuming the equation determines one unique motion: A second-order equation needs two initial conditions, commonly the starting displacement and velocity.