How to Find the Laplace Transform of a Mass-Spring-Damper System

A mass-spring-damper system models motion when a mass is connected to a spring and a viscous damper. The Laplace transform turns its differential equation into an algebraic equation, making it easier to find the displacement caused by a specified force.

The key is to keep the initial position and velocity when transforming derivatives. They disappear only when you explicitly assume the system starts at rest.

The equation of motion

Let $x(t)$ be the mass’s displacement from equilibrium, $F(t)$ the applied force, $m$ the mass, $c$ the viscous damping coefficient, and $k$ the spring constant. Newton’s second law gives

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

The spring force opposes displacement, the damper force opposes velocity, and the inertial term is mass times acceleration. This sign convention treats positive $F(t)$ as acting in the positive direction of $x(t)$.

Use the unilateral Laplace transform, which transforms functions for $t\geq 0$. Write $X(s)=\mathcal{L}\{x(t)\}$ and $F(s)=\mathcal{L}\{F(t)\}$. The derivative rules are

$$\mathcal{L}\{x'(t)\}=sX(s)-x(0), \qquad \mathcal{L}\{x''(t)\}=s^2X(s)-s x(0)-x'(0).$$

Transform and solve

Set $x_0=x(0)$ and $v_0=x'(0)$. Transforming each term in the motion equation gives

$$m\bigl(s^2X-sx_0-v_0\bigr)+c\bigl(sX-x_0\bigr)+kX=F(s).$$

Collect the terms containing $X(s)$, then divide by their coefficient:

$$\boxed{X(s)=\frac{F(s)+(ms+c)x_0+m v_0}{ms^2+cs+k}}.$$

This is the displacement transform for arbitrary initial position and velocity. If the mass starts at rest, so $x_0=v_0=0$, it reduces to

$$X(s)=\frac{F(s)}{ms^2+cs+k}.$$

The force-to-displacement transfer function is defined with zero initial conditions:

$$H(s)=\frac{X(s)}{F(s)}=\frac{1}{ms^2+cs+k}.$$

A transfer function describes the system’s response relationship, not the response to one particular input. To find a particular motion, multiply $H(s)$ by the chosen force transform and take the inverse Laplace transform.

Worked example: a constant force

Suppose $m=2\,\mathrm{kg}$, $c=6\,\mathrm{N\,s/m}$, and $k=8\,\mathrm{N/m}$. The system starts at rest and a constant $10\,\mathrm{N}$ force is switched on at $t=0$. A force that stays constant for $t\geq0$ has transform $F(s)=10/s$. Therefore,

$$X(s)=\frac{10}{s(2s^2+6s+8)} =\frac{5}{s(s^2+3s+4)}.$$

To invert it, complete the square in the quadratic: $s^2+3s+4=(s+3/2)^2+7/4$. Partial fractions give

$$X(s)=\frac{5}{4s}-\frac{5}{4}\frac{s+3/2}{(s+3/2)^2+7/4}-\frac{15}{8}\frac{1}{(s+3/2)^2+7/4}.$$

Using the standard inverse transforms for shifted sine and cosine terms, the displacement is

$$x(t)=\frac{5}{4}-\frac{5}{4}e^{-3t/2}\cos\!\left(\frac{\sqrt{7}}{2}t\right)-\frac{15}{4\sqrt{7}}e^{-3t/2}\sin\!\left(\frac{\sqrt{7}}{2}t\right), \qquad t\geq0.$$

The long-term displacement is $5/4=1.25\,\mathrm{m}$, which agrees with the static result $F/k=10/8$. The exponential terms describe the transient oscillation as the system approaches that equilibrium.

Common mistakes

• Dropping initial conditions without justification. The derivative transforms contain $x(0)$ and $x'(0)$. Omit them only when the system starts at rest, or when calculating the zero-initial-condition transfer function.
• Using the wrong force transform. A constant force switched on at zero is a step input, $F(s)=F_0/s$. A brief impulse of strength $J$ has transform $J$; these produce different responses.
• Confusing a transfer function with an output. The transfer function is $X(s)/F(s)$ under zero initial conditions. For a specified force, calculate $X(s)=H(s)F(s)$ before taking the inverse transform.
• Mixing units or signs. In SI units, $m$ is in kilograms, $c$ in newton-seconds per metre, and $k$ in newtons per metre. A different coordinate direction changes force signs, so use one convention consistently.

A useful form of the denominator

The denominator can also be written in terms of the undamped natural angular frequency and damping ratio:

$$\omega_n=\sqrt{\frac{k}{m}}, \qquad \zeta=\frac{c}{2\sqrt{mk}}, \qquad ms^2+cs+k=m\left(s^2+2\zeta\omega_n s+\omega_n^2\right).$$

These parameters help classify the response: $\zeta<1$ is underdamped, $\zeta=1$ is critically damped, and $\zeta>1$ is overdamped. They do not change the Laplace-transform method; they provide a convenient way to interpret the roots of the denominator.