A mass m is connected to a spring of stiffness k and a damper of coefficient c, both attached to a fixed wall, and is driven by an applied force F(t). The coordinate x is measured from the equilibrium position, with positive x defined to the right. All forces are written relative to this convention.
In the workspace, x is declared as the coordinate variable and m, c, k as constants, matching the symbols already on the page.
Isolate the mass and replace the spring, damper, and applied load with the forces they exert on it. The spring force opposes displacement and is written -kx. The damper force opposes velocity and is written -c ẋ. The applied force F(t) is already aligned with the positive direction and requires no sign change.
(1) spring: -kx. (2) damper: -c ẋ. (3) applied force: +F(t). These three signed terms are exactly what gets entered under Force Accounting.
Declare x as a coordinate variable and m, c, k as constants in the workspace. Set the inertia to m. In Force Accounting, add the three terms above with their signs: k·x negative, c·x' negative, F positive.
Summing forces and setting the result equal to ma gives m ẍ + c ẋ + kx = F(t), the equation of motion for the system. This is the same result obtained by applying ΣF = ma directly to the free body diagram.
Taking the Laplace transform of the equation of motion under zero initial conditions gives (ms² + cs + k) X(s) = F(s). Solving for the ratio of output to input yields the transfer function X(s)/F(s) = 1 / (ms² + cs + k).
By hand, this step is a table lookup: ẍ maps to s²X(s), ẋ maps to sX(s), under zero initial conditions. The segmented switch in Results carries out the same substitution and lets you read off Laplace or transfer function without rewriting the equation each time.
This is not a black box that guesses your equation of motion from a picture. It builds the equation the way you would by hand: from a list of forces you supply, each one signed relative to a direction you pick. The tool's job is arithmetic and bookkeeping, not judgment. Deciding what forces act on the mass, and which way they push, is still yours to do.
That division of labor matters. A tool that hides the free body diagram from you teaches you nothing and gives you no way to catch your own mistake. This one keeps the physics in your hands and takes over only once the physics is settled: summing terms, transforming to the Laplace domain, and arranging the result as a transfer function.
A mechanical system with one coordinate has one equation of motion. For a mass on a spring and damper, Newton's second law states that the sum of the forces acting on the mass equals its mass times acceleration. Collecting the inertia term, the damping term, and the stiffness term on the left and the applied forcing on the right gives the standard second order form.
m ẍ + c ẋ + k x = F(t)
Here m is the inertia, c the viscous damping coefficient, k the stiffness, and F(t) the applied force. The overdots denote derivatives with respect to time, so ẋ is velocity and ẍ is acceleration. Any single degree of freedom mechanical system reduces to this same shape once the forces are summed, which is why the classification below depends only on the three coefficients and not on the particular springs, dampers, or geometry that produced them.
A restoring force such as a spring opposes the coordinate, so kx enters the force balance with a negative sign when the positive direction points along increasing x. Damping opposes velocity, so c ẋ is also negative. The applied force keeps whatever sign it has relative to the chosen positive direction. Moving the coordinate dependent terms to the left of the equals sign flips their signs, which is how the familiar all positive left hand side arises. Getting a sign wrong does not produce an error message from an equation solver. It produces a physically wrong model that looks correct, which is why the tool asks you to state each sign explicitly rather than inferring it.
The Laplace transform turns the differential equation into an algebraic one. Under zero initial conditions each time derivative becomes a factor of s: acceleration ẍ maps to s² X(s), velocity ẋ maps to s X(s), and position x maps to X(s). Applying this to the equation of motion gives the transformed relation.
(m s² + c s + k) X(s) = F(s)
The polynomial multiplying X(s) is the characteristic polynomial of the system. Solving for the ratio of the transformed output to the transformed input gives the transfer function, which describes how the system responds to any input without re solving the differential equation each time.
X(s) / F(s) = 1 / (m s² + c s + k)
When the three coefficients are numeric, the second order system has two standard descriptors. The undamped natural frequency and the damping ratio follow directly from the coefficients.
ωn = √(k / m) ζ = c / (2 √(k m))
The natural frequency sets how fast the system oscillates in the absence of damping, and the damping ratio sets whether and how quickly those oscillations decay. The roots of the characteristic polynomial, the poles of the transfer function, determine the response type:
The tool reports these values automatically once every coefficient is a number. With symbolic coefficients it leaves the results in symbolic form, since a damping ratio has no numeric value until m, c, and k do.
Leaving the force list empty models free vibration: the system moves only under its own inertia, stiffness, and damping. In that case the transfer function is replaced by the characteristic equation, whose roots are the poles that govern the natural response. Adding a force term models forced response, where the constant step input F₀ transforms to F₀/s. Both cases run through the same force accounting; the only difference is whether a right hand side term is present.
Many systems apply force to the mass through another moving coordinate: a second mass connected by a spring, or a vibrating base under the suspension. Those force terms are written the same way they appear on paper, as a coefficient times a parenthesized difference, for example k₂(x₂ − x₁). The tool distributes the product, moves the terms in the model coordinate to the left hand side, and keeps the terms in the other coordinate on the right hand side as a known input. For a base excited system this reproduces the standard form m ẍ + c ẋ + kx = c ż + kz, and when that other coordinate is the only input, the transfer function view shows the ratio of the two transformed coordinates. A full multi mass solution still requires one equation of motion per inertia; this tool builds them one at a time.
This tool handles one inertia per equation: the equation of motion for a single mass, with linear stiffness and damping, where any other coordinates appear as prescribed inputs. Solving coupled equation sets simultaneously, nonlinear terms, and schematic entry are not supported. Every term must be linear in the coordinates and their derivatives, and derivatives above second order are rejected, since a second order system is the intended scope.
Enter every force term acting on the mass, tag its sign relative to your chosen positive direction, and build up the equation of motion the same way you would on paper. The tool sums the terms, applies the Laplace transform, forms the transfer function, and reports the natural frequency, damping ratio, and poles when the coefficients are numeric. It runs entirely in the browser with no account required.
The equation of motion is a differential equation in the time domain: it relates the coordinate and its derivatives directly, as in m ẍ + c ẋ + kx = F(t). The transfer function is the same relationship after a Laplace transform, expressed as a ratio of polynomials in s. Both describe the same system; the transfer function form is what you use for block diagram algebra, frequency response, and control design.
A transfer function is defined as an output over an input. If no force term is entered, there is no input, so there is nothing to divide by. The tool falls back to the characteristic equation, ms² + cs + k = 0, whose roots are still the poles that set the system's natural response.
The natural frequency and damping ratio are only numbers once m, c, and k are all numbers. If any constant is left as a symbol, the tool keeps the equation of motion and transfer function in symbolic form and does not attempt to compute a natural frequency or damping ratio from an unknown.
No. The Laplace transform used here assumes zero initial conditions, which is the standard assumption for deriving a transfer function. A nonzero initial condition problem is solved differently, by carrying the initial position and velocity terms through the transform, and is outside the scope of this tool.
Only as a coupled input. A second coordinate connected to the modeled mass, such as a base excitation term, is accepted as a prescribed input on the right hand side of the equation. Solving two or more coupled equations of motion simultaneously is not supported; each inertia needs its own equation of motion, built separately.
For summing forces or moments before setting up an equation of motion, see the vector resultant calculator. For finding the mass moment of inertia of a composite shape, see the centroid and inertia tool.