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Natural Frequency Calculator — Spring-Mass & Beam Vibration | f = (1/2π)·√(k/M) | Metric & Imperial

Calculate the natural frequency, angular frequency and period of a spring-mass system or a beam/cantilever carrying a concentrated mass. Uses ω = √(k/M) and f = ω/2π, with beam stiffness k = 48·E·I/L³ (simply supported, central load) or 3·E·I/L³ (cantilever, end load). Supports metric (SI) and American (imperial) units.

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Fill in the form on the left and click "Calculate" to see the natural frequency, angular frequency, and period.

Natural Frequency Calculator

Our natural frequency calculator helps you find the frequency at which objects vibrate in an unperturbed situation. From simple springs to structural elements, we will explain the math and the physics behind this fundamental quantity. Keep reading to learn what the natural frequency is, the formula to calculate the natural frequency of a coil spring, or how to calculate the natural frequency of a beam or cantilever in many different ways! The calculator supports both the metric (SI) and the US customary (Imperial) unit systems.


What is the natural frequency: definition and explanation

The natural frequency is a property of every object — a vibrational frequency at which the object oscillates in the absence of external forces (though we'll see that there are systems where the natural frequency has a definition even when external forces are involved).

The natural frequency of an object is at the origin of the sound emitted by the object itself (which is perceivable by humans only if it falls in the audible range). Simple objects (or well-designed ones!) have a single natural frequency, which causes them to produce a pure tone. Complex objects have multiple natural frequencies, their intensity depending on how the vibration was initially prompted. The resulting sound is a mixture of these tones, which results in what we call "noise".


Natural frequency vs. resonant frequency

Don't mistake the natural frequency for the resonant frequency. The former is the frequency at which a system oscillates in the absence of external forces. The latter is the frequency at which the system vibrates in case of a harmonic excitation (a periodic force) acting on it with a frequency equal to the natural frequency.

The result of a force acting in such a way is, in case the vibration of the system and the force are in phase, a dramatic increase in the amplitude of the oscillations, which may well be at the origin of structural failures such as the one of the Tacoma Narrows bridge.


How do you calculate the natural frequency of a coiled spring?

A coiled spring is the most straightforward system for which you can calculate the natural frequency. In this case, consider a spring-mass system, where a body with mass M is connected to a spring with elastic constant k.

Imagine a setup where you can ignore external forces, such as keeping the spring horizontal and placing the mass on a frictionless surface. In this situation, after an initial perturbation, the system would oscillate with a fixed frequency given by the natural frequency formula for the angular frequency:

ω = √(k / M)

Notice that we computed the angular frequency: it's a matter of conventions. You can easily convert the angular frequency to the "ordinary" frequency f (measured in hertz) by dividing it by 2π:

f = ω / 2π = (1 / 2π) · √(k / M)
  • k — spring (elastic) constant, in N/m (metric) or lbf/in (imperial)
  • M — mass attached to the spring, in kg (metric) or lb (imperial)
  • ω — angular natural frequency, in rad/s
  • f — natural frequency, in hertz (Hz)

The period of the oscillation — the time for a full cycle — is simply the inverse of the frequency:

T = 1 / f

Calculating the natural frequency: engineering approach

In engineering, many vibrating systems are not literal springs, but they behave like one: any elastic element that deflects under a load acts as an equivalent spring with an effective stiffness k. Once you know that stiffness and the vibrating mass M, you compute the natural frequency with the very same formula, ω = √(k / M).

A practical shortcut uses the static deflection δ produced by the weight of the mass. Since k = Mg / δ, the natural frequency becomes:

f = (1 / 2π) · √(g / δ)

The formulas to calculate the natural frequency of a beam

A beam carrying a concentrated mass behaves like a spring whose stiffness depends on the beam's geometry and material. The effective stiffness combines Young's modulus E, the area moment of inertia I of the cross-section, and the length L of the beam. The two most common configurations are:

Simply supported beam, central load:  k = 48 · E · I / L³
Cantilever beam, end load:  k = 3 · E · I / L³

Plug the effective stiffness into f = (1 / 2π) · √(k / M) to obtain the natural frequency. A higher Young's modulus, a stiffer cross-section (larger I), or a shorter beam all raise the natural frequency; adding mass lowers it.

  • E — Young's modulus, in GPa (metric) or Mpsi/ksi (imperial)
  • I — area moment of inertia, in mm⁴ (metric) or in⁴ (imperial)
  • L — beam length, in m (metric) or ft/in (imperial)

How to use our natural frequency calculator

  1. Choose your unit system — metric (SI) or imperial (US).
  2. Select the system type: a spring-mass system or a beam / cantilever.
  3. For a spring, enter the spring constant k. For a beam, choose the support type and enter Young's modulus E, the moment of inertia I, and the length L.
  4. Enter the vibrating mass M.
  5. Read off the natural frequency f (in Hz and rpm), the angular frequency ω, the period T, and the effective stiffness.

How to calculate the natural frequency of a beam: supported beam with central load

Consider a simply supported steel beam of length L = 1 m, with Young's modulus E = 200 GPa and area moment of inertia I = 1000 mm⁴, carrying a central mass M = 50 kg. The effective stiffness is:

k = 48 · (200 × 10⁹ Pa) · (1000 × 10⁻¹² m⁴) / (1 m)³ = 9600 N/m

The natural frequency is then:

f = (1 / 2π) · √(9600 / 50) ≈ 2.21 Hz

Switch the calculator to cantilever, end load and the stiffness coefficient drops from 48 to 3, lowering the natural frequency accordingly.


FAQs

What is the natural frequency of a spring with k = 200 N/m and M = 1 kg?

About 2.25 Hz. Using ω = √(k/M) = √(200/1) = 14.14 rad/s, then f = ω / 2π ≈ 2.25 Hz, with a period T ≈ 0.444 s.

Is natural frequency the same as resonant frequency?

No. Natural frequency is the frequency of free vibration with no external force. Resonance occurs when an external periodic force drives the system at (or near) its natural frequency, producing very large amplitudes.

How can I increase a system's natural frequency?

Increase the stiffness (a stiffer spring, a stiffer or shorter beam, a larger moment of inertia) or reduce the vibrating mass. Both raise the natural frequency.

What units does the natural frequency use?

The natural frequency f is measured in hertz (Hz) — cycles per second — and is the same regardless of the unit system you chose for the inputs. The angular frequency ω is in rad/s, and the period T is in seconds.

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