Natural Frequency Calculator

Calculate the natural frequency of vibrating systems using this Natural Frequency Calculator. Choose your system type — spring-mass, beam, or structure — then enter parameters like stiffness, mass, length, and elastic modulus. Get back the angular frequency (ω) in rad/s and the natural frequency (f) in Hz for your unperturbed vibration scenario.

Select the type of mechanical system to analyze.

N/m

Stiffness of the spring or structural element in N/m.

kg

Mass attached to the spring or structure in kg.

m

Length of the beam in meters. Required for beam calculations.

GPa

Young's modulus of the beam material in GPa. Steel ≈ 200 GPa.

m⁴

Second moment of area of the beam cross-section in m⁴.

kg/m

Mass per unit length in kg/m. Required for distributed mass systems.

Results

Natural Frequency (f)

--

Angular Frequency (ω)

--

Period (T)

--

System Type

--

Frequency Components

Frequently Asked Questions

What is natural frequency?

Natural frequency is the rate at which an object vibrates when disturbed from its equilibrium position and left to oscillate freely, without any external driving force. Every physical object or system has one or more natural frequencies determined by its material properties, mass, and geometry. It is typically measured in hertz (Hz) or radians per second (rad/s).

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

For a simple spring-mass system, the angular natural frequency is ω = √(k/M), where k is the spring stiffness in N/m and M is the mass in kg. The natural frequency in Hz is then f = ω / (2π). For example, a spring with k = 5000 N/m and a mass of 10 kg gives ω ≈ 22.36 rad/s and f ≈ 3.56 Hz.

What is the difference between natural frequency and resonant frequency?

Natural frequency is the frequency at which a system oscillates freely without external forces. Resonant frequency is the frequency at which an external driving force causes maximum amplitude vibration. For undamped systems these are identical, but in damped systems the resonant frequency is slightly lower than the natural frequency. In engineering, resonance can cause catastrophic failures if the driving frequency matches the natural frequency.

How do I calculate the natural frequency of a beam with a concentrated mass?

For a simply supported beam with a central concentrated mass M, the equivalent stiffness is k = 48·E·I / L³, where E is the elastic modulus, I is the second moment of area, and L is the beam length. The natural frequency is then f = (1/2π)·√(k/M). For a cantilever beam with a tip mass, the stiffness becomes k = 3·E·I / L³.

What is the natural frequency of a beam with distributed mass?

For a beam with distributed mass, the natural frequency depends on the boundary conditions and mode shape. For a simply supported beam, the fundamental natural frequency is f = (π/2L²)·√(E·I / μ), where μ is the mass per unit length. For a cantilever, the coefficient changes to approximately 0.5596/L². Higher modes have higher frequencies at integer multiples of the fundamental.

What units are used for natural frequency?

Natural frequency can be expressed in two common units: hertz (Hz), which counts complete cycles per second, and radians per second (rad/s), called angular frequency (ω). They are related by ω = 2π·f. The period T = 1/f gives the time for one complete oscillation in seconds. Engineering calculations often use rad/s internally, then convert to Hz for reporting.

Why is natural frequency important in engineering?

Understanding natural frequency is critical for structural safety and mechanical design. If an external force oscillates at or near the natural frequency of a structure, resonance occurs, causing dramatically amplified vibrations that can lead to fatigue, noise, or catastrophic failure. Engineers design bridges, buildings, aircraft wings, and rotating machinery to ensure their natural frequencies avoid expected operating or environmental excitation frequencies.

How does mass and stiffness affect natural frequency?

Natural frequency increases with higher stiffness (k) and decreases with higher mass (M), following f = (1/2π)·√(k/M). This means stiffer systems vibrate faster, while heavier systems vibrate slower. Engineers can tune the natural frequency of a system by modifying either the material stiffness (changing cross-section or material) or the effective mass (adding or removing material).

More Physics Tools