Natural Frequency: Formula and Calculation

Calculate a system’s natural frequency by using its physical model: for a simple mass–spring system, divide the spring stiffness by the mass, take the square root, then divide by 2π to get hertz. Enter stiffness in newtons per metre and mass in kilograms in the calculator below; it returns the frequency in cycles per second, plus the angular frequency in radians per second.

Formula

For an ideal, undamped mass–spring oscillator, the natural frequency is f = (1 / 2π)√(k / m). The angular natural frequency is ωₙ = √(k / m), and the relationship between them is f = ωₙ / 2π.

• f is natural frequency in hertz (Hz), or cycles per second.
• ωₙ is angular natural frequency in radians per second (rad/s).
• k is spring stiffness in newtons per metre (N/m).
• m is the oscillating mass in kilograms (kg).

The calculator uses the standard mass–spring formula. It assumes the spring behaves linearly, the mass is concentrated in one moving body, and damping is negligible. Its default values are k = 100 N/m and m = 1 kg.

Worked Example

With the calculator’s default stiffness of 100 N/m and mass of 1 kg:

1. Find the angular frequency: ωₙ = √(100 / 1) = 10 rad/s.
2. Convert radians per second to cycles per second: f = 10 / (2π) ≈ 1.5915 Hz.
3. Rounded to two decimal places, the natural frequency is 1.59 Hz; angular frequency is 10.00 rad/s.

In an ideal system, this corresponds to one complete oscillation about every 1 / 1.5915 ≈ 0.63 seconds. The period, T, is the time for one cycle and is related by T = 1/f.

What Natural Frequency Means

A natural frequency is a frequency at which a system tends to vibrate after it has been disturbed and then allowed to move according to its own properties. It depends on the system’s mass, stiffness, shape, and boundary conditions—not simply on how strongly or quickly someone initially pushes it.

A basic spring and mass has one natural frequency in this model. More complex objects, such as a beam, a vehicle, or a building, can have several natural modes. Each mode has its own frequency and characteristic pattern of motion: one part may move more while another remains near a node. A single mass–spring calculation cannot describe every mode of a flexible structure.

How to Choose the Right Calculation

First decide whether your system is reasonably represented by a single degree of freedom: one effective mass moving against one effective stiffness. If so, the calculator is useful for a first estimate. If not, use a formula that matches the physical model or a modal analysis that accounts for the structure’s geometry and supports.

• Mass on a spring: Use k and the mass that moves with the spring. Include attached components that move together, where appropriate.
• Small-angle pendulum: For a simple pendulum, f = (1 / 2π)√(g / L), where g is gravitational acceleration and L is the distance from pivot to the mass centre. This is a different model, not a spring calculation.
• Rotating system: A torsional oscillator uses rotational stiffness and moment of inertia: ωₙ = √(K / I), with K in N·m/rad and I in kg·m².
• Flexible structure: A beam or frame’s frequencies depend on material properties, dimensions, mass distribution, and how it is supported. Consult a suitable structural vibration model or analysis.

For a spring system, stiffness may be supplied by a manufacturer or measured from the force–deflection relationship. In the linear range, k = F/x, where F is force in newtons and x is deflection in metres. Do not substitute a material’s elastic modulus for spring stiffness; they are different quantities.

Using and Interpreting the Result

Use consistent units. In SI, newtons per metre and kilograms produce rad/s, which converts to Hz by dividing by 2π. If stiffness is given in pounds-force per inch and mass in pounds-mass, convert both to SI before using the calculator; mixing customary and metric values gives a meaningless result. The final frequency in hertz is the same physical frequency regardless of the units used in a correctly converted calculation.

Natural frequency helps compare a system’s tendency to vibrate with the frequency of a repeating force. If an external force acts near a natural frequency, resonance can produce a large response, especially when damping is low. The frequency alone does not predict the vibration’s amplitude or whether the system is safe: force size, damping, operating conditions, and structural details matter too.

Common Mistakes and Limits

• Confusing Hz with rad/s: They differ by a factor of 2π. The square root √(k/m) gives rad/s, not Hz.
• Using the wrong mass: Include the mass that participates in the motion. For a distributed spring or flexible component, its effective moving mass may require a more detailed model.
• Assuming every system is a spring: Pendulums, torsional systems, beams, and multi-part structures need their own equations or models.
• Ignoring damping and nonlinearity: Damping changes the observed free-vibration frequency, and strong nonlinear behavior can make frequency depend on motion amplitude. The calculator reports the ideal undamped value.
• Rounding too early: Keep intermediate values unrounded and round the final answer to a precision justified by the input data.

For a lightly damped, single-degree-of-freedom spring system, the damped frequency is lower than the undamped natural frequency. For heavier damping the motion may not oscillate, so describing it with an ordinary oscillation frequency is no longer appropriate. Real structures may also have multiple interacting modes, requiring measurement or a more complete model.

Frequently Asked Questions

What is the formula for natural frequency?
For an ideal mass–spring oscillator, f = (1 / 2π)√(k/m), where k is stiffness in N/m and m is mass in kg. The corresponding angular frequency is √(k/m) rad/s.

How do I calculate natural frequency in hertz?
Calculate √(k/m) using consistent SI units, then divide by 2π. The result is in hertz, meaning cycles per second.

What is the difference between natural frequency and resonance?
Natural frequency is a property of a system’s free vibration. Resonance is a large response that can occur when an external repeating force is near a natural frequency; damping and force strength affect how large that response becomes.

Does adding mass increase or decrease natural frequency?
For a spring with unchanged stiffness, adding mass lowers frequency: f varies with 1/√m. Increasing stiffness with mass unchanged raises it, with f varying as √k.

Can a system have more than one natural frequency?
Yes. A flexible or multi-part system can have multiple vibration modes, each with its own natural frequency and motion pattern. The simple calculator describes only a single mass–spring mode.

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in k | Spring stiffness | N/m | 100 | 0
in m | Mass | kg | 1 | 0
out f | Natural frequency | Hz | sqrt(k / m) / (2 * pi) | 2
out w | Angular frequency | rad/s | sqrt(k / m) | 2