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# Natural Frequency Calculator

What is the natural frequency: definition and explanationNatural frequency vs. resonant frequencyHow do you calculate the natural frequency of a coiled spring?Calculating the natural frequency: engineering approachThe formulas to calculate the natural frequency of a beamHow to use our natural frequency calculatorHow to calculate the natural frequency of a beam: supported beam with central loadFAQs

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!

## 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 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 $k$. We covered the spring-mass systems in detail at the Hooke's law calculator.

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:

$\omega = \sqrt{\frac{k}{M}}$

Notice that we computed the angular frequency: it's a matter of conventions. You can easily convert the angular frequency to the "ordinary" frequency $f$:

$f = \frac{\omega}{2\cdot\pi}$

The spring-mass system is an easy introduction to the definition of natural frequency. However, we can find many more useful applications for this concept when considering objects like beams.

## Calculating the natural frequency: engineering approach

In the case of solid structures, we will consider them to be subjected to gravitational force, even though it's, of course, an external force. Without it, any system resting in place would never oscillate! Let's learn how to calculate the natural frequency in several situations!

#### Calculate the natural frequency with concentrated mass

In this case, the load on the system is assumed to be concentrated at a point. It can be either the dead load (the structure's mass) or an additional load. The formula for the natural frequency is:

$f=\frac{1}{2\cdot\pi}\sqrt{\frac{g}{\delta}}$

where:

• $g$Acceleration due to gravity. We often take $g = 9.81\ \mathrm{m/s^2}$; and
• $\delta$Deflection due to the static load.

#### Natural frequency formula for a distributed mass

In this case, the formula takes an approximate form:

$f=\frac{a}{\sqrt{\delta}}$

where $a$ is a parameter that often takes the value $a=18$.

## The formulas to calculate the natural frequency of a beam

Let's move to better-defined systems: beams. When calculating the natural frequency of a beam, we encounter many situations depending on the type of load and the way the beam is supported. Let's meet them, starting from the most straightforward system.

In the following formulas, we'll meet some common quantities:

• $M$Mass of the beam and/or the load.
• $I$Moment of inertia: you can find this quantity tabulated for many shapes, or you can calculate it yourself, maybe with our moment of inertia calculator!
• $E$Young modulus or stiffness of the material.
• $L$Length of the beam.

Now that you know the characters, let's learn how to calculate the natural frequency of a beam!

#### Beam supported on two ends, mass concentrated in the center

In this case, the formula for the natural frequency becomes:

$f= \frac{1}{2\cdot\pi}\sqrt{\frac{48\cdot E \cdot I}{M\cdot L^3}}$

#### Beam supported on two ends, mass distributed uniformly over its length

In this case, which truthfully models the behavior of an actual beam, we calculate the natural frequency with the formula:

$f = \frac \pi 2\sqrt{\frac{E\cdot I}{q \cdot L^4}}$

where $q$ is the linear density of the material. In case the beam is under lateral load, and we can identify a contraflexure (a point where the curvature of the beam changes sign), we modify the previous formula to:

$f =2\cdot\pi\sqrt{\frac{E\cdot I}{q \cdot L^4}}$

Let's consider now cantilevers. In this case, one end of the beam is free to swing: you can imagine drastic changes in the formulas for the natural frequency! For these natural frequency formulas, we switch from talking of mass to talking of force, as, in these situations, it's natural to identify a torque (if you need a refresh, visit the torque calculator!).

#### Cantilever: mass concentrated at the free end

In this case, imagine a beam fixed on one side, with the force acting on the free-swinging end: think of a springboard in a diving contest. The formula for the natural frequency becomes:

$f =\frac{1}{2\cdot\pi}\sqrt{\frac{3\cdot E\cdot I}{ F\cdot L^3}}$

#### Cantilever with mass uniformly distributed

If the load is distributed over the entire length of the beam, the formula seen before changes to:

$f = 0.56 \sqrt{\frac{E\cdot I}{q\cdot L^4}}$

Here we meet the linear density $q$ again.

#### Beam with fixed ends and distributed load

The last system for which we give you the natural frequency formula is a beam fixed at both ends, with distributed load over its entire length. The situation differs slightly from the supported beam we've seen before. In this case, the formula is:

$f = 3.56 \sqrt{\frac{E\cdot I}{q\cdot L^4}}$

🙋 The beam load calculator is a dedicated tool to model the behavior of a beam under load!

## How to use our natural frequency calculator

Our natural frequency calculator allows you to choose the desired system among the ones we listed above and quickly calculate its natural frequency.

For every system, we will show the appropriate variables. You can change the units to the one of your liking and proceed to input the known data. You can also calculate the parameters of your system starting from a desired natural frequency: our tools work flawlessly also in reverse!

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

Take a steel beam with length $L = 10\ \mathrm{m}$, and lay it over two supports so that it's free to sag. The steel you chose has Young modulus $E = 178 \times 10^9 \ \mathrm{Pa}$ and momentum of inertia $I = 2.140\times 10^{-5}\ \mathrm{m^4}$. Assume you load it in the center with a mass $M = 500\ \mathrm{kg}$. What is the natural frequency of this beam?

Let's use the formula for the supported beam with central load and substitute the known values:

$\begin{split} f&= \frac{1}{2\cdot\pi}\sqrt{\frac{48\cdot E \cdot I}{M\cdot L^3}}\\[1em] &=\frac{1}{2\cdot\pi}\sqrt{\frac{48\cdot 178\ \mathrm{GPa}}{500\ \mathrm{kg}\cdot (10\ \mathrm{m})^3}}\\[1.2em] &\quad \cdot\sqrt{2.140\times 10^{-5}\ \mathrm{m^4}} \end{split}$

For good measure, check the units in the square root. If they simplify to $1/\mathrm{s}^2$, the quantities you used in the formula are correct. They do so in this situation, so we can proceed:

$f = 3.0435\ \mathrm{Hz} \approx 3.04\ \mathrm{Hz}$
FAQs

### What is the natural frequency?

The natural frequency is the frequency at which an object vibrates in the absence of external forces. Every object has at least a natural frequency: complicated objects may have more than one, though. Knowing the natural frequency of an object is fundamental in engineering, as this quantity is an intrinsic weakness of a system that can lead to catastrophic failures.

### What is the difference between natural and resonant frequency?

The natural frequency is the frequency at which an object vibrates in the absence of external forces. This frequency is identical, or very similar in value, to the resonant frequency. The resonant frequency is the frequency at which the object vibrates when stimulated by a periodic force with a frequency matching the natural one. Vibrating at the resonant frequency may cause the amplitude of the motion to increase uncontrollably.

### How do I calculate the natural frequency for a spring-mass system?

To calculate the natural frequency of a spring-mass system, follow these simple steps:

1. Identify or measure the spring constant k.
2. Measure the mass connected to the spring, M.
3. Calculate the ratio k/M, and take the square root of the result.
4. The result is the spring-mass system's natural angular frequency.
5. To calculate the natural frequency for the spring-mass system, divide the previous result by .

### What is the natural frequency of a spring with k = 64 N/m loaded with M = 0.5 kg

The natural frequency of a spring-mass system where the spring constant is k = 64 N/m and the mass is M = 0.5 kg is f = 1.8 Hz. To find this result:

1. Calculate the ratio between the spring constant and the mass:

k/M = 82 N/m/0.5 kg = 164 N/(kg · m) = 164 1/s2

2. Take the square root of the result to calculate the natural angular frequency:

ω =√k/M = √164 1/s2 = 12.81 Hz/rad

3. Divide the result by to find the natural frequency:

f = 12.81/2π = 2.04 Hz