Engineering · Mechanics/Statics

Undamped natural frequency of a single-degree-of-freedom system

The undamped natural frequency of a linear single-mass oscillator depends on stiffness and mass.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

ω_{n} = \frac{k}{m}

ωn in rad/s; k in N/m; m in kg

Variables & units

Variables & units
Symbol	Meaning	Unit
$ωn$	Undamped natural angular frequency	rad/s
$k$	Linear spring stiffness	N/m
$m$	Oscillating mass	kg

LaTeX

\omega_n=\sqrt{\frac{k}{m}}

QUANTA STUDYquanta-study.de

Derivation & background

This follows from m x¨+kx=0 for a linear spring and lumped mass. Frequency in hertz is fn=ωn/(2π). Damping, multiple degrees of freedom, distributed mass and nonlinear stiffness change real modal behavior; forced resonance is not automatically exactly the undamped natural frequency.

Exam blueprint

Validity range

Derivation steps

The undamped natural frequency of a linear single-mass oscillator depends on stiffness and mass.

1State the model assumptions and list quantities with units.
2Substitute into ωn = √(k/m); convert quantities to SI first.
3For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

Rearrangements

Base equation for the target quantity

\omega_n=\sqrt{\frac{k}{m}}

Preserve signs, boundary conditions and units when rearranging.

Task variant

Work through the numerical example for “Undamped natural frequency of a single-degree-of-freedom system” and check its unit and magnitude.

For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

Common mistakes

Substituting a related quantity that shares the same symbol.

Use variable definitions and check: ωn in rad/s; k in N/m; m in kg.

Exam context

Estimate the fundamental frequency of an idealized mass-spring system.

Formula cluster

Mechanics/Statics

The undamped natural frequency of a linear single-mass oscillator depends on stiffness and mass.

ζ = c/(2√km)

Damping ratio of a single-degree-of-freedom oscillator

k = Gd⁴/(8D³N)

Rate of a cylindrical helical spring

x(t) = A₀ e^(−βt) cos(ωdt+φ)

Damped harmonic oscillation

Worked example

For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

Applications

Estimate the fundamental frequency of an idealized mass-spring system.

How do you calculate with Undamped natural frequency of a single-degree-of-freedom system?

Here is how to work through a typical Undamped natural frequency of a single-degree-of-freedom system (ωn = √(k/m)) task step by step:

1
Task
Work through the numerical example for “Undamped natural frequency of a single-degree-of-freedom system” and check its unit and magnitude.
Solution path
For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

Frequently asked questions about Undamped natural frequency of a single-degree-of-freedom system

What is the effect of greater mass?+

At fixed stiffness, greater mass lowers natural angular frequency as 1/√m. The system oscillates more slowly because the same restoring force accelerates more inertia. Doubling mass does not halve frequency; it multiplies it by 1/√2. This is an undamped linear lumped-mass model; convert angular frequency to hertz by dividing by 2π. Application: Estimate the fundamental frequency of an idealized mass-spring system. The numerical example For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz anchors the idea in a concrete measurement case.

Why are ωn and fn reported differently?+

Angular frequency counts radians per second; ordinary frequency counts cycles per second. One cycle is 2π radians, hence fn=ωn/(2π). Radians are dimensionless but are written to distinguish angular frequency from hertz. This is an undamped linear lumped-mass model; convert angular frequency to hertz by dividing by 2π. The corresponding units are ωn in rad/s; k in N/m; m in kg. For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz. This checks that the reference quantities are compatible. Scope: This follows from m x¨+kx=0 for a linear spring and lumped mass. Frequency in hertz is fn=ωn/(2π). Damping, multiple degrees of freedom, distributed mass and nonlinear stiffness change real modal behavior; forced resonance is not automatically exactly the undamped natural frequency. A practical application is Estimate the fundamental frequency of an idealized mass-spring system.

Does the formula apply to every vibration?+

No. It applies to a linear lumped mass and spring, with damping excluded from this frequency definition. Distributed structures have multiple modes; nonlinear springs make frequency amplitude-dependent. Real modes depend on boundary conditions and mass distribution. This is an undamped linear lumped-mass model; convert angular frequency to hertz by dividing by 2π. Scope: This follows from m x¨+kx=0 for a linear spring and lumped mass. Frequency in hertz is fn=ωn/(2π). Damping, multiple degrees of freedom, distributed mass and nonlinear stiffness change real modal behavior; forced resonance is not automatically exactly the undamped natural frequency. A practical application is Estimate the fundamental frequency of an idealized mass-spring system.

How do I check the result and units for Undamped natural frequency of a single-degree-of-freedom system?+

Write the quantity and unit from the variable list next to every value: ωn in rad/s; k in N/m; m in kg. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. This is an undamped linear lumped-mass model; convert angular frequency to hertz by dividing by 2π.

Which model limit matters for Undamped natural frequency of a single-degree-of-freedom system?+

The background states this starting condition: This follows from m x¨+kx=0 for a linear spring and lumped mass. Frequency in hertz is fn=ωn/(2π). Damping,. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. This is an undamped linear lumped-mass model; convert angular frequency to hertz by dividing by 2π.

Quanta exam set

Curated exam set for "Undamped natural frequency of a single-degree-of-freedom system":

Question (front)

Which formula describes Undamped natural frequency of a single-degree-of-freedom system?

Answer in your set

Question (front)

How do you rearrange Undamped natural frequency of a single-degree-of-freedom system for Base equation for the target quantity?

Answer in your set

Question (front)

Which common mistake happens with Undamped natural frequency of a single-degree-of-freedom system?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]MIT OpenCourseWare, Mechanical Vibrations, natural frequency of a single-degree-of-freedom system.

More formulas: Mechanics, strength and materials

E = \frac{\sigma}{\varepsilon} = \frac{F L_0}{A_0 \Delta L}

Young's modulus

\Delta\vec p=\int_{t_1}^{t_2}\vec F_{\mathrm{ext}}dt, \qquad \Delta\vec L=\int_{t_1}^{t_2}\vec\tau_{\mathrm{ext}}dt

Linear and angular impulse–momentum balance

M(x)=EI\,\kappa(x)\approx EI\frac{d^2w}{dx^2}

Euler–Bernoulli beam moment–curvature relation

\sum_i \vec F_i=\vec 0,\quad \sum_i M_{i,O}=0

Static equilibrium conditions

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Undamped natural frequency of a single-degree-of-freedom system for exams

Create a curated FSRS exam set for ωn = √(k/m): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Quanta Study editorial team

Symbol

Meaning

Unit

ωn

Undamped natural angular frequency

rad/s

k

Linear spring stiffness

N/m

m

Oscillating mass

Exam blueprint

Validity range

Derivation steps

The undamped natural frequency of a linear single-mass oscillator depends on stiffness and mass.

1State the model assumptions and list quantities with units.
2Substitute into ωn = √(k/m); convert quantities to SI first.
3For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

How do you calculate with Undamped natural frequency of a single-degree-of-freedom system?

Here is how to work through a typical Undamped natural frequency of a single-degree-of-freedom system (ωn = √(k/m)) task step by step:

Task

Work through the numerical example for “Undamped natural frequency of a single-degree-of-freedom system” and check its unit and magnitude.

Solution path

For k=200 N/m and m=2 kg, ωn=10 rad/s and fn≈1.59 Hz.

Frequently asked questions about Undamped natural frequency of a single-degree-of-freedom system

What is the effect of greater mass?+

Why are ωn and fn reported differently?+

Does the formula apply to every vibration?+

How do I check the result and units for Undamped natural frequency of a single-degree-of-freedom system?+

Which model limit matters for Undamped natural frequency of a single-degree-of-freedom system?+