Physics · Oscillations and waves

Mass-spring oscillator: period and harmonic motion

Q: How do you calculate the period of a spring pendulum?

For an ideal linear undamped spring oscillator use $T=2\pi\sqrt{\frac mk}$. Enter total effective moving mass in kilograms and stiffness in newtons per metre to obtain seconds. With $m = 0.25 kg$ and $k =\frac{ 100 N}{m}$ the period is about 0.314 s. Reversing $k=\frac{4\pi^2m}{T^2}$ must return $\frac{100 N}{m}$. Displacement is measured from equilibrium, which for a vertical spring is different from its unloaded length. A heavy spring, moving holders, nonlinear extension or appreciable damping can require an extended model. A complete period ends at the same position with the same direction of motion.

Q: Why does the period not depend on the amplitude?

Amplitude independence follows from the linear equation $m\ddot x=-kx$: substituting a cosine solution cancels amplitude and leaves $\omega_0^2=\frac km$. Larger displacement creates a larger restoring force but also a proportionally longer path. This is a property of that model, not a statement about every real spring. Doubling amplitude doubles maximum speed $A\omega_0$ and quadruples energy $\frac12kA^2$ while keeping T unchanged. If stiffness varies with extension, motion reaches a stop, or resistance grows nonlinearly, the actual period change needs a new calculation or measurement. Always state the linear range when applying amplitude independence to experimental data.

Q: How do you determine the spring constant from an oscillation measurement?

Rearrange the ideal period equation to $k=\frac{4\pi^2m}{T^2}$. Time several complete cycles and divide elapsed time by their number. Include moving hooks and holders in the mass; a heavy spring can add effective inertia. For 0.480 kg and 20 periods in 7.54 s, $T = 0.377 s$ and $k \approx{}\frac{ 133.33 N}{m}$. A static cross-check uses $k=\frac{mg}y$ with extension from unloaded length. If estimates disagree beyond their uncertainties, examine moving mass, cycle counting, spring linearity and damping. Substituting the inferred k back into the period equation checks the algebra, but does not independently prove experimental accuracy.

Q: How does the energy move in an oscillating spring pendulum?

In the ideal undamped model, oscillation energy $E_{\mathrm{osc}}=\frac12mv^2+\frac12kx^2=\frac12kA^2$ is conserved, with x measured from equilibrium. At turning points velocity is zero and varying energy is potential; at equilibrium crossing it is kinetic. A vertical spring remains extended by $\frac{mg}k$ there, so it is not relaxed. Effective potential $\frac12kx^2$ then includes spring and gravitational energy relative to equilibrium. With $k =\frac{ 144 N}{m}$ and $A = 0.025 m$, $E_{\mathrm{osc}}=0.045\,\mathrm J$. At $x=\frac A{\sqrt2}$, kinetic and potential contributions are each 0.0225 J. Their sum checks the calculation without confusing oscillation energy with all absolute stored spring energy.

In the ideal linear undamped mass–spring model, natural period depends on m and k; amplitude does not change it within that model.

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Formula

Calculate

T = 2 π \frac{m}{k}

T in seconds [s] · m in kg · k in N/m

Variables & units

Variables & units
Symbol	Meaning	Unit
$T$	Oscillation period	s
$m$	Oscillating mass	kg
$k$	Spring constant	N/m

LaTeX

T = 2\pi \sqrt{\frac{m}{k}}

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Diagram: a cosine curve of displacement x over time t; the amplitude A and the period T between two crests are marked. — Ideal undamped linear mass-spring model: x is measured from stable equilibrium. The illustrated cosine phase starts at the positive maximum; T = 2π√(m/k).

Spring oscillator period calculator

Find period, frequency, mass or spring constant of an ideal mass on a spring without friction or driving force.

Find

Mass m

Mass oscillating on the spring, e.g. 100 g for a slotted mass hung on it.

Spring stiffness k

Spring constant k: force per metre of extension, e.g. 10 N/m for a soft school spring.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for T, substitute, evaluate

1. Rearrange for T
$T=2\pi\sqrt{\frac mk}$
2. Substitute values (in SI units)
- m = 100 g = 0.1 kg
$T=2\pi\sqrt{\frac{0.1}{10}}=0.628319$
3. Result
T = 0.6283 s

Result: Period T

T = 0.6283 s

With m = 100 g and k = 10 N/m, T = 0.6283 s.

Quick tips

Four times the mass, twice the period; four times the spring constant, half the period. How far you pull it out does not change T.
Unit check: N/m is the same as kg/s², so √(m ÷ k) is in seconds. By hand, use the mass in kg, so 100 g becomes 0.1 kg.

Assumptions and limits

Linear net restoring force about stable equilibrium, constant effective mass m > 0 and k > 0. Undamped and unforced. Gravity shifts equilibrium in the ideal vertical model, not period.

Derivation & background

Measured from equilibrium, $m\ddot x=-kx$ . The solution $x=A\cos(\omega_0t+\varphi)$ has $\omega_0=\sqrt{\frac km}$ , giving $T=\frac{2\pi}{\omega_0}=2\pi\sqrt{\frac mk}$ . A and φ follow from initial conditions. For a vertical linear spring, equilibrium is extended by $\frac{mg}k$ from the unloaded length; constant g shifts this position without changing natural period when m and k are unchanged. Nonlinear stiffness and damping can alter observed frequency.

Exam blueprint

Validity range

Ideal free undamped mass–spring system with constant effective mass $m > 0$ and linear net restoring force −kx with $k>0$ . Measure x from equilibrium. Spring inertia, nonlinearity, damping and driving require extensions.

Derivation steps

Linear restoring force sets natural angular frequency; a full phase advance gives period.

1Measure x from stable equilibrium and assume constant mass m with linear net force $F=-kx$ .
2Newton gives $m\ddot x=-kx$ , hence $\ddot x+\frac kmx=0$ .
3Use $x(t)=A\cos(\omega_0t+\varphi)$ , with A and φ matching initial position and velocity.
4Differentiating twice gives $\ddot x=-\omega_0^2x$ ; substitution requires $\omega_0^2=\frac km$ .
5A full period advances phase by 2π: $\omega_0T=2\pi$ .
6Thus $T=2\pi\sqrt{\frac mk}$ . In a vertical arrangement g only shifts equilibrium provided m and k remain unchanged.

Rearrangements

Spring constant from the period

k = \frac{4\pi^2 m}{T^2}

This is how k is determined experimentally from an oscillation measurement.

Mass from the period

m = \frac{k T^2}{4\pi^2}

The principle of mass measurement devices in weightlessness.

Frequency

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

A stiffer spring or smaller mass raises the frequency.

Task variant

A $0.5\,\mathrm{kg}$ mass oscillates freely without damping on an ideal linear spring with $T=1\,\mathrm s$ . Find stiffness $k$ .

$k=\frac{4\pi^2m}{T^2}=\frac{4\pi^2\cdot0.5}{1^2}\,\mathrm{N\,m^{-1}}\approx19.7\,\mathrm{N\,m^{-1}}$ .

How does T change when the mass is quadrupled?

$T\propto\sqrt m$ : quadrupling the mass doubles the period.

Common mistakes

Assuming a larger amplitude lengthens the period.

Period is amplitude independent in the ideal linear undamped model. This requires constant stiffness and motion within its valid linear range.

Using the simple pendulum formula $T=2\pi\sqrt{\frac lg}$ .

For the spring pendulum m and k appear under the root; g does not occur.

Swapping m and k under the root.

A heavier mass oscillates more slowly: m is in the numerator.

Exam context

Typical: determine k from a stretching experiment, compute T, then analyse energy or speed at the equilibrium crossing.

Formula cluster

Harmonic oscillations

Hooke supplies the force, the spring pendulum the time evolution, the wave the spatial propagation.

F = k·x

Hooke's Law (Spring Force)

c = λ·f

Wave Equation (c = λ·f)

Fz = m·v²/r

Centripetal Force (Circular Motion)

T = 2π·√(l/g)

Simple Pendulum (Period)

Worked example

For the ideal linear undamped model with $m=0.25\,\mathrm{kg}$ , $k=100\,\mathrm{N/m}$ , $T=2\pi\sqrt{\frac{0.25}{100}}\,\mathrm s\approx0.314159\,\mathrm s$ , hence $f=\frac1T\approx3.18310\,\mathrm{Hz}$ . Check: $\frac{4\pi^2m}{T^2}=100\,\mathrm{N/m}$ .

Applications

Estimating linear spring stiffness from a free period measurement; first estimates of natural frequency for mass–spring systems. Suspension and tuned mass dampers require additional damping, forcing and degrees of freedom.

How do you calculate with Spring Pendulum (Oscillation Period)?

Here is how to work through a typical Spring Pendulum (Oscillation Period) (T = 2π·√(m/k)) task step by step:

1
Task
A $0.5\,\mathrm{kg}$ mass oscillates freely without damping on an ideal linear spring with $T=1\,\mathrm s$ . Find stiffness $k$ .
Solution path
$k=\frac{4\pi^2m}{T^2}=\frac{4\pi^2\cdot0.5}{1^2}\,\mathrm{N\,m^{-1}}\approx19.7\,\mathrm{N\,m^{-1}}$ .
2
Task
How does T change when the mass is quadrupled?
Solution path
$T\propto\sqrt m$ : quadrupling the mass doubles the period.

Understand and apply the equation

A linear spring and movable mass form a harmonic oscillator. The restoring force sets angular frequency; initial position and velocity set the amplitude. Complete calculations connect motion, energy, and the correct equilibrium reference.

The spring pulls back while the mass keeps moving

Displace a mass from equilibrium and release it: the spring accelerates it back. At equilibrium the resultant force vanishes, but the mass has velocity and keeps moving. The motion therefore repeats. An ideal linear force law makes this motion sinusoidal.[1]

You need spring stiffness, Newton’s law, and the distinction between instantaneous displacement and maximum displacement. We measure displacement from equilibrium. For a vertical spring, that is not the unloaded spring position: weight first shifts the resting position.[1]

Which equilibrium position and linear range apply?

A constant mass m moves along a fixed axis. $x=0$ denotes its stable equilibrium. The net restoring force is −kx with $k>0$ ; its sign opposes displacement. Initially neglect friction, periodic external forces and the spring’s own inertia. x is a signed coordinate, whereas amplitude A is a nonnegative distance from equilibrium. Within this ideal linear, undamped model, period is independent of amplitude; do not extend that conclusion beyond the spring’s linear range.[1]

For a vertical spring, let y be downward extension from unloaded length. Net force is $mg-ky$ , and equilibrium is $y_{\mathrm{eq}}=\frac{mg}{k}$ . With $x=y-y_{\mathrm{eq}}$ , this becomes $F_{\mathrm{net}}=-kx$ . Weight thus shifts the resting position. With constant m and k the ideal natural period remains unchanged. Any preload-dependent stiffness requires an additional model.

\begin{aligned}m\ddot y&=mg-ky\\x&=y-\frac{mg}{k}\\m\ddot x&=-kx\end{aligned}

A constant shift of position zero does not change acceleration.

How do you calculate and derive the spring oscillator’s period?

For constant effective mass $m>0$ and linear spring stiffness $k>0$ , the undamped model gives $T=2\pi\sqrt{\frac{m}{k}}$ . Some problems write stiffness as $D$ rather than $k$ . Hooke’s law supplies the linear restoring force; Newton’s equation of motion then gives the period.[1]

Newton gives $m\ddot x=-kx$ , so acceleration is proportional and opposite to displacement. A cosine has that property: differentiating $A\cos(\omega_0t+\varphi)$ twice gives $-\omega_0^2x$ . Substitution requires $m\omega_0^2=k$ . This identifies natural angular frequency within the linear model; it does not establish that every real spring is linear.

\begin{aligned}&\ddot x+\omega_0^2x=0\\&\omega_0=\sqrt{\frac{k}{m}}\\&x=A\cos(\omega_0t+\varphi)\end{aligned}

φ sets initial phase and must match the initial conditions.

\begin{aligned}\omega_0T&=2\pi\\T&=2\pi\sqrt{\frac mk}\\f&=\frac1T\end{aligned}

A full period returns to the same position and direction of motion.

An equivalent general solution is $x_0\cos(\omega_0t)+\frac{v_0}{\omega_0}\sin(\omega_0t)$ . This keeps initial position and velocity separate. Amplitude is $A=\sqrt{x_0^2+\frac{v_0^2}{\omega_0^2}}$ ; only when $v_0=0$ does release occur at a turning point with $A=|x_0|$ . An arbitrary starting position is therefore not automatically the amplitude.

Where are speed, acceleration and energy maximal?

\begin{aligned}&v=-A\omega_0\sin(\omega_0t+\varphi)\\&a=-\omega_0^2x\\&v_{\max}=A\omega_0\end{aligned}

vmax is a magnitude; velocity sign depends on time.

E_{\mathrm{osc}}=\frac12mv^2+\frac12kx^2=\frac12kA^2

Oscillation energy uses potential energy relative to equilibrium.[2]

At turning points $v=0$ , so oscillation energy lies in the effective potential. Kinetic energy is maximal when crossing $x=0$ . A horizontal unloaded spring is relaxed there, but a vertical spring remains extended by $\frac{mg}{k}$ . Linear terms in spring and gravitational energy cancel about that equilibrium. The resulting relative potential is $\frac12kx^2$ . “Entirely kinetic” refers to this varying oscillation energy, not all absolute energy in a preloaded spring. A spring’s elastic potential energy instead refers to its actual extension from unloaded length. Especially for a vertical spring, keep this reference separate from oscillation energy measured about equilibrium.

A full period starts at a turning point

QUANTA STUDYquanta-study.de

For the first example,

x(t)=25\,\mathrm{mm}\cos(20\,\mathrm{s^{-1}}t)

. The horizontal axis shows

\frac tT

. At zero displacement the force vanishes, but velocity does not. The curve has its greatest slope magnitude there.[1][2]

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How do period, frequency, angular frequency and mass relate?

\begin{aligned}k&=\frac{4\pi^2m}{T^2}\\{} m&=\frac{kT^2}{4\pi^2}\\{} [\frac{m}{k}]&=\mathrm{s^2}\end{aligned}

Use kg for mass, N/m for stiffness and seconds for period.

$\mathrm{N\,m^{-1}}=\mathrm{kg\,s^{-2}}$ , so $\sqrt{\frac{m}{k}}$ has unit second. Quadrupling mass doubles T and quadrupling stiffness halves T. Amplitude does not appear in period, but enters vmax linearly and Eosc quadratically. Doubling amplitude at the same period doubles maximum speed and quadruples oscillation energy, provided the spring remains in the same linear range. Frequency $f=\frac1T$ counts oscillations per second in hertz; angular frequency $\omega_0=2\pi f$ is expressed in radians per second. A simple pendulum follows a different model: at small angles, its period depends on length and gravitational acceleration, not bob mass.[5]

Example: a complete motion state

A horizontal oscillator released from rest

Given: m = 0.360 kg, k = 144 N/m, x₀ = +25.0 mm, $v_0=0$ . Ideal model.
Find: Find period, maximum speed and state after $\frac T8$ .

Approach: Release from rest: initial position is the positive turning point of free motion.
$\begin{aligned}\omega_0&=\sqrt{\frac km}\\T&=\frac{2\pi}{\omega_0}\\x(t)&=x_0\cos\omega_0t\\v(t)&=-x_0\omega_0\sin\omega_0t\end{aligned}$
Substitute the given values with matching units.
$\begin{aligned}A&=0.0250\,\mathrm m\\{} \omega_0&=\sqrt{\frac{144}{0.360}}=20.0\,\mathrm{rad/s}\end{aligned}$
Calculate period and maximum speed.
$\begin{aligned}T&=\frac{2\pi}{20}=0.314159\,\mathrm s\\{} v_{\max}&=0.500\,\mathrm{m/s}\end{aligned}$
At $t=\frac T8$ , phase is $\frac\pi4$ and motion is toward equilibrium.
$\begin{aligned}x&=\frac{0.0250}{\sqrt2}=0.0176777\,\mathrm m\\{} v&=-\frac{0.500}{\sqrt2}=-0.353553\,\mathrm{m/s}\end{aligned}$
Use $a=-\omega_0^2x$ and $E_{\mathrm{osc}}=\frac12 kA^2$ .
$\begin{aligned}a&=-7.07107\,\mathrm{m/s^2}\\{} E_{\mathrm{osc}}&=0.0450\,\mathrm J\end{aligned}$

Result: T ≈ 0.314 s and vmax = 0.500 m/s; at T/8 x ≈ +17.68 mm and v ≈ −0.354 m/s.

Check: At $t=\frac T8$ , $\frac12kx^2$ and $\frac12mv^2$ are each $0.0225\,\mathrm J$ ; their sum remains $0.0450\,\mathrm J$ . Negative v with positive x agrees with returning motion.

How do you measure spring stiffness statically and dynamically?

Comparing dynamic and static estimates

Given: Total moving load m = 0.480 kg; 20 full periods take 7.54 s. Static extension y = 35.3 mm at g = 9.81 m/s². Light linear spring.
Find: Find both stiffness estimates and their relative difference.

Approach: Dynamic timing and static equilibrium give two separate stiffness estimates.
$\begin{aligned}T&=\frac{t_{\mathrm{series}}}N\\k_{\mathrm{dyn}}&=\frac{4\pi^2m}{T^2}\\k_{\mathrm{stat}}&=\frac{mg}{y_{\mathrm{eq}}}\end{aligned}$
Substitute the given values with matching units.
$T=\frac{7.54}{20}=0.377\,\mathrm s$
Rearrange the period equation for k.
$k_{\mathrm{dyn}}=\frac{4\pi^2\cdot0.480}{0.377^2}\approx133.327\,\mathrm{N/m}$
Use static equilibrium and the same load.
$k_{\mathrm{stat}}=\frac{0.480\cdot9.81}{0.0353}\approx133.394\,\mathrm{N/m}$

Result: Both give about 133 N/m; the static estimate is about 0.050% above the dynamic estimate.

Check: Using $k_{\mathrm{dyn}}$ gives $y=\frac{mg}{k_{\mathrm{dyn}}}\approx35.318\,\mathrm{mm}$ , close to 35.3 mm. Agreement checks the model; without uncertainties it does not establish 0.050% accuracy. Include moving hooks and holders in the load.

What do damping and periodic driving change?

The equation assumes a linear net restoring force and constant effective mass. Large extension can change k, while friction or drag reduces amplitude. A heavy spring carries kinetic energy, so attached mass alone is not a complete inertia model. Combine additional springs using the appropriate effective stiffness; a laterally moving suspension may add degrees of freedom.

For viscous underdamping, free angular frequency is $\omega_d=\sqrt{\frac km-\beta^2}$ , smaller than $\sqrt{\frac km}$ . Strong damping can prevent free oscillations altogether. With periodic driving, steady motion follows the driving frequency, so its observed period need not equal the natural period. Vehicle suspensions and tuned mass dampers require such extensions, although the ideal model provides an initial comparison. A damped harmonic oscillator therefore requires the appropriate damping model; a driving force is an additional model assumption.[3][4]

How do you find amplitude when starting at equilibrium with an initial push?

Initial displacement can be zero even when subsequent amplitude is not. An initial push supplies the energy and the sine term in the initial-value solution.[1][2]

Starting at equilibrium with an initial push

Given: An ideal horizontal spring oscillator has m=0.25 kg and k=100 N/m. Initially x₀=0 and v₀=+0.60 m/s. There is no friction or subsequent driving. Displacement is measured from equilibrium.
Find: Find amplitude, first positive turning time, and energy.

Approach: only the sine term remains in the general initial-value solution.
$\begin{aligned}\omega_0&=\sqrt{\frac km}\\x(t)&=\frac{v_0}{\omega_0}\sin\omega_0t\\A&=\frac{|v_0|}{\omega_0}\\t_{\mathrm{turn}}&=\frac{\pi}{2\omega_0}\\E&=\frac12mv_0^2=\frac12kA^2\end{aligned}$
The first turning point occurs after a quarter period.
$\begin{aligned}\omega_0&=20\,\mathrm{rad/s}\\A&=0.0300\,\mathrm m\\t_{\mathrm{turn}}&=\frac\pi{40}\,\mathrm s\approx0.0785398\,\mathrm s\\E&=\frac12(0.25)(0.60)^2\,\mathrm J=0.0450\,\mathrm J\end{aligned}$

Result: Although initial displacement is zero, the mass reaches 3.00 cm amplitude. After approximately 78.54 ms all initial mechanical energy is stored in the spring.

Check: $\frac12(100)(0.0300)^2\,\mathrm J=0.0450\,\mathrm J$ confirms the energy balance. Differentiation gives $v(0)=+0.60\,\mathrm{m/s}$ ; acceleration initially vanishes and is negative at the positive turning point.

Five solved exercises

Calculate · Foundation

m = 0.20 kg and k = 50 N/m. Find T.

Show hint

Form $\frac mk$ first.

Learning goal: Find oscillation period from mass and spring stiffness.

Show solution and calculation

$T=2\pi\sqrt{\frac mk}=2\pi\sqrt{0.004\,\mathrm{s^2}}\approx0.3974\,\mathrm s$ .

Check: $k=\frac{4\pi^2m}{T^2}=50\,\mathrm{N\,m^{-1}}$ .

\begin{aligned}T&=2\pi\sqrt{\frac mk}\\&=2\pi\sqrt{\frac{0.20}{50}}\,\mathrm s\approx0.3974\,\mathrm s\end{aligned}

Calculate · Application

$\omega_0=12\,\mathrm{rad/s}$ , $x_0=0.030\,\mathrm m$ and $v_0=0.480\,\mathrm{m/s}$ . Find A.

Show hint

Initial velocity is nonzero; include both contributions.

Learning goal: Reconstruct amplitude from initial displacement and velocity.

Show solution and calculation

$\frac{v_0}{\omega_0}=0.040\,\mathrm m$ .
Thus $A=\sqrt{x_0^2+\left(\frac{v_0}{\omega_0}\right)^2}=\sqrt{0.030^2+0.040^2}\,\mathrm m=0.050\,\mathrm m$ .
Amplitude exceeds initial displacement.

\begin{aligned}A&=\sqrt{x_0^2+\left(\frac{v_0}{\omega_0}\right)^2}\\&=\sqrt{0.030^2+\left(\frac{0.480}{12}\right)^2}\,\mathrm m\\&=0.050\,\mathrm m\end{aligned}

Interpret · Application

Mass doubles, stiffness halves and amplitude doubles. How do T and Eosc change?

Show hint

T scales with $\sqrt{\frac mk}$ , energy with $kA^2$ .

Learning goal: Distinguish period and energy scaling when parameters change together.

Show solution and calculation

$\frac{T_{\mathrm{new}}}{T_{\mathrm{old}}}=\sqrt{\frac{2}{\frac12}}=2$ .
$\frac{E_{\mathrm{new}}}{E_{\mathrm{old}}}=\frac12\cdot2^2=2$ .
Both assume the linear model remains applicable.

\begin{aligned}\frac{T_{\mathrm{new}}}{T_{\mathrm{old}}}&=\sqrt{\frac{m_{\mathrm{new}}k_{\mathrm{old}}}{m_{\mathrm{old}}k_{\mathrm{new}}}}=\sqrt{\frac2{\frac12}}=2\\\frac{E_{\mathrm{new}}}{E_{\mathrm{old}}}&=\frac{k_{\mathrm{new}}}{k_{\mathrm{old}}}\left(\frac{A_{\mathrm{new}}}{A_{\mathrm{old}}}\right)^2\\&=\frac12\cdot2^2=2\end{aligned}

Interpret · Application

An ideal horizontal spring oscillator has k=80 N/m and m=0.50 kg. At two passages through $x=+2.0\,\mathrm{cm}$ , velocity is positive once and negative once. Compare spring force, acceleration and motion direction in these states. In which state is speed increasing at that instant?

Show hint

Spring force depends on x, not on the sign of velocity. Speed increases when force and velocity point in the same direction.

Learning goal: Interpret an instantaneous state using displacement and velocity.

Show solution and calculation

Approach: obtain directed spring force from displacement, then find acceleration.
Both states have −1.6 N and −3.2 m/s².
Positive velocity means motion away from equilibrium and decreasing speed.
Negative velocity means motion toward equilibrium and increasing speed.

Check: power is positive in the latter state because force and velocity are both negative.

\begin{aligned}F_x&=-kx=-(80\,\mathrm{N/m})(0.020\,\mathrm m)=-1.6\,\mathrm N\\a_x&=\frac{F_x}{m}=-3.2\,\mathrm{m/s^2}\\P&=F_xv_x\end{aligned}

[1][2]

Error analysis · Application

A 0.40 kg mass hangs from an ideal vertical spring with k=100 N/m; g=10 m/s² and downward is positive. It is presently 5.0 cm below the unloaded spring position. A solution uses $a=-\frac{k(0.050\,\mathrm m)}m=-12.5\,\mathrm{m/s^2}$ . Correct this including gravity. How far is the mass from equilibrium?

Show hint

Determine static spring extension first. The oscillation equation uses the additional displacement relative to that equilibrium.

Learning goal: Separate displacement from equilibrium and spring extension in a vertical oscillator.

Show solution and calculation

Approach: balance gravity and spring force at the actual position.
Static extension is 4.0 cm, so the mass is 1.0 cm below equilibrium.
Net force is −1.0 N, giving −2.5 m/s².

Check: at equilibrium gravity and spring force must balance; the incorrect calculation would still predict acceleration there.

\begin{aligned}s_{\mathrm{eq}}&=\frac{mg}{k}=0.040\,\mathrm m\\x&=s-s_{\mathrm{eq}}=0.010\,\mathrm m\\a&=\frac{mg-ks}{m}=-\frac{kx}{m}=-2.5\,\mathrm{m/s^2}\end{aligned}

[1]

In brief

Choose equilibrium first. Stiffness and oscillating mass determine angular frequency; initial position and velocity determine amplitude and phase. Zero force does not mean zero velocity. Conserved oscillation energy alternates between kinetic and effective elastic potential energy about equilibrium.[1][2]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Clarified period, frequency, equilibrium reference and amplitude condition directly; exposed existing measurement and initial-push examples and linked appropriate spring, pendulum and damping models.
October 1, 2026Clarified equilibrium reference, force, and motion; added an equilibrium-start push example and a quantitative period graph. Preserved previous motion and measurement examples and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

Cite this reference page

Quanta. Spring Pendulum (Oscillation Period). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/harmonische-schwingung-federpendel

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Frequently asked questions about Spring Pendulum (Oscillation Period)

How do you calculate the period of a spring pendulum?+

For an ideal linear undamped spring oscillator use $T=2\pi\sqrt{\frac mk}$ . Enter total effective moving mass in kilograms and stiffness in newtons per metre to obtain seconds. With $m = 0.25 kg$ and $k =\frac{ 100 N}{m}$ the period is about 0.314 s. Reversing $k=\frac{4\pi^2m}{T^2}$ must return $\frac{100 N}{m}$ . Displacement is measured from equilibrium, which for a vertical spring is different from its unloaded length. A heavy spring, moving holders, nonlinear extension or appreciable damping can require an extended model. A complete period ends at the same position with the same direction of motion.

Why does the period not depend on the amplitude?+

Amplitude independence follows from the linear equation $m\ddot x=-kx$ : substituting a cosine solution cancels amplitude and leaves $\omega_0^2=\frac km$ . Larger displacement creates a larger restoring force but also a proportionally longer path. This is a property of that model, not a statement about every real spring. Doubling amplitude doubles maximum speed $A\omega_0$ and quadruples energy $\frac12kA^2$ while keeping T unchanged. If stiffness varies with extension, motion reaches a stop, or resistance grows nonlinearly, the actual period change needs a new calculation or measurement. Always state the linear range when applying amplitude independence to experimental data.

How do you determine the spring constant from an oscillation measurement?+

Rearrange the ideal period equation to $k=\frac{4\pi^2m}{T^2}$ . Time several complete cycles and divide elapsed time by their number. Include moving hooks and holders in the mass; a heavy spring can add effective inertia. For 0.480 kg and 20 periods in 7.54 s, $T = 0.377 s$ and $k \approx{}\frac{ 133.33 N}{m}$ . A static cross-check uses $k=\frac{mg}y$ with extension from unloaded length. If estimates disagree beyond their uncertainties, examine moving mass, cycle counting, spring linearity and damping. Substituting the inferred k back into the period equation checks the algebra, but does not independently prove experimental accuracy.

What is the difference between a spring pendulum and a simple pendulum?+

A spring oscillator obtains its linear restoring force from elastic stiffness and ideally has $T=2\pi\sqrt{\frac mk}$ . A simple pendulum is restored by tangential weight $-mg\sin\theta$ ; small angles give $T_0=2\pi\sqrt{\frac lg}$ . Spring period therefore depends on moving mass, whereas mass cancels from the ideal pendulum equation. Amplitude independence for the pendulum is only a small-angle approximation. A vertical spring needs g to locate equilibrium, but with constant stiffness g does not set its ideal natural period. Both models need the appropriate constraints and initially neglect damping and external driving. Using one period equation for the other mechanism would mix physically different restoring forces.

How does the energy move in an oscillating spring pendulum?+

In the ideal undamped model, oscillation energy $E_{\mathrm{osc}}=\frac12mv^2+\frac12kx^2=\frac12kA^2$ is conserved, with x measured from equilibrium. At turning points velocity is zero and varying energy is potential; at equilibrium crossing it is kinetic. A vertical spring remains extended by $\frac{mg}k$ there, so it is not relaxed. Effective potential $\frac12kx^2$ then includes spring and gravitational energy relative to equilibrium. With $k =\frac{ 144 N}{m}$ and $A = 0.025 m$ , $E_{\mathrm{osc}}=0.045\,\mathrm J$ . At $x=\frac A{\sqrt2}$ , kinetic and potential contributions are each 0.0225 J. Their sum checks the calculation without confusing oscillation energy with all absolute stored spring energy.

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[1]OpenStax, University Physics Vol. 1, §15.1 Simple Harmonic Motion

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Mass-spring oscillator: period and harmonic motion

Variables & units

Spring oscillator period calculator

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Harmonic oscillations

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How do you calculate with Spring Pendulum (Oscillation Period)?

The spring pulls back while the mass keeps moving

Which equilibrium position and linear range apply?

How do you calculate and derive the spring oscillator’s period?

Where are speed, acceleration and energy maximal?

A full period starts at a turning point

How do period, frequency, angular frequency and mass relate?

Example: a complete motion state

A horizontal oscillator released from rest

How do you measure spring stiffness statically and dynamically?

Comparing dynamic and static estimates

What do damping and periodic driving change?

How do you find amplitude when starting at equilibrium with an initial push?

Starting at equilibrium with an initial push

Five solved exercises

m = 0.20 kg and k = 50 N/m. Find T.

ω0=12 rad/s\omega_0=12\,\mathrm{rad/s}ω0​=12rad/s, x0=0.030 mx_0=0.030\,\mathrm mx0​=0.030m and v0=0.480 m/sv_0=0.480\,\mathrm{m/s}v0​=0.480m/s. Find A.

Mass doubles, stiffness halves and amplitude doubles. How do T and Eosc change?

An ideal horizontal spring oscillator has k=80 N/m and m=0.50 kg. At two passages through x=+2.0 cmx=+2.0\,\mathrm{cm}x=+2.0cm, velocity is positive once and negative once. Compare spring force, acceleration and motion direction in these states. In which state is speed increasing at that instant?

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$\omega_0=12\,\mathrm{rad/s}$ , $x_0=0.030\,\mathrm m$ and $v_0=0.480\,\mathrm{m/s}$ . Find A.

An ideal horizontal spring oscillator has k=80 N/m and m=0.50 kg. At two passages through $x=+2.0\,\mathrm{cm}$ , velocity is positive once and negative once. Compare spring force, acceleration and motion direction in these states. In which state is speed increasing at that instant?