Physics · Oscillations and waves

Spring Pendulum (Oscillation Period)

The oscillation period of a spring pendulum depends only on mass and spring constant, not on the amplitude.

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Formula

LaTeX: T = 2\pi \sqrt{\frac{m}{k}}
T in seconds [s] · m in kg · k in N/m
Diagram: a cosine curve of displacement x over time t; the amplitude A and the period T between two crests are marked.txAT
The harmonic oscillation with amplitude A and period T, which depends only on mass and spring constant.

Variables & units – Spring Pendulum (Oscillation Period)

SymbolMeaningUnit
TOscillation periods
mOscillating masskg
kSpring constantN/m

Derivation & background – Spring Pendulum (Oscillation Period)

The restoring force F = −k·x (Hooke) leads to the differential equation m·ẍ = −k·x with the solution x(t) = A·cos(ω·t), ω = √(k/m). From this follows T = 2π/ω = 2π·√(m/k). Remarkably, T is independent of the amplitude (isochronism) and, unlike the simple pendulum, also independent of location, since g does not appear.

Exam blueprint

Validity range

Holds as long as the spring follows Hooke law (elastic range) and friction is negligible. With strong damping the amplitude decays and the period shifts slightly.

Derivation steps

The restoring spring force is proportional to the displacement, which forces a sinusoidal oscillation.

  1. 1Newton + Hooke: m·ẍ = −k·x, solved by x(t) = A·cos(ωt) with ω = √(k/m).
  2. 2From T = 2π/ω follows T = 2π·√(m/k).

Rearrangements

Spring constant from the period

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

Mass from the period

The principle of mass measurement devices in weightlessness.

Frequency

A stiffer spring or smaller mass raises the frequency.

Task variant

A mass of 0.5 kg oscillates with T = 1 s. Find the spring constant k.

k = 4π²·m/T² = 4 × 9.87 × 0.5 / 1 ≈ 19.7 N/m.

How does T change when the mass is quadrupled?

T ∝ √m: quadrupling the mass doubles the period.

Common mistakes

Assuming a larger amplitude lengthens the period.

T is independent of amplitude (isochronism); only m and k matter.

Using the simple pendulum formula T = 2π√(l/g).

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.

These mistakes cost points in real exams. The set drills them until they stick.

Formula cluster

Harmonic oscillations

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

Worked example

A mass m = 0.25 kg hangs on a spring with k = 100 N/m: T = 2π × √(0.25/100) = 2π × 0.05 ≈ 0.31 s.

Applications

Tuned mass dampers in skyscrapers, vehicle suspension, force measurement, watchmaking (balance wheel)

Quanta exam set

Curated exam set for "Spring Pendulum (Oscillation Period)":

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Which formula describes Spring Pendulum (Oscillation Period)?

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How do you rearrange T = 2π·√(m/k) for Spring constant from the period?

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Which common mistake happens with Spring Pendulum (Oscillation Period)?

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Scientific sources

Common notations & search queries

T=2*pi*sqrt(m/k)T=2π√(m/k)Schwingungsdauer FederpendelFederpendel Formelharmonische Schwingung FormelPeriodendauer Federspring pendulum periodEigenfrequenz Feder

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Frequently asked questions about Spring Pendulum (Oscillation Period)

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

Insert mass and spring constant into T = 2π·√(m/k). Example: a mass of 0.25 kg on a spring with k = 100 N/m oscillates with T = 2π·√(0.25/100) = 2π·0.05 ≈ 0.31 s, a good three oscillations per second. The mass sits in the numerator under the root: more mass makes the oscillation more sluggish and slower. The spring constant is in the denominator: a stiffer spring pulls back harder and shortens the period. Because of the root, both influences are damped; four times the mass only doubles T. The frequency is the reciprocal f = 1/T, here about 3.2 Hz.

Why does the period not depend on the amplitude?+

This is the isochronism of harmonic oscillation, and it follows from the linear force law F = −k·x. With a larger displacement the path per oscillation gets longer, but the restoring force grows in exactly the same proportion, so the body moves correspondingly faster. Both effects compensate completely, and every oscillation takes the same time. Mathematically, the amplitude A appears in the solution x(t) = A·cos(ωt) only as a prefactor, while ω = √(k/m) depends solely on system properties. This property made springs and pendulums the timekeepers of clockmaking. It holds only while the spring stays in the linear Hooke range; overstretching makes the oscillation anharmonic.

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

Rearrange the period formula for k: k = 4π²·m/T². Measure the time for many oscillations and divide; ten oscillations in 10 s mean T = 1 s. Example: with m = 0.5 kg and T = 1 s, k = 4π² × 0.5 / 1 ≈ 19.7 N/m. Timing over many periods greatly reduces the stopwatch error, the most important practical tip in the experiment. Alternatively, the static extension gives the same k: attach a known mass and measure the stretch, k = m·g/x. If both methods agree, that is a nice confirmation of the model, a standard evaluation in physics labs and a popular exam task.

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

For the spring pendulum T = 2π·√(m/k) holds: the period depends on mass and spring constant but not on location; g does not appear, and the formula holds unchanged on the Moon. For the simple pendulum T = 2π·√(l/g) holds: the period depends on string length and local gravity but, surprisingly, not on mass. The reason: for the simple pendulum gravity provides the restoring force, which is itself proportional to the mass, so the mass cancels. Moreover, the simple pendulum is harmonic only for small angles (below about 10°), while the spring pendulum oscillates exactly harmonically throughout the Hooke range. Mixing up the two formulas is one of the most common exam mistakes in the oscillations chapter.

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

It swings losslessly between two forms: at the turning points everything is elastic energy of the spring, E = ½kA² with amplitude A, and the speed is zero. At the equilibrium crossing the spring is relaxed and all energy is kinetic, E = ½mv_max²; there the body is fastest, with v_max = ω·A. Equating the two gives the maximum speed directly: ½kA² = ½mv_max². Example: k = 100 N/m, m = 0.25 kg, A = 0.04 m gives v_max = A·√(k/m) = 0.04 × 20 = 0.8 m/s. Real pendulums lose a little energy per cycle to friction; the amplitude decays while the period stays nearly unchanged.

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Create a curated FSRS exam set for T = 2π·√(m/k): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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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. 1

    Task

    A mass of 0.5 kg oscillates with T = 1 s. Find the spring constant k.

    Solution path

    k = 4π²·m/T² = 4 × 9.87 × 0.5 / 1 ≈ 19.7 N/m.

  2. 2

    Task

    How does T change when the mass is quadrupled?

    Solution path

    T ∝ √m: quadrupling the mass doubles the period.