Simple Harmonic Motion Period

Equivalent forms

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

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

T = \frac{2\pi}{\omega}

Period is independent of amplitude — a deep result that makes clocks possible.

Symbol	Name	SI unit	Dimension	Range
$T$	Periodoutput Time for one complete oscillation	s	T	0.01 – 60
$m$	Mass Mass of the oscillating object	kg	M	0.01 – 100
$k$	Spring Constant Stiffness of the spring	N/m	MT⁻²	1 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$T$	s	s	s
$m$	kg	g	slug
$k$	N/m	dyn/cm	lbf/in

Where it holds

Valid for ideal springs obeying Hooke's law (small oscillations). Breaks down for large amplitudes, damped systems, or nonlinear restoring forces.

Dimensional analysis

T \to [T]

2\pi \sqrt{m/k} \to \sqrt{[M]/[MT^{-2}]} = \sqrt{[T^{2}]} = [T]

Both sides are

[T] =

seconds.

2\pi is

dimensionless.

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Discovery

Robert Hooke / Christiaan Huygens · 1673

Huygens analyzed pendulum motion in Horologium Oscillatorium; combined with Hooke's spring law, the SHM period formula emerged.

Try this

A 0.5 kg mass on a spring bounces with k = 200 N/m. How fast does it oscillate?

T = 2π√(m/k) = 2π√(0.5/200) = 2π × 0.05 = 0.314 s. That's about 3.2 oscillations per second — faster than you'd expect!

Research status: stable

Real-world applications

Quartz watch timekeeping
Seismometer design
Vibration isolation in buildings
Musical instrument tuning

Common misconceptions

Period does NOT depend on amplitude (for ideal SHM)
SHM is not limited to springs — pendulums, LC circuits, and molecular vibrations all exhibit it
Damping reduces amplitude but barely affects period (for light damping)

Experimental verification

Spring-mass lab experiments, quartz crystal oscillator frequency measurements, MEMS resonator calibration.

Derivation

From

F = -kx

and

F = ma

:

ma = -kx \to a = -(k/m)x

.

This is the SHM equation

x'' + \omega ^{2}x = 0

with

\omega = \sqrt{k/m}

.

The period

T = 2\pi /\omega = 2\pi \sqrt{m/k}

.

Limiting cases

m \to 0

⟶

T \to 0

A massless system oscillates infinitely fast (theoretical limit).

k \to \infty

⟶

T \to 0

An infinitely stiff spring snaps back instantly.

k \to 0

⟶

T \to \infty

No restoring force means the system never returns — period is infinite.

What if…

What if mass quadruples?

Period doubles $(\sqrt{4} = 2)$ . A 2 kg mass on the same 200 N/m spring: $T = 2\pi \sqrt{2/200} = 0.628\,\mathrm{s}$ .

What if you add damping?

Period barely changes for light damping: T_damped $= T/\surd (1 - (b/2m\omega )^{2})$ . Amplitude decays exponentially, but frequency is nearly the same.

What if amplitude doubles?

Period stays exactly the same — this is the key property of SHM. Amplitude independence is what makes pendulum clocks reliable.

1

Spring-mass oscillator

Given ·

m:: 0.5
k:: 200

Find · T and f

Steps

$T = 2\pi \sqrt{m/k} = 2\pi \sqrt{0.5/200}$
$\sqrt{0.0025} = 0.05$
$T = 2\pi \times 0.05 = 0.314\,\mathrm{s}$
$f = 1/T = 1/0.314 = 3.18$ oscillations per second

Result ·

T = 2\pi \sqrt{0.5/200} = 2\pi \times 0.05 = 0.314\,\mathrm{s} \to f = 3.18\,\mathrm{Hz}

2

Designing a 1 Hz oscillator

Given ·

f target:: 1
m:: 0.25

Find · k

Steps

Target: $T = 1\,\mathrm{s}$ , $so \omega = 2\pi rad/s$
From $T = 2\pi \sqrt{m/k}$ : $k = m(2\pi /T)^{2} = m\cdot \omega ^{2}$
$k = 0.25 \times (2\pi )^{2} = 0.25 \times 39.48 = 9.87\,\mathrm{N/m}$
A relatively soft spring for a 250 g mass

Result ·

k = m(2\pi f)^{2} = 0.25 \times (2\pi )^{2} = 0.25 \times 39.48 = 9.87\,\mathrm{N/m}

Hooke's Law→Kinetic Energy→Centripetal Acceleration→