Period Formula Physics: T = 1/f, Pendulum and Spring Period

The period formula physics students learn first is $T = \frac{1}{f}$ : the time period of an oscillation is one divided by its frequency. If a swing goes back and forth twice every second, each full swing takes half a second. The SI unit of period is the second (s).

For particular systems the period can be worked out from their physical properties. A simple pendulum has $T = 2\pi\sqrt{\frac{L}{g}}$ and a mass on a spring has $T = 2\pi\sqrt{\frac{m}{k}}$ . These are the forms behind the search for the formula of time period in Class 11 oscillations and in AP Physics 1. The calculator on this page uses the pendulum form, with g = 9.8 m/s².

Quick facts

General formula
$T = \frac{1}{f} = \frac{2\pi}{\omega}$
Simple pendulum
$T = 2\pi\sqrt{\frac{L}{g}}$
Spring and mass
$T = 2\pi\sqrt{\frac{m}{k}}$
SI unit
second (s)
Dimensional formula
[M⁰ L⁰ T¹]
Type
Scalar
From a stopwatch
$T = \frac{t}{N}$ : total time for N oscillations divided by N

What the period formula physics books give means

Quantities that set the time period
Symbol	Quantity	SI unit
$T$	Time period, the time for one complete cycle	s
$f$	Frequency, cycles per second	Hz
$\omega$	Angular frequency	rad/s
$L$	Length of the pendulum, pivot to centre of the bob	m
$g$	Acceleration due to gravity, 9.8 m/s²	m/s²
$m$	Mass attached to the spring	kg
$k$	Spring constant	N/m

A complete cycle means the object returns to the same position moving in the same direction. For a pendulum that is from the far left, to the far right and back to the far left. Passing the centre twice is still only one cycle.

Physics period formula for common oscillators

Each standard oscillator has its own result. They all come from the same idea: in simple harmonic motion $\omega^2$ equals restoring force per unit displacement divided by mass, and $T = \frac{2\pi}{\omega}$ .

Time periods to know for exams
System	Formula	Notes
Simple pendulum	$T = 2\pi\sqrt{\frac{L}{g}}$	Small swings only, under about 10°
Mass on a spring	$T = 2\pi\sqrt{\frac{m}{k}}$	Same on a table or hanging vertically
Wave at a point	$T = \frac{\lambda}{v}$	Time for one wavelength to pass
Uniform circular motion	$T = \frac{2\pi r}{v}$	Time for one revolution
Satellite in orbit	$T = 2\pi\sqrt{\frac{r^3}{GM}}$	Kepler's third law form

Look at what is missing. The bob's mass never enters the pendulum result, and nor does the amplitude while the swing stays small. Gravity never enters the spring result. These facts are favourite one mark questions. The satellite result is worked out in detail on the time period of satellite page.

Worked examples for the formula of time period

Example 1: from frequency

A tuning fork vibrates at 256 Hz. Find the time period of its vibration.

$T = \frac{1}{f} = \frac{1}{256} = 3.91 \times 10^{-3}\ \text{s}$

That is about 3.9 milliseconds per vibration.

Example 2: a simple pendulum

A pendulum is 1.0 m long. Find its period near Earth's surface.

Data: pendulum length 1.0 m, and take g as 9.8 m/s².

$T = 2\pi\sqrt{\frac{1.0}{9.8}} = 2\pi \times 0.3194 = 2.0\ \text{s}$

A metre long pendulum ticks once each way per second, a handy fact to remember.

Example 3: a spring with mass in grams

A 500 g block hangs from a spring of constant 200 N/m. Find the period of its vertical oscillations.

Convert: m = 500 g = 0.50 kg.

$T = 2\pi\sqrt{\frac{0.50}{200}} = 2\pi\sqrt{0.0025} = 2\pi \times 0.050 = 0.31\ \text{s}$

Example 4: timing in a lab

A student times 20 oscillations of a pendulum and gets 36.0 s. Find T and the length of the pendulum.

T = 36.0 ÷ 20 = 1.80 s. Rearranging the pendulum formula:

$L = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 1.80^2}{4\pi^2} = \frac{31.75}{39.48} = 0.80\ \text{m}$

Timing many swings and dividing reduces the effect of your reaction time.

Rearranging and checking with dimensions

Period is simply a duration, and a duration written with all three base letters reads [M⁰ L⁰ T¹]. Check the pendulum form: $\sqrt{L/g}$ has dimensions $\sqrt{\text{L} / \text{L}\,\text{T}^{-2}} = \text{T}$ , which works. For the spring, N/m is kg/s², so $\sqrt{m/k}$ is $\sqrt{\text{kg} \cdot \text{s}^2/\text{kg}}$ , again seconds.

Common rearrangements: $f = \frac{1}{T}$ , $L = \frac{gT^2}{4\pi^2}$ , $g = \frac{4\pi^2 L}{T^2}$ and $k = \frac{4\pi^2 m}{T^2}$ . The g form is how a school lab measures gravity with a pendulum. Plot $T^2$ against L: you get a line that starts at zero and climbs with gradient $\frac{4\pi^2}{g}$ .

Traps with time period questions

Forgetting the square root. Doubling the length of a pendulum does not double T; it multiplies T by √2 ≈ 1.41.
Using grams in the spring formula. Mass must be in kilograms for T in seconds.
Measuring L to the bottom of the bob. The length runs from the pivot to the centre of the bob.
Assuming a heavier bob swings more slowly. Mass cancels out for a pendulum.
Mixing up T and t. Capital T is the time for one cycle; small t is any time on the stopwatch.

Period and frequency are two views of the same thing. The formula of frequency page covers the reverse direction, and the waves and optics hub shows how T links to wavelength and wave speed. The amplitude of an ideal oscillator does not affect its period at all.

Frequently asked questions

What is the formula of time period?

In general T = 1/f, or T = 2π/ω. For a simple pendulum T = 2π√(L/g), and for a mass on a spring T = 2π√(m/k).

What is the SI unit of time period?

The second (s). It is the time taken for one complete oscillation or one full cycle of a wave.

Does the period of a pendulum depend on mass?

No. For a simple pendulum with small swings the period depends only on the length and on g. A heavy bob and a light bob of the same length swing together.

What is the relation between period and frequency?

They are reciprocals: T = 1/f and f = 1/T. Their product is always 1.

How does the period change on the Moon?

A pendulum swings more slowly there because g is about one sixth of its value on Earth. Since T ∝ 1/√g, the period becomes about √6 ≈ 2.45 times longer. A spring period stays the same.