Wave Speed, Frequency, and Wavelength: How to Find Each

Wave speed equals frequency times wavelength (v = f lambda). Rearrange for what you need: f = v / lambda, lambda = v / f, and T = 1 / f links period and frequency. When a wave enters a new medium, its speed and wavelength change but its frequency stays the same.

AP Physics: Unit 14 (topics 14.1 Properties of Wave Pulses and Waves, 14.2 Periodic Waves). This is AP Physics 2 material: Unit 14 (Waves, Sound, and Physical Optics) carries 12-15% of the multiple-choice section, and Topics 14.1 and 14.2 cover wave properties and periodic waves. The oscillators that drive waves appear in AP Physics 1 Unit 7.

The wave speed formula

Wave speed equals frequency times wavelength:

v = f\lambda

Here $v$ is wave speed in meters per second (m/s), $f$ is frequency in hertz (Hz), and $\lambda$ (lambda) is wavelength in meters (m). One hertz means one complete cycle per second, and wavelength is the distance between two identical points on neighboring cycles, crest to crest or trough to trough.

The logic behind the formula: if $f$ waves pass a point each second, and each wave is $\lambda$ meters long, then $f\lambda$ meters of wave move past that point every second. That is a speed.

The same equation covers every periodic wave: sound, water waves, waves on a string, and light. Only the typical numbers change. Sound in room-temperature air moves at about 343 m/s, while light in a vacuum moves at $3.0 \times 10^8$ m/s.

How to find frequency

Divide wave speed by wavelength:

f = \frac{v}{\lambda}

If a wave moves at 343 m/s and its wavelength is 0.50 m:

f = \frac{343 \text{ m/s}}{0.50 \text{ m}} = 686 \text{ Hz}

The units work out because meters cancel, leaving 1/s, which is the definition of a hertz.

Two other routes to frequency show up constantly:

From the period: $f = 1/T$ . If one cycle takes 0.020 s, then $f = 1/0.020 = 50$ Hz.
From counting: frequency is cycles divided by time. If 30 crests pass a dock post in 10 s, then $f = 30/10 = 3.0$ Hz.

Pick the route that matches the information the problem hands you. Given speed and wavelength, rearrange the wave speed formula. Given anything about time per cycle, start from the period.

How to find wavelength

Divide wave speed by frequency:

\lambda = \frac{v}{f}

A radio station broadcasting at 100 MHz ( $1.0 \times 10^8$ Hz) sends out electromagnetic waves traveling at $3.0 \times 10^8$ m/s, so

\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{1.0 \times 10^8 \text{ Hz}} = 3.0 \text{ m}

If you know the period instead of the frequency, multiply: $\lambda = vT$ . A wave traveling at 3.0 m/s with a 4.0 s period has $\lambda = (3.0 \text{ m/s})(4.0 \text{ s}) = 12$ m.

Sanity checks help here. At a fixed speed, higher frequency always means shorter wavelength, because the two multiply to the same $v$ . Audible sound in air runs from about 17 m (20 Hz) down to 1.7 cm (20,000 Hz), and visible light sits near half a micrometer. If your answer lands far outside the sensible range for that wave type, recheck the unit conversions first.

Period, T = 1/f, and reading graphs

Period $T$ is the time for one complete cycle, measured in seconds. Frequency counts cycles per second. They are reciprocals:

T = \frac{1}{f} \qquad f = \frac{1}{T}

A 100 Hz wave has a period of $1/100 = 0.010$ s. Substituting $f = 1/T$ into the wave speed formula gives a second useful version, $v = \lambda / T$ : the wave travels one wavelength per period.

Graphs are where students lose easy points. A displacement vs. position graph is a snapshot of the wave at one instant, so the distance between repeats is the wavelength. A displacement vs. time graph tracks one point in the medium as time passes, so the spacing between repeats is the period. The axis labels tell you which one you are reading.

The source of most periodic waves is an oscillator, often in simple harmonic motion, and the wave's frequency matches the oscillator's frequency.

What changes when a wave enters a new medium

Frequency is set by the source and does not change at a boundary. Speed is set by the medium. Wavelength adjusts to keep $v = f\lambda$ true.

Here is why frequency cannot change: wave fronts arrive at the boundary at some rate, and every front that arrives must continue on. If $5.0 \times 10^{14}$ fronts hit the boundary each second, $5.0 \times 10^{14}$ fronts enter the new medium each second. Anything else would mean waves piling up or being created at the boundary, which does not happen.

So when light passes from air into water, it slows down, and since $f$ is fixed, $\lambda = v/f$ shrinks by the same factor. When a wave on a thin string crosses onto a heavier rope, it also slows and its wavelength compresses. Memorize the pattern: a new medium changes $v$ and $\lambda$ , never $f$ .

Units, common mistakes, and the AP angle

Convert everything to base SI units before substituting:

Frequency in hertz (1 kHz = 1000 Hz, 1 MHz = $10^6$ Hz)
Wavelength in meters (1 nm = $10^{-9}$ m, 1 cm = 0.01 m)
Speed in m/s

The mistakes that cost the most points:

Plugging a wavelength in nanometers straight into $v = f\lambda$ without converting
Reading a wavelength off a displacement vs. time graph (that spacing is the period)
Assuming a taller or louder wave is faster (amplitude does not affect wave speed)
Changing frequency at a boundary instead of wavelength

On the AP side, this is AP Physics 2 territory: Unit 14 (Waves, Sound, and Physical Optics) is worth 12-15% of the multiple-choice section, and Topics 14.1 and 14.2 cover exactly this material. Keep the AP Physics 2 formula sheet open while you practice so the symbols become automatic.

Wavelength of a 440 Hz tone in air

A speaker plays a 440 Hz tone (concert A). Sound travels at 343 m/s in room-temperature air. Find the wavelength.

List knowns: $f = 440$ Hz, $v = 343$ m/s. Unknown: $\lambda$ .
Rearrange $v = f\lambda$ to solve for wavelength: $\lambda = \frac{v}{f}$ .
Substitute with units: $\lambda = \frac{343 \text{ m/s}}{440 \text{ Hz}} = 0.7795 \text{ m}$ .
Round to three sig figs: $\lambda = 0.780$ m.

$\lambda = 0.780$ m, about 78 cm. Middle-of-the-piano sound waves in air are roughly door-width long.

Speed and frequency of water waves from the period

Wave crests on a lake are 12 m apart. A buoy bobs through one full up-and-down cycle every 4.0 s. Find the wave's frequency and speed.

The buoy's cycle time is the period, $T = 4.0$ s. The crest-to-crest distance is the wavelength, $\lambda = 12$ m.
Convert period to frequency: $f = \frac{1}{T} = \frac{1}{4.0 \text{ s}} = 0.25$ Hz.
Apply the wave speed formula: $v = f\lambda = (0.25 \text{ Hz})(12 \text{ m}) = 3.0$ m/s.
Check with the period form: $v = \lambda / T = 12 \text{ m} / 4.0 \text{ s} = 3.0$ m/s. Same answer, good.

$f = 0.25$ Hz and $v = 3.0$ m/s.

Light entering water: what actually changes

Orange light with a wavelength of 600 nm travels through air at $3.00 \times 10^8$ m/s, then enters water, where it slows to $2.26 \times 10^8$ m/s. Find its frequency, then its wavelength in the water.

Convert the wavelength: 600 nm = $6.00 \times 10^{-7}$ m.
Find the frequency in air: $f = \frac{v}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{6.00 \times 10^{-7} \text{ m}} = 5.00 \times 10^{14}$ Hz.
Frequency does not change at the boundary, so in water $f$ is still $5.00 \times 10^{14}$ Hz.
Find the new wavelength from the new speed: $\lambda = \frac{v}{f} = \frac{2.26 \times 10^8 \text{ m/s}}{5.00 \times 10^{14} \text{ Hz}} = 4.52 \times 10^{-7} \text{ m} = 452$ nm.

The frequency stays at $5.00 \times 10^{14}$ Hz in both media. The wavelength shrinks from 600 nm to about 452 nm because the wave slowed down.

Frequently asked questions

Does frequency change when a wave enters a new medium?

No. Frequency is set by the source, and every wave front that reaches the boundary continues into the new medium at the same rate. The wave's speed changes because the medium changed, and the wavelength adjusts (lambda = v/f) so the wave speed formula still holds.

What is the difference between period and frequency?

They are reciprocals: T = 1/f. Period is the time for one complete cycle, in seconds. Frequency is the number of cycles per second, in hertz. A 50 Hz wave has a period of 1/50 = 0.020 s. If you know one, you always know the other.

Does a bigger amplitude make a wave faster?

No. Wave speed depends only on the medium: the tension and mass per length of a string, the temperature of air for sound, and so on. Amplitude is related to the wave's energy, not its speed. A loud sound and a quiet one arrive at the same time.

How do I find wavelength or period from a graph?

Check the horizontal axis first. On a displacement vs. position graph (a snapshot), the repeat distance is the wavelength. On a displacement vs. time graph (one point tracked over time), the repeat spacing is the period. Then connect them with v = f lambda or v = lambda/T.

Which AP course covers the wave speed formula?

AP Physics 2. Waves, sound, and physical optics make up Unit 14, worth 12-15% of the multiple-choice section, and Topics 14.1 and 14.2 cover wave properties and periodic waves. AP Physics 1 covers the related oscillations material in its Unit 7.