Wave parameters — amplitude, wavelength, frequency, period, wave speed: Worked examples

Every quantity we use here was already earned in the parent note. As a one-line reminder, so no symbol appears unexplained:

The scenario matrix

Every wave-parameter problem is one of these cells. The examples below are tagged with the cell(s) they cover.

#	Cell (case class)	What's tricky about it	Example
C1	Direct — given $f, λ$ → find $v$	just plug in	Ex 1
C2	Inverse — given $v, f$ → find $λ$	rearrange	Ex 2
C3	Same medium, change $f$	$v$ is FIXED, $λ$ moves	Ex 3
C4	Amplitude from a swing	peak-to-peak $= 2 A$ , must halve	Ex 4
C5	Period ↔ frequency reciprocal	$f = 1/ T$ , watch small/large numbers	Ex 4, Ex 5
C6	Unit-prefix trap (kHz, MHz, cm, nm)	convert before plugging in	Ex 6, Ex 7
C7	Two-route cross-check ( $v T$ vs $v / f$ )	confirm answer two ways	Ex 5
C8	Real-world word problem	dig the numbers out of a story	Ex 8
C9	Degenerate / limiting inputs ( $f \to 0$ , $λ \to \infty$ , $f \to \infty$ )	what breaks, what stays finite	Ex 9
C10	Exam twist — reading period off a graph, changing medium	$v$ does change when medium changes	Ex 10

Intuition Why there are no "negative" or "quadrant" cells

Unlike an angle (which can point into any of four quadrants and carries a sign), the five wave parameters are all non-negative magnitudes — you cannot have a $- 2 m$ wavelength or a $- 5 Hz$ frequency. So this matrix has no signed or quadrant cases. Two edge-of-the-sign-line situations still deserve a word, and we cover them explicitly:

Zero values — e.g. $f \to 0$ (no wave at all): handled in cell C9 / Example 9.
Negative displacement: the displacement $y$ of a particle can be negative (below the middle line) — that is a phase inversion, worth naming. But that is a sign on $y$ , not on $A, λ, T, f, v$ . Amplitude $A$ is the maximum $∣ y ∣$ , always $\geq 0$ . We flag this in the note after Example 9 so no reader wonders where the "minus signs" went.

Example 1 — Direct (cell C1)

Worked example Given

$f$ and $λ$ , find $v$ A guitar string vibrates at $f = 200 Hz$ with wavelength $λ = 1.6 m$ . Find the wave speed $v$ .

Forecast: each of the $200$ bobs per second launches a crest that is $1.6 m$ long, so guess: a few hundred m/s.

Write the master equation. $v = f λ$ . Why this step? We want $v$ ; we hold the two things ( $f, λ$ ) that multiply to give it. This is the only equation that ties these exact three together.
Substitute the numbers. $v = 200 \times 1.6$ . Why this step? We replace the symbols with the given values so the unknown $v$ becomes a plain arithmetic product we can evaluate.
Compute. $v = 320 m/s$ . Why this step? Multiplying out turns the expression into the single final number the question asks for.

Verify: units are $Hz \times m = s^{- 1} \cdot m = m/s$ ✓. In one second, $200$ crests each move $1.6 m$ : $200 \times 1.6 = 320 m$ per second ✓.

Example 2 — Inverse (cell C2)

Example 3 — Same medium, change frequency (cell C3)

Worked example The "

$v$ is fixed" trap On the same string as Example 1 ( $v = 320 m/s$ ), you now play a harmonic at $f = 800 Hz$ . Find the new $λ$ .

Forecast: we made $f$ four times bigger. The medium didn't change, so $v$ can't change → $λ$ must shrink to a quarter: from $1.6$ down to about $0.4 m$ .

State what is fixed. Same string ⇒ same tension and density ⇒ $v = 320 m/s$ unchanged. Why this step? Because speed is set by the medium, not by how fast you drive it. See Speed of Waves on a String. This is the whole trick of the problem.
Rearrange for $λ$ . $λ = \frac{v}{f} = \frac{320}{800}$ . Why this step? We want $λ$ and now hold both $v$ (fixed) and the new $f$ , so we isolate $λ$ and drop the values in.
Compute. $λ = 0.40 m$ . Why this step? Carrying out the division gives the new wavelength for the higher harmonic.

Verify: ratio check — $f$ went ×4, so $λ$ should go ÷4: $1.6/4 = 0.40$ ✓. And $f λ = 800 \times 0.40 = 320 = v$ ✓.

Look at the figure: both curves fit into the same box (same $v$ -per-second travel), but the burnt-orange high-frequency wave crams more crests in, so each crest is shorter. Same speed, different $λ$ .

Example 4 — Amplitude from a swing, plus reciprocal (cells C4, C5)

Worked example Peak-to-peak and period

A cork on the sea moves through a total vertical swing of $0.90 m$ , taking $T = 3.0 s$ for one complete up-and-down. Find the amplitude $A$ and the frequency $f$ .

Forecast: amplitude is half the full swing (the classic mistake is not halving) → about $0.45 m$ . And a slow $3 s$ bob means a small frequency, well under $1 Hz$ .

Halve the swing. The full swing top-to-bottom is peak-to-peak $= 2 A$ , so $A = \frac{0.90}{2} = 0.45 m$ . Why this step? Amplitude is measured from the middle line, not tip to tip — the cork rises $A$ above and falls $A$ below the calm surface. See Simple Harmonic Motion for that middle-line picture.
Invert the period. $f = \frac{1}{T} = \frac{1}{3.0}$ . Why this step? Frequency is "bobs per second"; if one bob takes $T$ seconds you fit $1/ T$ bobs in a second.
Compute. $f = 0.33 Hz$ (to 2 s.f.; exactly $\frac{1}{3} Hz$ ). Why this step? Evaluating $1/3.0$ gives the numerical frequency; we keep the exact $\frac{1}{3}$ in mind so the verify below is clean.

Verify: $A$ is below the full swing ✓ (must be, since $A = \frac{1}{2} \times$ swing). And $f \cdot T = \frac{1}{3} \times 3 = 1$ ✓ — exactly one cycle per period, as it must be.

The figure marks the plum peak-to-peak span ( $0.90 m$ ) and the teal amplitude ( $0.45 m$ ) from the dashed middle line — one is literally double the other.

Example 5 — Two-route cross-check (cells C5, C7)

Worked example From

$T$ and $v$ , find $λ$ two ways A water wave has period $T = 0.50 s$ and speed $v = 6.0 m/s$ . Find $λ$ using both $λ = v T$ and $λ = v / f$ , and check they agree.

Forecast: in half a second the crest moves half of $6 m$ → about $3 m$ .

Route A: distance in one period. In one period the crest advances exactly one wavelength, so $λ = v T = 6.0 \times 0.50 = 3.0 m$ . Why this step? This is the definition-level route: distance $=$ speed $\times$ time, and the time for one wavelength is $T$ .
Route B: via frequency. First $f = 1/ T = 1/0.50 = 2.0 Hz$ , then $λ = v / f = 6.0/2.0 = 3.0 m$ . Why this step? To prove the two forms are the same identity, since $1/ f = T$ .
Compare. Both give $3.0 m$ . Why this step? Agreement of two independent routes is our confidence check that no algebra slipped.

Verify: the two routes are algebraically identical: $v T = v \cdot \frac{1}{f} = v / f$ ✓. Numeric match $3.0 = 3.0$ ✓.

Example 6 — Unit-prefix trap: radio (cell C6)

Worked example Megahertz and the speed of light

An FM radio station broadcasts at $f = 100 MHz$ . Radio waves are electromagnetic, so they travel at the speed of light $v = 3.0 \times 1 0^{8} m/s$ (see Electromagnetic Spectrum). Find $λ$ .

Forecast: enormous speed, high frequency — the metre-scale wavelengths of FM are famous ("a few metres"), so guess around $3 m$ .

Convert the prefix first. $100 MHz = 100 \times 1 0^{6} Hz = 1.0 \times 1 0^{8} Hz$ . Why this step? The equation only works in base SI units; "M" means mega $= 1 0^{6}$ . Forgetting this is the #1 slip here.
Rearrange for $λ$ and substitute. $λ = \frac{v}{f} = \frac{3.0 \times 1 0 ^{8}}{1.0 \times 1 0 ^{8}}$ . Why this step? We want $λ$ and hold $v, f$ , so isolate $λ$ ; putting the numbers in now turns it into a single division.
Compute. $λ = 3.0 m$ . Why this step? The powers of ten cancel, leaving the plain metre answer the question asks for.

Verify: $f λ = (1.0 \times 1 0^{8}) (3.0) = 3.0 \times 1 0^{8} = v$ ✓. Matches the forecast ✓.

Example 7 — Unit-prefix trap: light (cell C6)

Example 8 — Real-world word problem (cell C8)

Worked example Counting waves at a pier

Standing on a pier, you count 12 wave crests passing a post in exactly $30 s$ . The crests are spaced $8.0 m$ apart. How fast are the waves moving?

Forecast: roughly one crest every couple of seconds, each $8 m$ long → a few m/s.

Extract frequency from the story. $12$ crests in $30 s$ means $f = \frac{12}{30} = 0.40 Hz$ . Why this step? "Crests per second" is the frequency; the story hands us a count and a time.
Identify wavelength. Crest spacing = one full space-repeat = $λ = 8.0 m$ . Why this step? Crest-to-crest distance is the definition of $λ$ .
Apply the master equation. $v = f λ = 0.40 \times 8.0 = 3.2 m/s$ . Why this step? Now that both $f$ and $λ$ are pinned down, $v = f λ$ delivers the speed the question wants.

Verify: period $T = 1/ f = 2.5 s$ per crest; in $2.5 s$ a crest moves $λ = 8.0 m$ , giving $8.0/2.5 = 3.2 m/s$ ✓. Matches the forecast ✓.

Example 9 — Degenerate and limiting inputs (cell C9)

Worked example What happens at the extremes

For a fixed-medium wave with $v = 20 m/s$ , examine each limit and say what happens to $λ$ or $T$ . (a) $f \to 0$ (never wiggles). (b) $f \to \infty$ (infinitely fast wiggling). (c) $A \to 0$ .

Forecast: slower shaking → longer waves; faster shaking → shorter waves; zero amplitude → no wave at all, even though the maths for $λ$ still "works."

Limit (a): $f \to 0$ . $λ = v / f = 20/ f \to \infty$ ; equivalently $T = 1/ f \to \infty$ . Why this step? If the source never completes a cycle, the "next crest" is infinitely far away — the pattern stretches without bound. Physically: no wave (flat medium).
Limit (b): $f \to \infty$ . $λ = 20/ f \to 0$ and $T = 1/ f \to 0$ . Why this step? Crank frequency and, since $v$ is pinned by the medium, the only free quantity $λ$ is squeezed toward zero. Real media break this eventually (particle spacing sets a floor), but in the ideal formula $λ \to 0$ .
Limit (c): $A \to 0$ . $A$ appears in none of $v = f λ$ , $f = 1/ T$ . So $v, f, λ, T$ are all unaffected — but the wave carries zero displacement, i.e. it vanishes as a physical disturbance. Why this step? To show amplitude is an independent knob: it sets the wave's size/energy, never its speed or spacing.

Verify: $λ \cdot f = 20$ holds for every finite $f$ (e.g. $f = 0.001 \Rightarrow λ = 20000$ ; $f = 1 0^{6} \Rightarrow λ = 2 \times 1 0^{- 5}$ ) ✓. Amplitude absent from both relations ✓.

Example 10 — Exam twist: read period off a graph, then change medium (cell C10)

Worked example Graph-reading + a real speed change

A displacement-vs-time graph of one particle shows it returns to the same state every $0.20 s$ , and the wave (in medium 1) has speed $v_{1} = 40 m/s$ . (a) Find $f$ and $λ_{1}$ . (b) The wave now passes into medium 2 where its speed becomes $v_{2} = 60 m/s$ . The frequency stays the same (the source sets it). Find $λ_{2}$ .

Forecast: faster medium at the same frequency → longer wavelength, so $λ_{2} > λ_{1}$ .

Part (a).

Read $T$ from the graph. The pattern repeats every $0.20 s$ , so $T = 0.20 s$ . Why this step? A displacement–time graph repeats every period (the horizontal axis is time), never every wavelength. Confusing the two axes is the classic exam trap.
Get $f$ from $T$ . $f = \frac{1}{T} = \frac{1}{0.20} = 5.0 Hz$ . Why this step? Frequency is the reciprocal of period; inverting the read-off time gives bobs per second.
Wavelength in medium 1. $λ_{1} = \frac{v _{1}}{f} = \frac{40}{5.0} = 8.0 m$ . Why this step? We now hold $v_{1}$ and $f$ , so $λ = v / f$ gives the space-repeat length; substituting and dividing yields the number.

Part (b). 4. Decide what stays and what changes. The source keeps beating at the same rate, so $f = 5.0 Hz$ is unchanged; but we changed the medium, so $v$ genuinely changes to $v_{2} = 60 m/s$ . Why this step? Contrast with Example 3: there we kept the medium and $v$ was fixed while $f$ changed. Here we swap the medium, so $v$ moves while $f$ holds. Getting this "what is fixed?" call right is the whole point of the twist. 5. Wavelength in medium 2. $λ_{2} = \frac{v _{2}}{f} = \frac{60}{5.0} = 12 m$ . Why this step? Same relation $λ = v / f$ , now with the new speed and the unchanged frequency; dividing gives the new wavelength.

Verify: $f$ is identical in both media ✓. Ratio check: $\frac{λ _{2}}{λ _{1}} = \frac{12}{8} = 1.5 = \frac{v _{2}}{v _{1}} = \frac{60}{40}$ ✓ — wavelength scales with speed at fixed frequency, matching the forecast that $λ_{2} > λ_{1}$ . See Doppler Effect for the related case where $f$ (not the medium) changes.

Active recall

Recall Which cell is which?

Match: "12 crests in 30 s" ::: word problem, C8 (extract $f$ from a count). Match: "500 nm green light, find f" ::: unit-prefix trap, C6/C7. Match: "same string, quadruple f" ::: same-medium, $v$ fixed, C3. Match: "wave crosses into faster medium" ::: exam twist, $v$ changes, $f$ fixed, C10. Match: "0.90 m total swing" ::: amplitude-from-swing, halve it, C4.

Recall Self-test

A displacement–time graph repeats every 0.25 s — is that $λ$ or $T$ ? ::: $T$ (period), because the axis is time. Fixed medium, $f$ tripled — what happens to $λ$ ? ::: It falls to one third ( $λ = v / f$ ). Does $A$ appear in $v = f λ$ ? ::: No — amplitude is independent of speed, frequency and wavelength. Wave enters a slower medium at same $f$ — does $λ$ grow or shrink? ::: Shrink, since $λ = v / f$ and $v$ dropped. Can any of $A, λ, T, f, v$ be negative? ::: No — they are non-negative magnitudes; only the displacement $y$ can go negative (phase inversion).

Connections

Wave parameters — amplitude, wavelength, frequency, period, wave speed — the parent note these examples drill.
Simple Harmonic Motion — the middle-line picture behind amplitude (Ex 4).
Speed of Waves on a String — why $v$ is medium-set (Ex 3).
Sound Waves — the $340 m/s$ air value (Ex 2).
Electromagnetic Spectrum — the speed of light in Ex 6–7.
The Wave Equation y(x,t) — packages all these parameters into $y = A sin (k x - ω t)$ , where negative $y$ (phase inversion) lives.
Doppler Effect — the case where $f$ itself changes, complementing Ex 10.