Mechanical waves — transverse and longitudinal: Worked examples

Before we start, one reminder of the symbols we reuse (all built in the parent):

$A$ = amplitude, the biggest displacement a particle reaches (metres).
$ω$ = angular frequency (radians per second), and $f = ω /2 π$ (cycles per second, hertz).
$k$ = angular wave number (radians per metre), and $λ = 2 π / k$ (metres per cycle).
$v = ω / k = f λ$ = speed the pattern moves.
$v_{p} = \partial y / \partial t$ = speed a single particle moves up/down.
$φ$ = phase constant (radians): a fixed head-start added inside the sine, $y = A sin (ω t - k x + φ)$ . It just shifts where the wave sits at $t = 0$ ; it never changes $A, ω, k,$ or $v$ .
$T$ = tension in a string (newtons): the pull along the string that acts as the restoring force. (Not to be confused with a period; on this page $T$ always means tension.)
$μ$ = linear mass density of a string (kilograms per metre): mass per unit length, the inertia the restoring pull has to move.
For bulk media: $Y$ = Young's modulus, $B$ = bulk modulus (both in pascals, $Pa$ ), $ρ$ = mass density ( $kg/m^{3}$ ) — the elastic and inertial properties of a solid/fluid.

See Mechanical waves — transverse and longitudinal for how each of these was born, and Simple Harmonic Motion for why the sine appears at all.

The scenario matrix

Cell	Case class	What's tricky	Example
C1	Read a $+ x$ wave equation	extract every quantity, get sign of travel right	Ex 1
C2	Read a $- x$ wave equation	sign flip $sin (ω t + k x)$ → direction reversed	Ex 2
C3	Particle velocity & the slope link	don't confuse $v_{p}$ with $v$ ; sign of $v_{p}$ at a chosen instant	Ex 3
C4	Speed from medium (string)	$v = T / μ$ , and scaling under $\times$ tension	Ex 4
C5	Speed from medium (solid vs fluid)	pick $Y / ρ$ vs $B / ρ$ ; which wave types are allowed	Ex 5
C6	Degenerate / zero inputs	$A = 0$ , $f = 0$ , $T = 0$ , $k = 0$ — what "wave" survives?	Ex 6
C7	Limiting behaviour	$f \to \infty$ ( $λ \to 0$ ), $μ \to \infty$ ( $v \to 0$ )	Ex 7
C8	Real-world word problem	echo / distance, must model then compute	Ex 8
C9	Exam-style twist	build the equation from graph + given direction	Ex 9
C10	Nonzero phase constant $φ$	read/build $y = A sin (ω t - k x + φ)$	Ex 10

Every numeric answer below is machine-checked in the verify block.

Ex 1 — C1 · Reading a wave moving in $+ x$

Worked example Statement

A wave is $y = 0.05 sin (400 t - 8 x)$ (SI units). Find $A, ω, f, k, λ, v$ , and the direction of travel.

Forecast: Guess — is $λ$ bigger or smaller than 1 metre? Is the wave going left or right?

Step 1. Compare $0.05 sin (400 t - 8 x)$ to the standard form $A sin (ω t - k x)$ . Why this step? The standard form is a template: each slot has a physical meaning, so lining them up hands us every quantity at once. $A = 0.05 m, ω = 400 rad/s, k = 8 rad/m$

Step 2. Frequency: $f = \frac{ω}{2 π} = \frac{400}{2 π} = 63.7 Hz$ . Why this step? $ω$ counts radians per second; dividing by $2 π$ (radians per full cycle) converts to cycles per second.

Step 3. Wavelength: $λ = \frac{2 π}{k} = \frac{2 π}{8} = 0.785 m$ . Why this step? $k$ counts radians per metre; $2 π$ radians is one full spatial cycle = one wavelength.

Step 4. Speed: $v = \frac{ω}{k} = \frac{400}{8} = 50 m/s$ . Why this step? A crest is where the phase $ω t - k x$ stays fixed; keeping it fixed forces $x$ to advance at $ω / k$ .

Step 5. Direction: the form is $ω t - k x$ (a minus), so the wave moves in $+ x$ . Why this step? To keep phase constant as $t$ grows, $x$ must grow too — the pattern slides in the $+ x$ direction.

Verify: cross-check $v = f λ = 63.7 \times 0.785 = 50 m/s$ ✓ — matches Step 4. Units: $(Hz) (m) = m/s$ ✓.

Ex 2 — C2 · Reading a wave moving in $- x$

Ex 3 — C3 · Particle velocity and the slope link

Figure — Mechanical waves — transverse and longitudinal

This figure shows a snapshot of the Ex 1 wave frozen at $t = 0$ (white curve, displacement $y$ up the page vs position $x$ across it). The yellow dot marks the particle at $x = 0$ , sitting on the rest line $y = 0$ . The pink arrow points straight up from it — that's the particle velocity $v_{p}$ , showing this particle is climbing at that instant. The separate blue arrow points along $+ x$ — that's the wave velocity, the whole pattern sliding sideways. Two different arrows, two different motions: read them as "the dot bobs along pink; the shape marches along blue."

Worked example Statement

For $y = 0.05 sin (400 t - 8 x)$ (the Ex 1 wave), find (a) the maximum particle speed, and (b) the particle velocity at $x = 0, t = 0$ .

Forecast: The wave moves at $50 m/s$ (Ex 1). Guess: is a particle's up-down speed bigger or smaller than that?

Step 1. Particle velocity is $v_{p} = \frac{\partial y}{\partial t}$ . Differentiate $y = A sin (ω t - k x)$ with respect to $t$ (holding $x$ fixed): $v_{p} = A ω cos (ω t - k x)$ Why this step? "How fast does one particle move up/down?" is exactly the rate of change of its displacement in time — that's the time-derivative, one particle at a time.

Step 2. Maximum particle speed = $A ω$ because $cos$ maxes at $1$ : $v_{p, m a x} = A ω = 0.05 \times 400 = 20 m/s$ Why this step? The cosine can't exceed $1$ ; the biggest speed happens as the particle whips through its rest line (crest/trough are where it momentarily stops).

Step 3. At $x = 0, t = 0$ : $v_{p} = A ω cos (0) = 20 m/s$ (moving up, positive). Why this step? Plug the instant in. Look at the yellow dot in the figure: at this instant the particle sits at $y = 0$ and is climbing — maximum upward speed.

Step 4 (sanity, the slope link — derived here, not assumed). Differentiate the same $y = A sin (ω t - k x)$ with respect to $x$ (holding $t$ fixed) to get the snapshot's slope: $\frac{\partial y}{\partial x} = - A k cos (ω t - k x)$ Compare it to the time-derivative $\partial y / \partial t = A ω cos (ω t - k x)$ from Step 1. Divide one by the other: $\frac{\partial y / \partial t}{\partial y / \partial x} = \frac{A ω cos}{- A k cos} = - \frac{ω}{k} = - v$ . Rearranged: $v_{p} = \frac{\partial y}{\partial t} = - v \frac{\partial y}{\partial x}$ Why this step? This is the "slope link": a particle's up-down speed equals $- v$ times the local steepness of the frozen shape. The minus sign is the physics — where the snapshot slopes downward to the right, the wave sliding into that point pushes the particle up. Plugging back reproduces $v_{p} = A ω cos$ , so the two routes agree.

Verify: $v_{p, m a x} = 20 m/s$ is smaller than the wave speed $50 m/s$ here — but note they measure different things (up-down vs pattern). Units: $(m) (rad/s) = m/s$ ✓.

Ex 4 — C4 · Speed set by the string, and the scaling

Worked example Statement

A string has linear mass density $μ = 4 \times 1 0^{- 3} kg/m$ and tension $T = 100 N$ . (a) Find $v$ . (b) If tension is quadrupled to $400 N$ , find the new $v$ .

Forecast: Quadruple the tension — does $v$ quadruple, double, or something else?

Step 1. Transverse string speed: $v = \frac{T}{μ}$ . Why this step? Speed is a tug-of-war: restoring pull $T$ (the tension, up) over inertia $μ$ (mass per length, down). The square root balances their units to give m/s.

Step 2. $v = \frac{100}{4 \times 1 0 ^{- 3}} = 25000 = 158 m/s$ . Why this step? Direct substitution of the given medium properties.

Step 3. Quadruple $T$ : $v_{new} = \frac{400}{4 \times 1 0 ^{- 3}} = 100000 = 316 m/s$ . Why this step? Since $v \propto T$ , multiplying $T$ by $4$ multiplies $v$ by $4 = 2$ .

Verify: $316/158 = 2$ exactly ✓ — the turned a $\times 4$ into a $\times 2$ . Units: $(N) / (kg/m) = m^{2} / s^{2} = m/s$ ✓.

Ex 5 — C5 · Solid vs fluid: pick the right elasticity

Worked example Statement

(a) A steel rod: Young's modulus $Y = 2.0 \times 1 0^{11} Pa$ , density $ρ = 8000 kg/m^{3}$ . Find the longitudinal wave speed. (b) Water: bulk modulus $B = 2.2 \times 1 0^{9} Pa$ , $ρ = 1000 kg/m^{3}$ . Find the sound speed. (c) Which of these media can carry a transverse bulk wave?

Forecast: Steel is far stiffer than water. Guess: is steel's speed ~2×, ~5×, or ~10× water's?

Step 1. Solid rod uses Young's modulus $Y$ : $v = Y / ρ$ . Why this step? A rod stretched along its length resists with $Y$ ; that's the restoring elasticity for a longitudinal rod wave. $v_{steel} = \frac{2.0 \times 1 0 ^{11}}{8000} = 2.5 \times 1 0^{7} = 5000 m/s$

Step 2. Fluid uses bulk modulus $B$ : $v = B / ρ$ . Why this step? A fluid has no shear stiffness, only resistance to compression — that's $B$ . $v_{water} = \frac{2.2 \times 1 0 ^{9}}{1000} = 2.2 \times 1 0^{6} = 1483 m/s$

Step 3. Transverse bulk waves: only the steel can carry them. Why this step? Transverse motion needs shear rigidity. Solids have it; the fluid (water bulk) does not — exactly the parent's rule. See Sound waves for why fluid sound is longitudinal.

Verify: $5000/1483 = 3.37$ — steel is ~3.4× faster ✓. Units of both: $(Pa) / (kg/m^{3}) = m/s$ ✓.

Ex 6 — C6 · Degenerate / zero inputs

Worked example Statement

Discuss (a) $A = 0$ , (b) $f = 0$ , (c) tension $T = 0$ on a string, (d) $k = 0$ . What "wave" survives each?

Forecast: Which of these four still leaves something moving?

Step 1. $A = 0$ : $y = 0 \cdot sin (ω t - k x) = 0$ for all $x, t$ . Why this step? Amplitude scales the whole displacement; zero amplitude = flat medium = no wave at all. Every particle stays home.

Step 2. $f = 0$ : then $ω = 2 π f = 0$ , so $y = A sin (- k x)$ , frozen in time. Why this step? Zero frequency means nothing oscillates in time — you get a static bent shape, not a travelling wave. No energy propagates.

Step 3. $T = 0$ : $v = T / μ = 0 = 0$ . Why this step? With no tension there is no restoring force, so a disturbance can't be pulled onward — speed collapses to zero; a "kink" just sits there.

Step 4. $k = 0$ : then $λ = 2 π / k \to \infty$ and $y = A sin (ω t)$ — no spatial variation, every particle oscillates in step (in phase). Why this step? Zero wave number means the "wave" has infinite wavelength — no crests or troughs anywhere along $x$ , so the whole medium bobs together like a single SHM oscillator. It's the opposite degenerate limit to $f = 0$ : here time varies but space doesn't. Nothing propagates (no phase lag between points), so it isn't a travelling wave either.

Verify: four degeneracies, four distinct failures of propagation — via $A \to 0$ (flat), $ω \to 0$ (frozen in time), $v \to 0$ (can't be carried), $k \to 0$ (no spatial pattern). $0 = 0$ ✓ and $2 π / k \to \infty$ as $k \to 0$ ✓.

Ex 7 — C7 · Limiting behaviour

Worked example Statement

On a fixed string (speed $v = 200 m/s$ fixed by the medium): (a) as $f \to \infty$ , what happens to $λ$ ? Compute $λ$ at $f = 100$ , $f = 1000$ , $f = 10000 Hz$ . (b) If instead $μ \to \infty$ at fixed $T$ , what happens to $v$ ?

Forecast: Which quantity shrinks toward zero in each limit?

Step 1. From $v = f λ$ with $v$ fixed: $λ = v / f$ . Why this step? Medium fixes $v$ ; the source picks $f$ ; so $λ$ is forced to adjust — it's the dependent variable here.

Step 2. Compute: $λ_{100} = \frac{200}{100} = 2 m, λ_{1000} = 0.2 m, λ_{10000} = 0.02 m$ Why this step? As $f \to \infty$ , $λ = v / f \to 0$ — waves crowd tighter and tighter while their speed never changes.

Step 3. $μ \to \infty$ limit: $v = T / μ \to 0$ . Why this step? An infinitely heavy string has infinite inertia, so a fixed restoring tension can barely accelerate it; the disturbance is passed on ever more slowly, and in the limit the wave crawls to a standstill ( $v \to 0$ ).

Verify: each $\times 10$ in $f$ gave a $\div 10$ in $λ$ ( $2 \to 0.2 \to 0.02$ ) ✓ — clean inverse proportionality, and $v$ stayed $200 m/s$ throughout.

Ex 8 — C8 · Real-world word problem (echo)

Ex 9 — C9 · Exam twist: build the equation from a snapshot

This figure is a snapshot of the wave you're about to build: displacement $y$ (up) vs position $x$ (across), amplitude $A = 0.03 m$ marked by the blue vertical arrow from the rest line to a crest, and the wavelength $λ = 0.5 m$ marked by the pink double-arrow spanning crest to crest (two yellow dots). The yellow arrow near the bottom shows the required travel direction ( $- x$ ). Everything you need to fill the template $A sin (ω t \pm k x)$ is read straight off this picture.

Worked example Statement

A transverse wave has amplitude $A = 0.03 m$ , wavelength $λ = 0.5 m$ , frequency $f = 20 Hz$ , travelling in the $- x$ direction. Write $y (x, t)$ and find $v$ .

Forecast: Given $- x$ travel, will your equation contain $- k x$ or $+ k x$ ?

Step 1. Angular frequency: $ω = 2 π f = 2 π (20) = 40 π \approx 125.66 rad/s$ . Why this step? The standard form needs $ω$ , not $f$ ; convert cycles/s to radians/s with $2 π$ .

Step 2. Wave number: $k = \frac{2 π}{λ} = \frac{2 π}{0.5} = 4 π \approx 12.57 rad/m$ . Why this step? The form needs $k$ (radians per metre), obtained from wavelength.

Step 3. Direction $- x$ ⇒ use $ω t + k x$ (from Ex 2's rule). So: $y (x, t) = 0.03 sin (40 π t + 4 π x) m$ Why this step? A plus sign inside makes phase constant require decreasing $x$ — i.e. travel toward $- x$ , as required.

Step 4. Speed: $v = f λ = 20 \times 0.5 = 10 m/s$ . Why this step? In one period the pattern advances one wavelength, so $v = f λ$ — independent of direction.

Verify: cross-check $v = ω / k = 40 π /4 π = 10 m/s$ ✓ — matches $f λ$ . The figure shows the snapshot with $λ = 0.5 m$ marked crest-to-crest.

Ex 10 — C10 · Nonzero phase constant $φ$

Worked example Statement

A wave is $y = 0.04 sin (50 t - 2 x + \frac{π}{2})$ (SI). (a) Find $A, ω, k, v$ . (b) What is $y$ at $x = 0, t = 0$ , and why is it not zero? (c) Rewrite this wave using cosine.

Forecast: With that extra $+ \frac{π}{2}$ inside, will $A$ , $ω$ , $k$ , or $v$ change compared with a $φ = 0$ wave?

Step 1. Compare to the full form $A sin (ω t - k x + φ)$ : read off $A = 0.04$ , $ω = 50$ , $k = 2$ , $φ = \frac{π}{2}$ . Why this step? The phase constant $φ$ is just one more slot in the template — the largest new idea here is recognising it as a fixed head-start, not a rate.

Step 2. Speed and everything geometric are untouched: $v = \frac{ω}{k} = \frac{50}{2} = 25 m/s$ . Why this step? $φ$ is a constant; it drops out when you track a crest (phase $=$ const still gives $v = ω / k$ ). So $φ$ shifts where the wave starts, never how fast it goes.

Step 3. At $x = 0, t = 0$ : $y = 0.04 sin (0 - 0 + \frac{π}{2}) = 0.04 sin \frac{π}{2} = 0.04 m$ . Why this step? Without $φ$ the wave would sit at $y = 0$ at the origin; the $\frac{π}{2}$ head-start advances it a quarter-cycle, so it starts at a crest ( $y = A$ ) instead. That's exactly what a phase constant does — it sets the starting point.

Step 4. Cosine form: since $sin (θ + \frac{π}{2}) = cos θ$ , we get $y = 0.04 cos (50 t - 2 x)$ . Why this step? A $+ \frac{π}{2}$ phase shift is the difference between sine and cosine — showing that "which trig function" and "what $φ$ " are the same choice, not two different physics.

Verify: $A, ω, k, v$ match the $φ = 0$ case's method exactly ( $v = 25 m/s$ ); only the $t = 0$ starting value moved from $0$ to $0.04 m$ ✓. And $0.04 sin (50 t - 2 x + \frac{π}{2}) = 0.04 cos (50 t - 2 x)$ is an identity ✓.

Recall Scenario matrix — cover the answers

Sign inside the sine is minus ( $ω t - k x$ ) ::: wave travels in $+ x$ Sign inside the sine is plus ( $ω t + k x$ ) ::: wave travels in $- x$ Maximum particle speed ::: $A ω$ Quadruple the tension changes $v$ by factor ::: $4 = 2$ Fluid bulk cannot carry which wave type ::: transverse (no shear rigidity) $A = 0$ , $f = 0$ , $T = 0$ , $k = 0$ each kill propagation via ::: $A \to 0$ (flat), $ω \to 0$ (frozen), $v \to 0$ (uncarried), $k \to 0$ (no spatial pattern) As $f \to \infty$ at fixed $v$ , $λ$ ::: $\to 0$ Echo: divide the round-trip distance by ::: 2 A phase constant $φ$ changes ::: only where the wave sits at $t = 0$ , not $A, ω, k, v$

Related deep tools: Wave equation, Superposition and Interference, Standing waves & resonance, Doppler effect.