Worked examples — Standing waves — formation, nodes, antinodes

Before anything, three reminders you will lean on the whole way down:

• $k$ is the wave number: it counts radians of the wave per metre. It ties to wavelength by $k=\frac{2\pi }{\lambda }$, so $\lambda =\frac{2\pi }{k}$. (See Wave number k and wavelength.)
• $\omega$ is the angular frequency: radians of the breathing per second. It ties to ordinary frequency $f$ by $\omega =2\pi f$, and to period $T$ by $\omega =\frac{2\pi }{T}$.
• $v$ is the wave speed: how many metres a travelling wave crest advances per second. It is fixed by the medium (tension and mass of a string, air for sound). Two ways to compute it: $v=\frac{\omega }{k}$ (from a wave equation) or $v=f\lambda$ (frequency times wavelength). Both must agree.

Both $k$ and $\omega$ come from comparing your given equation to the template $y=2A\sin (kx)\cos (\omega t)$: whatever multiplies $x$ inside the sine is $k$; whatever multiplies $t$ inside the cosine is $\omega$.

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The scenario matrix

Every standing-wave problem is one (or a blend) of these case classes. Each row is a distinct thing that can go wrong or surprise you; the examples below hit every cell.

Cell	Case class	What's tricky about it	Example
C1	Read $k,\omega ,A$ off a given standing wave	matching to the template; $\lambda ,T,f$	Ex 1
C2	Build a standing wave from two travelling waves	sum-to-product, correct final $A$	Ex 2
C3	Node/antinode positions (indexing $n$)	off-by-one; "3rd from origin"	Ex 3
C4	Zero / degenerate node at $x=0$	$n=0$ node sits at origin; not all setups do	Ex 4
C5	Cosine-in-$x$ standing wave (free end)	node/antinode conditions swap	Ex 5
C6	Limiting / snapshot in time	at $t$ where $\cos \omega t=0$ the whole string is flat	Ex 6
C7	Boundary-fit / normal mode (real-world)	only certain $\lambda$ fit a length $L$	Ex 7
C8	Exam twist: velocity of a point, energy at a node	node has max strain, antinode max speed	Ex 8

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Example 1 — Cell C1: reading everything off the equation

Step 1. Compare to the template $y=2A\sin (kx)\cos (\omega t)$. So $2A=0.02\Rightarrow A=0.01\text{m}$; $k=5\pi \text{rad/m}$; $\omega =200\pi \text{rad/s}$. Why this step? Because the template already separated shape from breathing, each coefficient sits in exactly one place — matching is just pattern recognition, no algebra.

Step 2. Get $\lambda$ from $k=\frac{2\pi }{\lambda }\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\pi }{5\pi }=0.4\text{m}$. Why this step? The question "how long is one full spatial repeat?" is exactly what $\lambda$ answers, and $k$ is the only thing that stores that info.

Step 3. Get $T$ and $f$ from $\omega =\frac{2\pi }{T}$: $T=\frac{2\pi }{200\pi }=0.01\text{s}$, and $f=\frac{1}{T}=100\text{Hz}$. Why this step? $\omega$ is the only carrier of timing information; $T$ answers "how long is one breath."

Verify: $A=0.01$ is half of $0.02$ ✓ (forecast right). $\lambda =0.4\text{m}<0.5\text{m}$ ✓. Units check: the ratio $\frac{[\omega ]}{[k]}=\frac{\text{rad/s}}{\text{rad/m}}=\text{m/s}$ is a speed — and in fact that ratio is the wave speed $v$ (defined above), $v=\frac{\omega }{k}=\frac{200\pi }{5\pi }=40\text{m/s}$. Cross-check the other way: $v=f\lambda =100\times 0.4=40\text{m/s}$ ✓ — both routes agree, so our $k,\omega$ are consistent.

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Example 2 — Cell C2: superposing two travelling waves

Step 1. Add them and apply $\sin P+\sin Q=2\sin \frac{P+Q}{2}\cos \frac{P-Q}{2}$ (this is the Superposition principle plus a trig identity): with $P=2x-6t,Q=2x+6t$, we get $\frac{P+Q}{2}=2x$ and $\frac{P-Q}{2}=-6t$. $y=2(0.3)\sin (2x)\cos (-6t)=0.6\sin (2x)\cos (6t)$ Why this step? Using $\cos (-\theta )=\cos \theta$ tidies the sign; sum-to-product is the only identity that turns "$x$ and $t$ tangled in one bracket" into "$x$ alone times $t$ alone."

Step 2. Max amplitude is the front number: $2A=0.6\text{m}$. Why this step? $|\sin (2x)|$ can reach 1, so the biggest swing a point ever gets is $0.6$.

Step 3. $k=2\Rightarrow \lambda =\frac{2\pi }{2}=\pi \text{m}$.

Verify: These came from Travelling waves each of amplitude $0.3$; two of them stacking to $0.6$ at antinodes is exactly $2A$ ✓. The $t$-part is pure $\cos (6t)$ with no $x$ inside — space and time separated ✓.

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Example 3 — Cell C3: node & antinode positions (indexing carefully)

Step 1. Nodes: $\sin (2x)=0\Rightarrow 2x=n\pi \Rightarrow x=\frac{n\pi }{2},n=0,1,2,\dots$ Why this step? A node is "never moves," i.e. the shape factor is zero, and $\sin$ is zero exactly at integer multiples of $\pi$.

Step 2. Index them: $n=0\to x=0$ (the origin itself, the hollow dot in Figure 1), $n=1\to x=\pi /2$, $n=2\to x=\pi$. The 2nd node after the origin is $n=2$: $x=\pi \approx 3.14\text{m}$. Why this step? "After the origin" means we don't count $n=0$ as one of them — this is the classic off-by-one trap the hollow dot in the figure warns against.

Step 3. Antinodes: $|\sin (2x)|=1\Rightarrow 2x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{4}\approx 0.785\text{m}$ (1st one — the orange square in Figure 1). Why this step? Max swing needs the shape factor at $\pm 1$; $\sin$ first hits $1$ at $\pi /2$.

Verify: Node(0)$\to$antinode($\pi /4$) $=\pi /4=\lambda /4$ since $\lambda =\pi$ ✓. Node$\to$node $=\pi -\pi /2=\pi /2=\lambda /2$ ✓. Matches the forecast and the spacings drawn in Figure 1.

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Example 4 — Cell C4: the degenerate node at the origin

Step 1. (a) $R(0)=2A\sin (k\cdot 0)=2A\sin 0=0$. So $x=0$ is a node for the sine form. Why this step? The amplitude function evaluated at the origin decides it — no motion means node.

Step 2. (b) A fixed end cannot move, ever: physically its displacement is clamped to zero. So the boundary demands $R(0)=0$. The sine form automatically satisfies this — that's why reflection at a fixed end (see Reflection of waves at boundaries) produces a $\sin (kx)$ standing wave, not a cosine one. Why this step? We are matching the maths to a real constraint; the boundary picks the form, not the other way round.

Step 3. Degenerate check: if instead $A=0$ (no wave at all), then $R(x)=0$ everywhere — every point is trivially a "node," but there's no wave. This is the degenerate limit.

Verify: $\sin 0=0$ exactly ✓. And a fixed end with $R(0)\ne 0$ would contradict "clamped," so the sine form is the only consistent one ✓. The $A=0$ case gives the flat, wave-free string ✓.

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Example 5 — Cell C5: the cosine form (free end) — conditions swap

Step 1. Antinodes (max swing): $|\cos (4x)|=1\Rightarrow 4x=n\pi \Rightarrow x=\frac{n\pi }{4},n=0,1,2,\dots$ The candidate at $n=0$ is $x=0$ — the free end — but the domain is $x>0$, so we discard it and take $n=1$: the first antinode with $x>0$ is $x=\frac{\pi }{4}\approx 0.785\text{m}$. Why this step? $\cos$ hits $\pm 1$ at integer multiples of $\pi$ — that's where cosine is extreme, unlike sine; and the restriction "$x>0$" is a genuine constraint that rules out the $n=0$ endpoint.

Step 2. Nodes ($R=0$): $\cos (4x)=0\Rightarrow 4x=\frac{\pi }{2}+n\pi \Rightarrow x=\frac{\pi }{8}+\frac{n\pi }{4}$. First node ($n=0$) with $x>0$: $x=\frac{\pi }{8}\approx 0.393\text{m}$. Why this step? Cosine crosses zero a quarter-period after its peak, so the node is offset by $\lambda /4$ from the free-end antinode — the geometry is unchanged, only which point is at $x=0$ flips.

Step 3. Wavelength: $k=4\Rightarrow \lambda =\frac{2\pi }{4}=\frac{\pi }{2}$.

Verify: Free-end antinode(0)$\to$node($\pi /8$) $=\pi /8=\lambda /4$ ✓, and node($\pi /8$)$\to$next antinode($\pi /4$) $=\pi /8=\lambda /4$ ✓ (spacing law identical to the sine case — only the labels swapped). This is exactly the Sound in pipes situation for an open end.

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Example 6 — Cell C6: the limiting snapshot when the string is flat

Step 1. (a) $t=0$: $\cos (0)=1$, so $y(x,0)=0.02\sin (5\pi x)$ — the full spatial pattern, antinodes at max displacement. Why this step? The breathing factor is at its peak, so we see the shape at full size.

Step 2. (b) $t=T/4$: $\omega t=200\pi \cdot 0.0025=0.5\pi$, and $\cos (0.5\pi )=0$. So $y(x,T/4)=0$ for every $x$ — the string is dead flat. Why this step? When the breathing factor passes through zero, the whole standing pattern momentarily vanishes in displacement.

Step 3. Where did the energy go? At this instant every point is at equilibrium displacement but moving fastest — the energy is all kinetic. (Antinodes zip through with maximum speed here.) Why this step? Zero displacement $\ne$ zero energy; the parent note's mistake-callout warned exactly this.

Verify: $\cos (0.5\pi )=0$ exactly ✓. $\cos (0)=1$ ✓. The flat snapshot is a real, well-known instant — consistent with energy sloshing KE$\leftrightarrow$PE, never leaving the string.

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Example 7 — Cell C7: boundary-fit / normal mode (word problem)

Step 1. Both ends fixed $\Rightarrow$ node at $x=0$ and node at $x=L$. Node spacing is $\lambda /2$, so a whole number of half-wavelengths must fill $L$ (count the humps in Figure 2): $L=n\frac{{\lambda }_n}{2}\Rightarrow {\lambda }_n=\frac{2L}{n},n=1,2,3,\dots$ Why this step? Only wavelengths obeying this fit the boundary; every other wave would need the ends to move, which they can't. This is the origin of Resonance and normal modes.

Step 2. Fundamental ($n=1$, orange string in Figure 2): ${\lambda }_1=2L=2\times 0.65=1.3\text{m}$. Why this step? The lowest note is the longest wave that fits — that is $n=1$, one single hump.

Step 3. Fundamental frequency from $v=f\lambda \Rightarrow f_1=\frac{v}{{\lambda }_1}=\frac{260}{1.3}=200\text{Hz}$. Why this step? The wave speed $v$ is fixed by the string; dividing it by the wavelength converts "metres per repeat" into "repeats per second," which is frequency.

Step 4. 3rd harmonic ($n=3$, plum string in Figure 2): ${\lambda }_3=\frac{2L}{3}=\frac{1.3}{3}\approx 0.4333\text{m}$, and its frequency $f_3=\frac{v}{{\lambda }_3}=3f_1=600\text{Hz}$. Why this step? Harmonics are integer multiples of the fundamental because $n$ counts how many half-loops fit — three loops means three times the frequency.

Verify: $f_1=200$Hz, ${\lambda }_1=1.3$m give $v=f_1{\lambda }_1=260$m/s ✓. $f_3=600$Hz, ${\lambda }_3=0.4333$m give $600\times 0.4333=260$m/s ✓. Matches the Waves on a string harmonic pattern and the humps counted in Figure 2.

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Example 8 — Cell C8: exam twist (speed of a point; node strain)

Step 1. First antinode: $\sin (5\pi x)=1\Rightarrow 5\pi x=\pi /2\Rightarrow x=0.1\text{m}$. There the amplitude is the full $0.02$m. Why this step? We need a point that actually moves; the antinode is the maximum-swing point.

Step 2. Transverse velocity is the rate of change of $y$ with time at fixed $x$: $v_y=\frac{\partial y}{\partial t}=-0.02\sin (5\pi x)\cdot (200\pi )\sin (200\pi t).$ Why a derivative, and why $\partial /\partial t$? "Speed of this point" asks how its height changes per unit time while holding $x$ fixed — that is exactly the time-derivative at constant position. We use the partial derivative because $x$ is frozen (we're watching one point, not moving along the string). Note this transverse speed $v_y$ is a different thing from the wave speed $v$ defined at the top — one is how fast the material flicks up and down, the other is how fast the pattern would travel.

Step 3. At the antinode $\sin (5\pi x)=1$, and $|\sin (200\pi t)|$ peaks at $1$, so $v_{y,\max }=0.02\times 200\pi =4\pi \approx 12.57\text{m/s}.$ Why this step? Both sine factors at their max give the largest possible speed — the antinode's fastest flick.

Step 4. The force / strain at a node: slope is $\frac{\partial y}{\partial x}=0.02(5\pi )\cos (5\pi x)\cos (200\pi t)$. At a node, $\sin (5\pi x)=0$ so $\cos (5\pi x)=\pm 1$ — the slope factor is largest exactly where displacement is zero. Big slope means the string is most stretched/kinked there, so restoring tension peaks at the node. Why this step? This is the parent note's "node has max strain" fact, now derived: displacement and slope of a sine are a quarter-cycle out of step, so zeros of one line up with peaks of the other.

Verify: $v_{y,\max }=4\pi \approx 12.566$m/s ✓. Units: $[\text{m}][\text{rad/s}]=\text{m/s}$ ✓. The node's slope factor $\cos (5\pi x)=\pm 1$ where $\sin =0$ confirms max strain at nodes ✓.

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Active recall

Recall Which cell is this? (cover answers)

Given $y=0.4\sin (3x)\cos (5t)$, "3rd node after origin" ::: $\sin (3x)=0\Rightarrow x=n\pi /3$; 3rd after origin is $n=3\Rightarrow x=\pi \approx 3.14$ m (Cell C3). Free end at $x=0$: what's the shape factor? ::: $\cos (kx)$ — the origin is a displacement antinode (Cell C5). Both-fixed string, allowed wavelengths? ::: ${\lambda }_n=2L/n$ (Cell C7). Why is the string flat at $t=T/4$? ::: $\cos (\omega t)=0$ there; all energy is kinetic, not gone (Cell C6). Max transverse speed of an antinode of amplitude $R$? ::: $R\omega$ (Cell C8).

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Connections

• Superposition principle — every "add the two waves" step.
• Travelling waves — the $y_1,y_2$ building blocks (Ex 2).
• Reflection of waves at boundaries — why the origin is forced to be a node/antinode (Ex 4–5).
• Resonance and normal modes / Waves on a string / Sound in pipes — the boundary-fit example (Ex 7).
• Wave number k and wavelength — $k=2\pi /\lambda$ used in every extraction.
• Parent: Standing waves — formation, nodes, antinodes.