Standing waves — formation, nodes, antinodes: Exercises

Definition Symbols we will keep using (define once, use forever)

$x$ — position along the string, in metres. Think "how far along the rope."
$t$ — time, in seconds.
$λ$ (lambda) — wavelength: length of one full wiggle. Picture one complete hill+valley.
$k$ — wave number: $k = \frac{2 π}{λ}$ . It counts how many radians of "wave phase" you march through per metre. Big $k$ = tightly packed wiggles.
$T$ — period: the time (in seconds) for one full up-down-up breath of a point.
$f$ — frequency: how many full breaths per second, in hertz (Hz). It is just $f = \frac{1}{T}$ .
$ω$ (omega) — angular frequency: $ω = \frac{2 π}{T} = 2 π f$ , radians of breathing per second.
$A$ — amplitude of each travelling wave; the standing pattern swings up to $2 A$ .
$R (x)$ — the position-dependent amplitude $2 A sin (k x)$ : how big the swing is at each point $x$ .
$μ$ (mu) — mass per unit length of the string, in kg/m (how heavy each metre of rope is). It sets the wave speed via $v = F / μ$ ; see Waves on a string.
$F$ — the tension in the string (the pulling force along it, in newtons).

The picture below is the map for the whole page: it shows the fixed envelope $R (x)$ (blue), a single snapshot in time (yellow dashed), the red nodes that never move, and the green antinodes that swing hardest. Notice the labelled $λ /4$ gap from a node to the next antinode and the $λ /2$ gap between neighbouring nodes — those two spacings drive almost every problem on this page, so glance back at it whenever you compute a position.

Figure — Standing waves — formation, nodes, antinodes

Figure 1 — The standing-wave envelope $R (x) = 2 A sin (k x)$ : red = nodes, green = antinodes, labelled $λ /4$ and $λ /2$ spacings.

L1 · Recognition

Recall Solution 1

WHAT to look for: a standing wave splits into (something in $x$ only) $\times$ (something in $t$ only). $x$ and $t$ never sit inside the same bracket. This is exactly the shape-times-breathing split drawn in Figure 1.

(a) $x$ and $t$ are tangled in one bracket $\to$ this travels. Not standing.
(b) Factored as $sin (3 x) \cdot cos (5 t)$ $\to$ shape $\times$ breathing $\to$ standing wave ✓.
(c) Two opposite travellers; by sum-to-product this becomes $0.4 sin (3 x) cos (5 t)$ $\to$ also standing. Answer: (b) and (c). Why (c) counts: the Superposition principle lets us add them, and the trig identity collapses the sum to the same factored form as (b).

Recall Solution 2

Compare with $y = 2 A sin (k x) cos (ω t)$ . Here $2 A = 0.2$ , $k = 3$ , $ω = 5$ .

$R (x) = 0.2 sin (3 x)$ metres.
Maximum swing $= 2 A = 0.2$ m, reached wherever $∣ sin (3 x) ∣ = 1$ (the green antinode points in Figure 1).

L2 · Application

Recall Solution 3

Wavelength: $k = 2 π$ , and $k = \frac{2 π}{λ} \Rightarrow λ = \frac{2 π}{k} = 1$ m. Nodes: $R (x) = 0 \Rightarrow sin (2 π x) = 0 \Rightarrow 2 π x = nπ \Rightarrow x = \frac{n}{2}$ . First three at $x > 0$ : $x = 0.5, 1.0, 1.5$ m. Why $sin$ zeros? a node never moves, so its swing $R (x)$ must be exactly zero, and $sin$ vanishes at whole multiples of $π$ . Spacing $= 0.5$ m $= λ /2$ ✓ — this is exactly the red-dot spacing in Figure 1 (drawn there for $λ = 1$ m).

Recall Solution 4

Antinodes: an antinode swings hardest, so its amplitude factor $∣ sin (2 π x) ∣$ must hit its largest possible value, $1$ . Why $∣ sin ∣ = 1$ ? the sine of an angle never exceeds $1$ in size, so the swing $R (x) = 2 A sin (2 π x)$ is biggest exactly where $sin$ reaches its peak $\pm 1$ — that is the point flapping with the full $2 A$ . $∣ sin (2 π x) ∣ = 1 \Rightarrow 2 π x = (n + \frac{1}{2}) π \Rightarrow x = \frac{2 n + 1}{4} .$ First two: $x = 0.25$ m and $x = 0.75$ m. Max swing $= 2 A = 0.5$ m. Check: node at $x = 0$ , antinode at $x = 0.25$ ; gap $= 0.25 = λ /4$ ✓ — the yellow $λ /4$ arrow in Figure 1.

Recall Solution 5

Add using the sum-to-product identity stated at the top of the page, $sin P + sin Q = 2 sin \frac{P + Q}{2} cos \frac{P - Q}{2}$ , with $P = 4 x - 2 t$ and $Q = 4 x + 2 t$ : $y = 2 (3) sin (4 x) cos (2 t) = 6 sin (4 x) cos (2 t) .$ Antinode: $sin (4 x) = 1 \Rightarrow 4 x = \frac{π}{2} \Rightarrow x = \frac{π}{8} \approx 0.3927$ m. Check $λ = \frac{2 π}{4} = \frac{π}{2}$ , so $λ /4 = \frac{π}{8}$ ✓ (first antinode sits one quarter-wave from the node at origin).

L3 · Analysis

Recall Solution 6

WHY the boundary matters: fixed ends can't move, so both ends must be nodes. A pattern with nodes at both ends fits only if a whole number $n$ of half-waves spans $L$ (see Resonance and normal modes): $L = n \frac{λ}{2} \Rightarrow λ = \frac{2 L}{n} .$ (a) $λ = \frac{2 ( 1.2 )}{3} = 0.8$ m. (b) The $n = 3$ mode has $n + 1 = 4$ nodes (both ends + 2 interior) and $n = 3$ antinodes. Sketch check: three "bellies" (antinodes) separated by four nodes — exactly the $n = 3$ picture in Figure 2 below.

The figure below draws precisely this $n = 3$ mode: white walls mark the two fixed ends (forced nodes), the blue curves are the $+ R (x)$ and $- R (x)$ envelope the string swings between, red dots are the four nodes and green dots the three antinodes. Count them off the picture to confirm part (b).

Figure 2 — Third harmonic ( $n = 3$ ) on a string fixed at both ends: 4 red nodes (including the two walls) and 3 green antinodes, with $λ = 0.8$ m.

Recall Solution 7

$R (0.25) = 0.5 sin (2 π \cdot 0.25) = 0.5 sin (\frac{π}{2}) = 0.5$ m (full swing available here). Breathing factor: $cos (100 π \cdot \frac{1}{300}) = cos (\frac{π}{3}) = 0.5$ . $y = 0.5 \times 0.5 = 0.25 m .$ Interpretation: the antinode can reach $0.5$ m, but at this instant the breathing factor is only $0.5$ , so it sits halfway up at $+ 0.25$ m and is on its way down.

Recall Solution 8

WHY a derivative: velocity is "how fast displacement changes with time," i.e. $v = \frac{\partial y}{\partial t}$ . The $sin (2 π x)$ factor is a constant in time, so only $cos (100 π t)$ gets differentiated: $v = \frac{\partial y}{\partial t} = - 0.5 sin (2 π x) (100 π) sin (100 π t) = - 50 π sin (2 π x) sin (100 π t) .$ At $x = 0.25$ , $sin (2 π \cdot 0.25) = 1$ .

$t = 0$ : $sin (0) = 0 \Rightarrow v = 0$ . (Antinode is at the top, momentarily still — max displacement, zero speed.)
$t = \frac{1}{200}$ : $100 π /200 = \frac{π}{2}$ , $sin (\frac{π}{2}) = 1 \Rightarrow v = - 50 π \approx - 157.1$ m/s. (Passing through equilibrium at top speed.)

L4 · Synthesis

Recall Solution 9

Fundamental: $L = \frac{λ}{2} \Rightarrow λ = 2 L = 4$ m $\Rightarrow k = \frac{2 π}{λ} = \frac{π}{2}$ rad/m. Time part: $ω = 2 π f = 2 π (25) = 50 π$ rad/s. Amplitude: max swing $2 A = 0.04 \Rightarrow A = 0.02$ m. Put it together (node at $x = 0$ so we use $sin$ ): $y = 0.04 sin ((\frac{π}{2}) x) cos (50 π t) m .$ Check: at $x = 1$ m (the middle, $L /2$ ), $sin ((\frac{π}{2}) \cdot 1) = sin (\frac{π}{2}) = 1$ $\to$ full swing $0.04$ m ✓ (that is the single antinode of the fundamental).

Recall Solution 10

WHY $v = f λ$ : speed = (wiggles per second) $\times$ (length per wiggle). The standing pattern is built from travellers of this speed (see Waves on a string). (a) $v = f λ = 25 \times 4 = 100$ m/s. (b) $v = F / μ \Rightarrow F = μ v^{2} = 5 \times 1 0^{- 3} \times 10 0^{2} = 50$ N.

L5 · Mastery

Recall Solution 11

Mode $n$ has $λ_{n} = \frac{2 L}{n}$ , so $k_{n} = \frac{2 π}{λ _{n}} = \frac{nπ}{L}$ and $R_{n} (x) = 2 A sin (\frac{nπ x}{L})$ . $n = 2$ antinode: $sin (\frac{2 π x}{L}) = \pm 1 \Rightarrow \frac{2 π x}{L} = \frac{π}{2} \Rightarrow x = \frac{L}{4}$ . $n = 4$ node at that $x$ ? $sin (\frac{4 π}{L} \cdot \frac{L}{4}) = sin (π) = 0$ ✓ — yes, it is a node. Answer: $x = \frac{L}{4}$ is an antinode of $n = 2$ and a node of $n = 4$ . (By symmetry $x = \frac{3 L}{4}$ works too.) Why it works: $n = 4$ has twice the spatial frequency of $n = 2$ , so every $n = 2$ antinode lands on an $n = 4$ node.

Recall Solution 12

First, recall the two symbols used here (both are in the panel at the top): $F$ is the string tension in newtons — the same $F$ from Problem 10 — and $P$ is the instantaneous power, the rate (in watts) at which energy flows past the point $x_{0}$ at a given instant. Positive $P$ means energy moving in the $+ x$ direction, negative means $- x$ . The formula $P = - F \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}$ is "force times velocity" for the transverse motion of the string.

Compute the two slopes: $\frac{\partial y}{\partial x} = 2 A k cos (k x) cos (ω t), \frac{\partial y}{\partial t} = - 2 A ω sin (k x) sin (ω t) .$ Multiply: $P = - F [2 A k cos (k x) cos (ω t)] [- 2 A ω sin (k x) sin (ω t)]$ $= 4 F A^{2} k ω sin (k x) cos (k x) sin (ω t) cos (ω t) .$ The time part is $sin (ω t) cos (ω t) = \frac{1}{2} sin (2 ω t)$ . Its average over a full cycle is zero (a sine averages to zero). Hence $⟨ P ⟩ = 0$ at every $x_{0}$ . Meaning: power oscillates back and forth but never nets out — energy stays trapped, swapping between kinetic and potential, exactly as the parent note claimed.

Recall Solution 13

Antinodes of mode $n$ sit at $x = \frac{( 2 m - 1 ) L}{2 n}$ for $m = 1, 2, \dots$ (odd multiples of $\frac{L}{2 n}$ ). Require $\frac{( 2 m - 1 ) ( 1.2 )}{2 n} = 0.3 \Rightarrow (2 m - 1) = \frac{0.3 \cdot 2 n}{1.2} = 0.5 n$ . The left side is a positive odd integer, so $0.5 n$ must be a positive odd integer: $n = 2 (\to 0.5 n = 1)$ , $n = 6 (\to 3)$ , $n = 10 (\to 5)$ , … Lowest is $n = 2$ . Then $λ = \frac{2 L}{n} = \frac{2 ( 1.2 )}{2} = 1.2$ m. Check: the $n = 2$ antinodes are at $x = \frac{( 2 m - 1 ) L}{2 n} = \frac{( 2 m - 1 ) ( 1.2 )}{4}$ , giving $x = 0.3$ m ( $m = 1$ ) and $x = 0.9$ m ( $m = 2$ ) ✓. So the lowest harmonic that places an antinode at $x = 0.3$ m is $n = 2$ , with $λ = 1.2$ m.

Recall Solution 14

Sum-to-product with $P = k x - ω_{1} t$ , $Q = k x + ω_{2} t$ : $y = 2 A sin (k x + \frac{ω _{2} - ω _{1}}{2} t) cos (\frac{ω _{1} + ω _{2}}{2} t) .$ The first factor now contains both $x$ and $t$ , so it is not a fixed shape — space and time did not fully separate. The zeros of that factor sit where $k x + \frac{ω _{2} - ω _{1}}{2} t = nπ$ , i.e. $x = \frac{nπ}{k} - \frac{ω _{2} - ω _{1}}{2 k} t,$ which drifts at speed $v_{drift} = \frac{ω _{1} - ω _{2}}{2 k} .$ Direction — read the sign carefully:

If $ω_{1} > ω_{2}$ (the right-mover is faster), $v_{drift} > 0$ : the whole pattern of nodes creeps toward $+ x$ (in the direction of the faster wave). This makes physical sense — the faster wave "wins" and drags the interference pattern along with it.
If $ω_{1} < ω_{2}$ , $v_{drift} < 0$ : the nodes crawl toward $- x$ , following the (now faster) left-mover.
If $ω_{1} = ω_{2}$ , $v_{drift} = 0$ : the drift vanishes and the nodes lock in place — we recover the true standing wave. This is exactly why equal frequencies are required for a genuine standing wave: only then is the drift speed zero.

Active recall

Recall Rapid-fire self-test (cover the answers)

Wavelength if $k = 2 π$ ? ::: $λ = 1$ m. Node positions of $sin (2 π x)$ ? ::: $x = n /2$ (i.e. $0, 0.5, 1.0, \dots$ ). First antinode of $sin (2 π x)$ at $x > 0$ ? ::: $x = 0.25$ m. Standing wave from $3 sin (4 x - 2 t) + 3 sin (4 x + 2 t)$ ? ::: $y = 6 sin (4 x) cos (2 t)$ . Antinode nearest origin for $6 sin (4 x) cos (2 t)$ ? ::: $x = π /8 \approx 0.39$ m. $λ$ of the 3rd harmonic on an $L = 1.2$ m string fixed both ends? ::: $0.8$ m. How many nodes and antinodes in that $n = 3$ mode (ends included)? ::: 4 nodes, 3 antinodes. Fundamental wavelength for an $L = 2$ m string fixed both ends? ::: $4$ m. Full standing wave for $L = 2$ m, $2 A = 0.04$ m, $f = 25$ Hz? ::: $y = 0.04 sin ((\frac{π}{2}) x) cos (50 π t)$ m. Wave speed and tension for that string ( $μ = 5 \times 1 0^{- 3}$ )? ::: $v = 100$ m/s, $F = 50$ N. Why does a real standing wave transport zero net power? ::: the time factor $sin ω t cos ω t = \frac{1}{2} sin 2 ω t$ averages to zero over a cycle. When do two opposite waves fail to make a true standing wave? ::: when their frequencies differ ( $ω_{1} \neq = ω_{2}$ ), so the nodes drift at $v = (ω_{1} - ω_{2}) /2 k$ .

Connections

Standing waves — formation, nodes, antinodes — the parent note these exercises drill.
Superposition principle — used to add the two waves in every derivation.
Travelling waves — the $y_{1}, y_{2}$ building blocks.
Reflection of waves at boundaries — why one end fixes a node (or antinode).
Resonance and normal modes — the $L = nλ /2$ mode-fitting used in L3–L5.
Waves on a string — $v = f λ$ and $v = F / μ$ links.
Sound in pipes — the free-end/open-end variant of the boundary trap.
Wave number k and wavelength — $k = 2 π / λ$ throughout.