In Physics 2 you met standing waves on a guitar string: fixed nodes, swinging antinodes, frequencies $f_n = nv/2L$. Now we can see where they come from. A wave traveling right and its reflection traveling left are the same size and overlap all along the string. Their sum, using the sum formulas, is a standing wave.
🔑 Key idea
$$A\sin(kx - \omega t) + A\sin(kx + \omega t) = 2A\sin(kx)\cos(\omega t).$$ Every point oscillates in time as $\cos\omega t$, with its own amplitude $2A|\sin kx|$. Nodes (amplitude 0) sit at $kx = n\pi$, that is $x = 0, \frac{\lambda}{2}, \lambda, \ldots$; antinodes (amplitude $2A$) at $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \ldots$ The nodes never move.
💡 Why it works
Use $\sin a + \sin b = 2\sin\frac{a + b}{2}\cos\frac{a - b}{2}$ with $a = kx - \omega t$ and $b = kx + \omega t$: the half-sum is $kx$ and the half-difference is $-\omega t$, and $\cos(-\omega t) = \cos\omega t$. The result is a product of a function of $x$ and a function of $t$. That is the mark of a standing wave: the shape $\sin kx$ stays put and only grows and shrinks. Where $\sin kx = 0$, the product is zero at every moment, so those points never move.
Two traveling waves (dashed) and their sum. The sum's nodes never move, while its antinodes swing between $+2A$ and $-2A$.
On a string fixed at both ends, $x = 0$ and $x = L$ must both be nodes. That requires $kL = n\pi$, so $\lambda_n = 2L/n$ and $f_n = \frac{v}{\lambda_n} = \frac{nv}{2L}$: the harmonics you already know, now derived.
Worked examples
✏️ Worked example: Read a standing wave
A string vibrates as $$y = (0.060\ \text{m})\sin(0.50\pi x)\cos(40\pi t)$$ ($x$ in m, $t$ in s). Find the distance between nodes, the amplitude and speed of the two traveling waves.
$k = 0.50\pi$ rad/m, so $\lambda = 2\pi/k = 4.0$ m. Nodes are $\lambda/2 = 2.0$ m apart.
$2A = 0.060$ m, so each traveling wave has $A = 0.030$ m.
$v = \omega/k = \frac{40\pi}{0.50\pi} = 80$ m/s.
Answer: Nodes 2.0 m apart; traveling waves of amplitude 3.0 cm moving at 80 m/s
✏️ Worked example: How much does a point move?
For the same string, how far from the middle does the point at $x = 0.50$ m swing, and what is its top speed?
Amplitude there: $0.060|\sin(0.50\pi \times 0.50)| = 0.060\sin(\pi/4) = 0.0424$ m.
$\frac{\partial y}{\partial t} = -(0.060)\sin(kx)(40\pi)\sin(40\pi t)$, so the top speed is $0.0424 \times 40\pi$.
$= 5.33$ m/s.
Answer: Swings 4.2 cm each way, top speed 5.3 m/s
⚠️ Common mistake
Thinking the nodes drift along the string, the way crests do in a traveling wave. In $2A\sin(kx)\cos(\omega t)$ the position factor $\sin kx$ never changes, so a point with $\sin kx = 0$ is still at every instant. No energy flows past a node on average: that is why the wave is called standing.
⚠️ Common mistake
Using the standing wave's amplitude $2A$ as the amplitude of each traveling wave. The traveling waves each have half: $A$.
The third harmonic: 4 nodes (red) that never move and 3 antinodes (green) that swing the most.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.