Superposition of waves — derive equation for stationary wave

Question

Two transverse waves of the same amplitude $A$ , frequency $\omega$ , and wave number $k$ travel in opposite directions along a string. Derive the equation of the resulting stationary wave. Identify the positions of nodes and antinodes.

(NCERT Class 11, Chapter 15)

Solution — Step by Step

Wave moving in the +x direction: $y_1 = A\sin(kx - \omega t)$

Wave moving in the -x direction: $y_2 = A\sin(kx + \omega t)$

Both have the same amplitude, frequency, and wavelength — this is the condition for a perfect standing wave.

By the principle of superposition: $y = y_1 + y_2$

y = A\sin(kx - \omega t) + A\sin(kx + \omega t)

Using $\sin P + \sin Q = 2\sin\left(\frac{P+Q}{2}\right)\cos\left(\frac{P-Q}{2}\right)$ :

y = 2A\sin(kx)\cos(\omega t)

\boxed{y = 2A\sin(kx)\cos(\omega t)}

This is NOT a travelling wave — there is no $(kx \pm \omega t)$ term. The spatial part $\sin(kx)$ and temporal part $\cos(\omega t)$ are separated. Every point on the string oscillates with the same frequency but with an amplitude that depends on position: $A_{eff}(x) = 2A|\sin(kx)|$ .

Nodes (zero amplitude): $\sin(kx) = 0 \Rightarrow kx = n\pi \Rightarrow x = n\lambda/2$ where $n = 0, 1, 2, \ldots$

Antinodes (maximum amplitude $2A$ ): $|\sin(kx)| = 1 \Rightarrow kx = (n + \frac{1}{2})\pi \Rightarrow x = (2n+1)\lambda/4$

Nodes are separated by $\lambda/2$ . Between consecutive nodes lies one antinode.

Why This Works

When two identical waves travel in opposite directions, they interfere constructively at some points (antinodes) and destructively at others (nodes). At nodes, the two waves always cancel — the string never moves. At antinodes, they always reinforce — the string oscillates with maximum amplitude $2A$ .

The key difference from a travelling wave: in a standing wave, energy does not flow along the string. Energy oscillates between kinetic (at the antinodes) and potential (in the string tension) forms, but stays localised between nodes.

Alternative Method

You can also start with complex exponentials. Writing $y_1 = A\,\text{Im}(e^{i(kx - \omega t)})$ and $y_2 = A\,\text{Im}(e^{i(kx + \omega t)})$ , add them to get $y = A\,\text{Im}(e^{ikx}(e^{-i\omega t} + e^{i\omega t})) = 2A\sin(kx)\cos(\omega t)$ . Same result, but the algebra is more transparent if you are comfortable with complex notation.

In JEE, standing wave questions often ask: “what is the amplitude of the particle at $x = \lambda/6$ ?” Just substitute into $2A\sin(kx) = 2A\sin(2\pi/\lambda \cdot \lambda/6) = 2A\sin(\pi/3) = A\sqrt{3}$ . Practice evaluating $\sin(kx)$ at common fractions of $\lambda$ .

Common Mistake

Students sometimes write the standing wave as $y = 2A\sin(kx - \omega t)$ or include a phase velocity term. A standing wave has no phase velocity — it does not travel. The correct form always has $\sin(kx)$ and $\cos(\omega t)$ as separate, multiplied factors. If your answer has $kx$ and $\omega t$ inside the same trig function, you have written a travelling wave, not a standing wave.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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