Equation of standing wave formed by y₁=Asin(kx-ωt) and y₂=Asin(kx+ωt)

Question

Two waves travelling in opposite directions are given by:

y_1 = A\sin(kx - \omega t)

y_2 = A\sin(kx + \omega t)

Find the equation of the resultant standing wave. Identify the positions of nodes and antinodes.

Solution — Step by Step

When two waves overlap, the net displacement at any point is the algebraic sum of individual displacements:

y = y_1 + y_2 = A\sin(kx - \omega t) + A\sin(kx + \omega t)

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

Here, $P = kx - \omega t$ and $Q = kx + \omega t$ :

\frac{P+Q}{2} = \frac{(kx - \omega t) + (kx + \omega t)}{2} = kx

\frac{P-Q}{2} = \frac{(kx - \omega t) - (kx + \omega t)}{2} = -\omega t

Therefore:

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

(Using $\cos(-\omega t) = \cos(\omega t)$ )

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

This is the standing wave equation. Key observation: the amplitude at any position $x$ is $2A\sin(kx)$ — a function of position alone, not time. The entire wave oscillates in time as $\cos(\omega t)$ , but the amplitude pattern is fixed in space. That’s why it’s called a “standing” wave — it doesn’t travel.

Nodes are points where amplitude = 0 always. Amplitude = $2A\sin(kx) = 0$ when:

\sin(kx) = 0 \Rightarrow kx = n\pi, \quad n = 0, 1, 2, 3, ...

x = \frac{n\pi}{k} = \frac{n\lambda}{2} \quad \left(\text{since } k = \frac{2\pi}{\lambda}\right)

Nodes occur at $x = 0, \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2}, ...$

They are spaced $\lambda/2$ apart.

Antinodes are points of maximum amplitude ( $= 2A$ ). These occur when $|\sin(kx)| = 1$ :

kx = \frac{(2n-1)\pi}{2}, \quad n = 1, 2, 3, ...

x = \frac{(2n-1)\lambda}{4}

Antinodes occur at $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, ...$

They are spaced $\lambda/2$ apart, and located midway between nodes.

Why This Works

The superposition of two identical waves travelling in opposite directions produces a standing wave. The mathematics shows why: adding $\sin(kx-\omega t)$ and $\sin(kx+\omega t)$ separates the spatial and temporal dependencies. In a travelling wave, position and time are always coupled as $(kx - \omega t)$ — the pattern moves. In the standing wave $2A\sin(kx)\cos(\omega t)$ , they’re separated — the spatial pattern $\sin(kx)$ is fixed, and the whole thing just pulsates in time.

This happens physically in strings fixed at both ends, organ pipes, and microwave cavities. The standing wave represents resonance — the wave bouncing back and forth constructively interferes with itself.

Distance between adjacent node and antinode = $\lambda/4$ . Distance between two adjacent nodes (or antinodes) = $\lambda/2$ . These are the key numbers for string vibration problems — knowing the boundary conditions (fixed/free ends), you can count half-wavelengths to find allowed frequencies.

Common Mistake

Students often apply the identity incorrectly and get $y = 2A\cos(kx)\sin(\omega t)$ or other wrong forms. The error usually comes from mixing up which functions the identity applies to. Double-check: $y_1 + y_2$ uses $\sin P + \sin Q = 2\sin(\frac{P+Q}{2})\cos(\frac{P-Q}{2})$ . The first factor (2A × sin) contains the spatial part, and the second (cos) contains the temporal part. If you get them swapped, check your substitution for P and Q.

Question

Solution — Step by Step

Why This Works

Common Mistake

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