f₁ = v/(2L). Second harmonic f₂ = 2f₁. Nodes and antinodes.

Standing Waves on a String — Fundamental and Harmonics

Question

A string of length 80 cm is fixed at both ends. The speed of transverse waves on the string is 320 m/s. Find the frequency of the fundamental mode and the second harmonic. How many nodes and antinodes does the second harmonic have?

(JEE Main 2024, similar pattern)

Solution — Step by Step

Both ends are fixed, so both ends must be nodes. This is the key constraint that forces standing waves into specific allowed patterns — only wavelengths that “fit” between two nodes are permitted.

For a string of length $L$ , the condition is: $L = n \cdot \dfrac{\lambda}{2}$ , where $n = 1, 2, 3, \ldots$

For $n = 1$ (fundamental mode), we get the longest possible wavelength:

\lambda_1 = 2L = 2 \times 0.80 = 1.6 \text{ m}

Using $f = v/\lambda$ :

f_1 = \frac{v}{\lambda_1} = \frac{v}{2L} = \frac{320}{1.6} = \mathbf{200 \text{ Hz}}

This formula $f_1 = \dfrac{v}{2L}$ is what you must memorise cold — it’s the foundation for everything that follows.

The $n^\text{th}$ harmonic has frequency $f_n = n \cdot f_1$ . So the second harmonic:

f_2 = 2f_1 = 2 \times 200 = \mathbf{400 \text{ Hz}}

The wavelength shortens by half: $\lambda_2 = L = 0.80$ m.

In the $n^\text{th}$ harmonic:

Nodes = $n + 1$ (the two fixed ends are always nodes)
Antinodes = $n$

For $n = 2$ : 3 nodes (both ends + 1 middle) and 2 antinodes.

Draw it out: N — A — N — A — N. That’s the second harmonic pattern.

Why This Works

A standing wave forms when two identical waves travel in opposite directions — the reflected wave interferes with the incident wave. At fixed ends, displacement is always zero (nodes). Between nodes, the string oscillates with maximum amplitude (antinodes).

The “fit” condition ( $L = n\lambda/2$ ) simply says: we need a whole number of half-wavelengths to fit between the two fixed nodes. Any other wavelength would destructively cancel itself out over time and die away. Only the allowed harmonics survive.

The fundamental ( $n = 1$ ) has one antinode at the centre — the string bulges out at the middle. The second harmonic ( $n = 2$ ) splits the string into two equal loops, with a node exactly at the midpoint. Each higher harmonic adds one more node and one more antinode.

Alternative Method — Using the Harmonic Number Directly

Instead of finding $\lambda$ first, use the direct formula:

f_n = \frac{nv}{2L}

For $n = 1$ : $f_1 = \dfrac{1 \times 320}{2 \times 0.80} = 200$ Hz

For $n = 2$ : $f_2 = \dfrac{2 \times 320}{2 \times 0.80} = 400$ Hz

In JEE, when a question says “third overtone” — watch out. The third overtone is the fourth harmonic ( $n = 4$ ), not the third. Overtone count starts from 1 after the fundamental, so: 1st overtone = 2nd harmonic, 2nd overtone = 3rd harmonic, and so on.

Common Mistake

The most frequent error: students write $f_1 = v/L$ instead of $f_1 = v/(2L)$ . This happens when you confuse the standing wave condition with the formula for an open pipe or a string fixed at one end.

For a string fixed at both ends (or a pipe open at both ends), it’s $v/(2L)$ . For a pipe closed at one end (or string fixed at one end, free at the other), it’s $v/(4L)$ .

The physical reason: with both ends fixed, the minimum “fit” is a half-wavelength ( $\lambda/2 = L$ ), not a full wavelength. If you sketch the wave shape, this becomes obvious — one arch fits between two nodes.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Harmonic Number Directly

Common Mistake

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