Standing Waves on Strings

Standing waves: where two travelling waves meet

When two waves of the same frequency and amplitude travel in opposite directions on a string, they superpose to create a pattern that does not move along the string. Some points stay perfectly still (nodes); others vibrate with maximum amplitude (antinodes). This is the standing wave.

Every guitar string, sitar wire, and tabla skin uses this principle. The string vibrates at frequencies determined by its length, tension, and mass per unit length — not by how hard we pluck it. The pluck just sets the amplitude; the harmonics are pre-determined.

For JEE and NEET, this topic carries a steady 4-6 mark weightage every year. CBSE Class 11 expects derivations.

Key Terms & Definitions

Node — a point of zero displacement. Adjacent nodes are separated by $\lambda/2$ .

Antinode — a point of maximum displacement. Sits exactly halfway between two nodes.

Fundamental frequency ( $n_1$ or $f_1$ ) — the lowest natural frequency of vibration. Corresponds to one antinode in the middle, two nodes at the ends.

Harmonic — an integer multiple of the fundamental. The $n$ -th harmonic has frequency $n f_1$ .

Overtone — a frequency higher than the fundamental that the string actually produces. For a string fixed at both ends, all harmonics are overtones (with the first overtone being the second harmonic).

Frequency of vibration

For a string of length $L$ , fixed at both ends, the wavelength of the $n$ -th mode is:

\lambda_n = \frac{2L}{n}, \quad f_n = \frac{n v}{2L} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

Here $v = \sqrt{T/\mu}$ is the wave speed, $T$ is tension, and $\mu$ is mass per unit length.

The fundamental ( $n = 1$ ) gives $f_1 = (1/2L)\sqrt{T/\mu}$ . Higher modes are integer multiples — $f_2 = 2 f_1$ , $f_3 = 3 f_1$ , and so on.

Why integer multiples?

The string is fixed at both ends, so both ends must be nodes. The only wavelengths that fit are those where an integer number of half-wavelengths span $L$ : $L = n \lambda/2$ .

This boundary condition is the entire reason musical instruments produce discrete tones rather than a continuous spectrum.

Worked Examples

Easy (CBSE Class 11)

Example 1. A string of length $0.5\text{ m}$ has a wave speed of $200\text{ m/s}$ . Find its fundamental frequency.

f_1 = \frac{v}{2L} = \frac{200}{1} = 200\text{ Hz}

The second harmonic is $400\text{ Hz}$ , the third is $600\text{ Hz}$ , and so on.

Medium (JEE Main)

Example 2. A wire of length $1\text{ m}$ and mass $4\text{ g}$ is stretched with a tension of $100\text{ N}$ . Find the frequency of the third harmonic.

$\mu = 4\times 10^{-3}/1 = 4\times 10^{-3}\text{ kg/m}$ . Wave speed:

v = \sqrt{T/\mu} = \sqrt{100/(4\times 10^{-3})} = \sqrt{25000} = 158.1\text{ m/s}

Third harmonic: $f_3 = 3v/(2L) = 3 \times 158.1/2 = 237.2\text{ Hz}$ .

Hard (JEE Advanced)

Example 3. Two strings of equal length and tension but different mass per unit length ( $\mu_1 = 4\,\mu_2$ ) are joined end-to-end. The fundamental frequency of the combined system is $f$ . Find $f$ in terms of $f_2$ (the fundamental of string 2 alone).

The thicker string has wave speed $v_1 = \sqrt{T/\mu_1} = v_2/2$ . For a junction of two strings, the standing wave forms with a node at the junction (in many setups), and the frequencies must match across both segments. This typically gives $f$ between $f_1$ and $f_2$ — JEE Advanced 2019 has the full setup.

Exam-Specific Tips

CBSE awards full marks if you derive $f_n = nv/(2L)$ from the boundary conditions. Memorising the formula without derivation costs 1-2 marks.

For JEE, $f_1 \propto 1/L$ , $f_1 \propto \sqrt{T}$ , $f_1 \propto 1/\sqrt{\mu}$ . Most numerical questions test these proportions. If $T$ becomes $4T$ , $f$ doubles. Faster than re-computing.

NEET 2024 had a question where they specified a string fixed at one end (e.g. attached to a wall, free at the other). For that case, only odd harmonics exist: $f_n = (2n-1)v/(4L)$ . Watch the boundary conditions carefully.

Common Mistakes to Avoid

Using $L = \lambda$ for the fundamental. Wrong — for a string fixed at both ends, $L = \lambda/2$ for the fundamental, so $\lambda = 2L$ .

Mixing up “harmonic” and “overtone.” For a string fixed at both ends, the first overtone IS the second harmonic. They are not the same numbering.

Forgetting that $\mu$ is mass per unit length, not total mass. Always divide total mass by length first.

Treating velocity as if it depends on frequency. For a given string, $v = \sqrt{T/\mu}$ is fixed. Frequency adjusts via $\lambda$ .

Using $v$ from sound-in-air ( $340\text{ m/s}$ ) for waves on a string. The string has its own wave speed, much faster, set by tension and density.

Practice Questions

Q1. A 1 m string has $f_1 = 100\text{ Hz}$ . What is its wave speed?

$v = 2L f_1 = 2 \times 1 \times 100 = 200\text{ m/s}$ .

Q2. Tension is doubled. New $f_1$ ?

$f_1 \propto \sqrt{T}$ . New frequency $= \sqrt{2} \times$ old.

Q3. A string vibrates at $300\text{ Hz}$ in its third harmonic. What is the fundamental?

$f_1 = f_3 / 3 = 100\text{ Hz}$ .

Q4. Find the position of antinodes in the second harmonic of a 1 m string.

For $n = 2$ : nodes at $0, 0.5, 1$ m. Antinodes at $0.25\text{ m}$ and $0.75\text{ m}$ .

Q5. A string of $\mu = 0.01\text{ kg/m}$ under $T = 40\text{ N}$ . Wave speed?

$v = \sqrt{40/0.01} = \sqrt{4000} \approx 63.2\text{ m/s}$ .

Q6. Why does a thicker string sound lower?

Thicker means larger $\mu$ , so $v$ is smaller, so $f_1 = v/(2L)$ is smaller — lower pitch.

Q7. A guitarist presses the string at the midpoint. What harmonic do they get?

Pressing creates a node — the effective length halves, so the new fundamental is $2 f_1$ , the second harmonic of the open string.

Q8. How many nodes (excluding endpoints) in the fifth harmonic?

Total nodes including ends $= n + 1 = 6$ . Excluding ends: $4$ nodes.

FAQs

Q. Why don’t standing waves transport energy?

Energy is exchanged between adjacent string segments, but no net energy flows along the string — the waves moving in opposite directions carry equal and opposite energy fluxes.

Q. Can a string vibrate at multiple harmonics simultaneously?

Yes — a real plucked string contains the fundamental plus several overtones. The combination gives the instrument its timbre.

Q. What changes when one end is free?

The free end becomes an antinode, not a node. Only odd harmonics ( $1, 3, 5, \ldots$ ) exist; the formula becomes $f_n = (2n-1)v/(4L)$ .

Q. Why do nodes feel still even though the string is vibrating?

Nodes are points where the two travelling waves interfere destructively at all times — the displacements always cancel exactly there.

Q. How do real instruments enrich the basic standing-wave model?

Real strings have stiffness, finite end conditions, and air-loaded boundaries. These cause slight detuning of high harmonics — important for piano tuning but ignorable for JEE/NEET.