Standing Waves in Strings and Pipes
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Direct answer
Two identical waves travelling in opposite directions superpose to form a stationary wave: the medium vibrates in loops between nodes (zero displacement) and antinodes (maximum displacement). A string fixed at both ends and an open pipe sustain every harmonic f_n = n v/(2L), while a pipe closed at one end sustains only odd harmonics f_n = (2n − 1) v/(4L) — the boundary conditions alone decide the spectrum.
What you must remember
- String fixed at both ends: nodes at the ends, all harmonics, f_n = n v/(2L), with v = sqrt(T/mu) for tension T and mass per unit length mu.
- Sonometer laws: frequency varies as 1/L, as sqrt(T) and as 1/sqrt(mu).
- Closed pipe: displacement node at the closed end, antinode at the open end; f_n = (2n − 1) v/(4L), fundamental v/(4L), only odd harmonics.
- Open pipe: antinodes at both ends, f_n = n v/(2L), all harmonics; for equal length its fundamental is twice the closed pipe's.
- End correction: each open end lengthens the effective column by about 0.6 times its radius; in a resonance tube the difference between successive resonance lengths is lambda/2, the standard route to the speed of sound.
- In air columns a displacement node is a pressure antinode: pressure fluctuates most where air stands still, at the closed end.
- Beats: two close frequencies give loudness maxima at |f1 − f2| per second; loading a tuning fork with wax lowers its frequency.
Common confusion
The classic error is carrying "all harmonics" from strings and open pipes into closed pipes — the mismatched end conditions there admit only odd harmonics, which is why a closed pipe sounds characteristic. Students also swap harmonic and overtone counts: the first overtone of a closed pipe is the third harmonic, not the second. Remember finally that it is the displacement wave which has a node at a closed end; the pressure wave has an antinode there.
Exam-focused takeaway
JEE Main asks harmonic frequencies, resonance-tube sound-speed measurement, end-correction arithmetic and beat frequency as numerical-value questions. JEE Advanced prefers assemblies: a pipe closed by a rising water level (find the resonating lengths), two pipes sounded together, beats deciding an unknown fork frequency after waxing, temperature dependence (v grows as the square root of absolute temperature), and strings with shifted bridges or loads. Draw the allowed standing-wave pattern first; the frequency follows.
Frequently asked questions
Why does a closed pipe produce only odd harmonics?
The closed end must be a node and the open end an antinode, so only an odd number of quarter-wavelengths fits; even harmonics cannot satisfy both conditions.
What is end correction?
The antinode actually sits just outside each open end, about 0.6 times the radius beyond the rim, so the effective pipe is longer than its physical length.
How do a harmonic and an overtone differ?
Harmonics count from the fundamental (first harmonic); overtones count above it — the first overtone is the second harmonic for a string or open pipe, but the third for a closed pipe.
How does a resonance tube measure sound speed?
Water sets the column length; successive resonances differ by lambda/2, and v = frequency × wavelength, with the end correction cancelling in the subtraction.
What are beats used for?
Two nearby frequencies give loudness oscillations at |f1 − f2|; musicians tune by nulling beats, and JEE problems use them to deduce unknown frequencies when a fork is waxed or filed.
Why must a node form at a closed end?
Air at the rigid wall cannot move along the pipe, forcing zero displacement amplitude there — while the pressure fluctuation is maximum.