Physics · Waves · NEET
A string is fixed at BOTH ends, so both ends must be nodes. The longest wave that fits with a node at each end has one antinode in the middle, and the string length holds exactly HALF a wavelength: L = lambda/2, so lambda = 2L. Then f = v/lambda = v/2L. The v/4L formula belongs to a closed organ pipe (one end open, one end closed), which is a different boundary condition. Always start from the boundary: string fixed both ends → half wavelength → 2L in the denominator.
Harmonics are counted from the fundamental: 1st harmonic = f, 2nd harmonic = 2f, 3rd harmonic = 3f, and so on (n·f). Overtones are counted from the first tone ABOVE the fundamental: 1st overtone = 2nd harmonic = 2f, 2nd overtone = 3rd harmonic = 3f. So for a string, (n)th overtone = (n+1)th harmonic. Students lose marks by mixing these up. For a string, ALL harmonics are present, so there is a clean +1 shift between overtone number and harmonic number.
f = (1/2L)·sqrt(T/mu). Frequency depends on the SQUARE ROOT of tension. If you double the tension T, the frequency becomes sqrt(2) times larger (about 1.41×), not double. To DOUBLE the frequency you must make the tension 4 times larger. This is why guitar tuning pegs make only small pitch changes for large tension changes.
Because both ends are nodes, the allowed patterns are 1 loop, 2 loops, 3 loops, 4 loops... any whole number of half-wavelengths fits. That gives f, 2f, 3f, 4f — every integer multiple, both even and odd. A closed organ pipe (node at one end, antinode at the other) only allows odd harmonics f, 3f, 5f. So a string is richer: it contains all harmonics.
In the fundamental (1st harmonic), the string vibrates as a single loop: 2 nodes (one at each fixed end) and 1 antinode (in the middle). For the nth harmonic there are (n+1) nodes and n antinodes. So the 2nd harmonic has 3 nodes and 2 antinodes, and so on.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
f = v/2L = (1/2L)·sqrt(T/mu), where v is wave speed, L is length, T is tension, and mu is linear mass density (mass per unit length).
f1 = v/2L (fundamental), f2 = 2·(v/2L) = v/L, f3 = 3·(v/2L) = 3v/2L. In general f_n = n·v/2L.
A string fixed at both ends produces ALL harmonics (1f, 2f, 3f, 4f...), both even and odd. Only a closed organ pipe is limited to odd harmonics.
f is inversely proportional to L: f ∝ 1/L. A shorter string gives a higher pitch. Halving the length doubles the fundamental frequency (if tension and mu stay fixed).
For a string, the nth overtone equals the (n+1)th harmonic, with frequency (n+1)·f. Example: 1st overtone = 2nd harmonic = 2f.
