Physics · Waves · NEET
An air column closed at one end must have a node at the closed (water) end and an antinode at the open top. The shortest column that fits this is L1 = lambda/4. The next one that still fits a node-at-bottom, antinode-at-top pattern adds one more half-loop, giving L2 = 3 lambda/4. The difference L2 - L1 = 3 lambda/4 - lambda/4 = lambda/2. Each new resonance adds exactly one half-wavelength of air, so every gap is lambda/2. This is why you can find lambda without knowing the end correction — the correction cancels when you subtract.
With a node fixed at the closed end and an antinode at the open end, the allowed lengths are L = lambda/4, 3 lambda/4, 5 lambda/4, ... = (2n-1) lambda/4. The frequencies are v/(4L), 3v/(4L), 5v/(4L), ... i.e. odd multiples of the fundamental only. Even harmonics would need an antinode at both ends or a node at both ends, which a one-end-closed pipe cannot provide. So a closed pipe (and a resonance tube) gives 1st, 3rd, 5th harmonics — never the 2nd or 4th.
Take any two successive resonating lengths L1 and L2 for the same tuning fork of frequency f. The wavelength is lambda = 2(L2 - L1). Then v = f x lambda = 2f(L2 - L1). You do not need the tube's exact length or the end correction because subtracting two lengths removes it. Example: f = 320 Hz, L1 = 20 cm, L2 = 73 cm gives lambda = 2(0.73 - 0.20) = 1.06 m, so v = 320 x 1.06 = 339 m/s.
Because resonances follow lambda/4, 3 lambda/4, 5 lambda/4 — an odd-multiple pattern, not a doubling. If the first resonance is at 50 cm (= lambda/4, so lambda = 200 cm), the next is at 3 lambda/4 = 150 cm, not 100 cm. Students who write '2 x 50 = 100 cm' fall for the trap. Multiply by 3 (then 5, 7...) not by 2.
The antinode does not sit exactly at the open rim; it forms a little above it by a distance e (about 0.3 x diameter). So the true resonance condition is L + e = (2n-1) lambda/4. But when you subtract two successive lengths, the +e appears in both and cancels: (L2 + e) - (L1 + e) = lambda/2. That is exactly why using two successive lengths gives an end-correction-free value of the speed of sound — a favourite NEET point.
An air column closed at one end and open at the other resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column that resonates with the same tuning fork is:
A tuning fork produces resonance in a glass tube whose air-column length is varied by a piston. At 27 C, two successive resonances occur at column lengths 20 cm and 73 cm. If the tuning fork frequency is 320 Hz, the speed of sound in air at 27 C is:
A tuning fork of frequency 800 Hz produces resonance in a resonance-column tube (upper end open, lower end closed by the water surface). Successive resonances occur at lengths 9.75 cm, 31.25 cm and 52.75 cm. The speed of sound in air is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
L + e = (2n-1) lambda/4, where n = 1, 2, 3... and e is the end correction. Without end correction the resonating lengths are lambda/4, 3 lambda/4, 5 lambda/4.
Exactly lambda/2. So lambda = 2 x (difference between any two successive resonating lengths), independent of the end correction.
Because the gap between two successive resonances equals lambda/2 and cancels the unknown end correction. With a known fork frequency f, v = 2f(L2 - L1) gives the speed of sound directly.
Yes. Higher temperature raises the speed of sound, which increases lambda for the same frequency, so the resonating lengths get longer. v is proportional to the square root of absolute temperature.
No. Being closed at the water surface, it supports only odd harmonics (1st, 3rd, 5th...). Even harmonics require an open-open or closed-closed configuration.
