Physics · Waves · NEET
The first resonance length L1 is NOT a clean quarter wavelength because of the end correction — the antinode sits slightly ABOVE the open top, so the real column is a bit longer than what you measure. If you use lambda = 4L1 you carry that error into your answer. But BOTH resonances have the same end correction, so when you subtract, L2 - L1 = lambda/2 and the error cancels. That is exactly why the difference method is the accurate one, and why NEET numericals give you two lengths.
An open end is a displacement antinode, but the air just outside the tube also vibrates, so the antinode forms a small distance 'e' (about 0.3 times the tube diameter) OUTSIDE the tube. So the true acoustic length = measured length + e. First resonance: L1 + e = lambda/4. Second: L2 + e = 3 lambda/4. Subtract: L2 - L1 = lambda/2. The end correction e drops out completely.
The water surface is a rigid boundary, so it forces a displacement NODE there (air cannot move into water). The open top is an ANTINODE. A closed-open pipe supports only odd multiples of lambda/4: lambda/4, 3 lambda/4, 5 lambda/4. That is why resonances appear at these specific lengths and spacing lambda/2 apart. If it were open at both ends the pattern (and formulas) would change.
The tuning fork forces the air at one fixed frequency f. The air column has its own natural frequencies set by its length. When the column length makes one of its natural frequencies equal to f, the two match and energy builds up into a large-amplitude standing wave — you hear a sudden loud sound. At in-between lengths the frequencies do not match, so the response is weak.
No correction is needed for the number you calculate — v = f x lambda already gives the speed of sound at whatever temperature the experiment was run. The temperature only matters if the question asks you to COMPARE with another temperature, because v is proportional to the square root of absolute temperature (v ~ sqrt(T)). The 2018 PYQ states 27 C just to fix the value; the calculation itself is temperature-free.
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.
v = f x lambda, where f is the tuning fork frequency and lambda = 2(L2 - L1), with L1 and L2 the first and second resonance lengths. Together: v = 2 f (L2 - L1).
It behaves as a pipe closed at one end (water surface = node) and open at the other (top = antinode). So it resonates only at odd multiples of lambda/4: lambda/4, 3 lambda/4, 5 lambda/4.
End correction e = (L2 - 3 L1) / 2. It comes from L1 + e = lambda/4 and L2 + e = 3 lambda/4. Roughly e is about 0.3 times the inner diameter of the tube.
The shortest resonating length is the fundamental mode where the tube holds a quarter wavelength (lambda/4). It is the first (smallest) length at which the air column's natural frequency matches the fork, so you hear the first loud sound as you slowly raise the water level down.
No. The tuning fork frequency f is fixed and known. You change the air-column LENGTH (by moving the water level) until the column's natural frequency matches this fixed f, giving resonance.
