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A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:
Options
1
2f
2
f/2
3
f
4
3f/2
Correct Answer
f
Solution
1

Original: Open pipe, length L.

Fundamental frequency f = v/2L (antinodes at both ends, L = lambda/2).

2

After dipping to L/2 depth: lower end blocked by water surface (rigid closed end).

Becomes CLOSED pipe of length L/2.

New frequency = v / (4 x L/2) = v/2L = f (same!)

Open pipe L: f = v/2L
Closed pipe L/2: f = v/(4 x L/2) = v/2L = same f
Halving length + changing open to closed = no net change in frequency
Theory: Waves
1. Open Pipe vs Closed Pipe

Open pipe (both ends open): antinodes at both ends. Fundamental f = v/2L. All harmonics present: f, 2f, 3f... Closed pipe (one end closed): node at closed end, antinode at open end. Fundamental f = v/4L. Only odd harmonics: f, 3f, 5f. For same length, closed pipe has half the frequency of open pipe.

2. This Problem Analysis

Key insight: when open pipe (length L) is dipped to half its length, it becomes a closed pipe of length L/2. Open pipe: f = v/2L. Closed pipe (L/2): f = v/4(L/2) = v/2L. These are equal! The 2x reduction in length exactly compensates for the 2x reduction in fundamental frequency when changing from open to closed configuration.

3. Standing Waves in Pipes

Standing waves form by interference of incident and reflected sound waves. Boundary conditions: closed end = displacement node (rigid reflection, phase inversion). Open end = displacement antinode (free reflection, no phase change). Resonance occurs at frequencies satisfying these boundary conditions simultaneously at both ends.

4. Speed of Sound

Speed of sound in air: v = sqrt(gamma RT/M). At 0 deg C: 331 m/s. At 20 deg C: 343 m/s. Increases with temperature as v proportional to sqrt(T). Frequency determined by source (vibrating body). Wavelength changes with medium: lambda = v/f. Musical instruments detune in cold weather (lower v, lower f).

5. Harmonics and Overtones

Harmonics are integer multiples of fundamental. First harmonic = fundamental. Overtones are harmonics above fundamental. Open pipe has all harmonics giving richer sound. Closed pipe has only odd harmonics giving hollow, mellow sound. Flute = open pipe. Clarinet = closed pipe (reed). Organ uses both types for different tonal qualities.

6. Resonance Tube Experiment

Water-filled tube with variable air column length. Tuning fork held at open end. Resonance (loud sound) when L = lambda/4, 3lambda/4, 5lambda/4. From two consecutive resonating lengths L1 and L2: lambda = 2(L2-L1). Speed of sound v = f x lambda. End correction: open end antinode is slightly outside tube, effective length = L + 0.3d where d = tube diameter.

7. Doppler Effect

Apparent frequency change when source and observer have relative motion. f = f0(v + v_obs)/(v - v_src). Source approaching: higher pitch (f > f0). Source receding: lower pitch (f < f0). Applications: police speed guns, SONAR, medical ultrasound (blood flow), astronomical redshift (expanding universe).

8. Beats

When two sound waves of slightly different frequencies f1 and f2 superpose: beat frequency = |f1 - f2|. Intensity varies periodically with beat frequency. Used to tune instruments: adjust string tension until beats disappear (f1 = f2). Audible as periodic loudness variation. Above about 10 Hz beats merge into rough dissonant sound.

Frequently Asked Questions
1. Why does the water surface act as a closed end? ⌄
Sound is a longitudinal pressure wave. Water is essentially incompressible and rigid to sound. At the water surface, sound waves reflect with phase reversal, creating a displacement node (zero displacement). This is exactly the boundary condition at a closed pipe end. The air below the water surface is no longer part of the resonating column.
2. Would the frequency change if dipped to 1/3 of its length? ⌄
If dipped to L/3: remaining air = 2L/3 (closed pipe). New f = v/(4 x 2L/3) = 3v/8L. Original f = v/2L = 4v/8L. New frequency = 3f/4. Different from original. Only dipping to exactly L/2 gives the same frequency. This is a special case.
3. What is the end correction in pipe resonance? ⌄
The displacement antinode at an open end is not exactly at the pipe opening but slightly outside. Effective length = actual length + end correction e, where e = 0.3d (d = diameter). This shifts all resonant frequencies slightly lower. For precision experiments, end correction must be accounted for. In the resonance tube experiment, using two resonances cancels the end correction.
4. How does this relate to musical instruments? ⌄
This problem illustrates why instrument design matters. A wind instrument that is partially filled with water changes its resonant frequency in a complex way depending on how much is filled and the open/closed nature of its ends. Pipe organs require careful temperature control because pitch shifts with temperature (v changes with T). In cold concerts halls, instruments play flat and need warming up time.
5. What if the pipe were dipped 3/4 of its length? ⌄
If dipped to 3L/4: remaining air = L/4 (closed pipe of length L/4). New f = v/(4 x L/4) = v/L = 2f. Frequency doubles! So: dip to L/2 → same frequency f. Dip to 3L/4 → double frequency 2f. Dip to L/4 → remaining L = 3L/4 closed → f = v/3L = 2f/3.
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