A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vert

Q: Why does the water surface act as a closed end?

Water is much denser than air, making it act as a rigid boundary. Air molecules cannot easily displace the water surface, forcing a displacement node (zero motion) to form at the interface, which is the defining characteristic of a closed end.

Q: If the fundamental frequency doesn't change, does anything change about the sound?

Yes! While the fundamental frequency remains $f$, the overtones (harmonics) change completely. An open pipe produces all integer harmonics ($f, 2f, 3f, \dots$), whereas a closed pipe only produces odd harmonics ($f, 3f, 5f, \dots$). The quality or timbre of the sound will be noticeably different.

Q: What would happen if we dipped the tube to a depth other than half its length?

If dipped to a depth $d$, the active air column length becomes $L' = L - d$. The new fundamental frequency would be $f' = \frac{v}{4(L-d)}$. This will only equal $f$ when $L-d = L/2$, which is exactly the halfway point.

Q: Does the speed of sound $v$ change when the tube is dipped in water?

No, the speed of sound $v$ depends only on the properties of the medium inside the active column (which is still air at the same temperature and pressure), so $v$ remains constant.

Q: Is end correction considered in this problem?

In standard introductory problems like this one, end correction is neglected unless explicitly mentioned. If end correction $e$ were considered, the frequencies would not match perfectly, but the ideal theoretical answer remains $f$.

Question

A cylindrical tube, open at both ends, has a fundamental frequency $f$ in air. The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now

Options:

(A) $f/2$

(B) $3f/4$

(C) $f$

(D) $2f$

Sigma Percentile · Accepted Answer

### The Magic of Standing Waves

Imagine holding a simple hollow cylindrical tube in your hands.

To the untrained eye, it is just a piece of plastic or metal.

But to a physicist, it is a **resonant cavity**—a playground where air molecules dance in perfect harmony to create sound.

When you blow across one end of this tube, you generate a spectrum of sound waves.

Most of these waves quickly die out, but a select few—those whose wavelengths perfectly match the geometry of the tube—reflect back and forth, reinforcing themselves to form **standing waves**.

### The Open Pipe Symphony

Initially, our tube is completely open at both ends.

Because both ends are open to the atmosphere, the air molecules there are free to move with maximum freedom.

This means that at both boundaries, we must have **displacement antinodes**.

For the simplest, most fundamental mode of vibration, the wave must transition from an antinode at one end, pass through a single point of zero motion (a **node**) in the center, and return to an antinode at the other end.

This beautiful, symmetric pattern represents exactly half of a full wave cycle.

Therefore, the length of the tube $L$ is equal to half the wavelength:

$$L = \frac{\lambda}{2} \implies \lambda = 2L$$

Using the fundamental wave relation $v = f \lambda$, we can write the fundamental frequency $f$ of this open pipe as:

$$f = \frac{v}{2L}$$

This is our starting point.

### Dipping into the Water: A Metamorphosis

Now, let's perform a fascinating experiment.

We dip this tube vertically into water until exactly half of its length is submerged.

What happens to the air column inside?

First, the water enters the lower half of the tube, completely filling it.

This reduces the length of the active air column to exactly half of its original value:

$$L' = \frac{L}{2}$$

Second, and more importantly, the water surface acts as a rigid, unyielding barrier.

Air molecules right at the water's surface cannot move at all.

This forces a **displacement node** to form at the water interface.

The top of the tube, however, remains open to the air, preserving a **displacement antinode**.

Our open-ended tube has undergone a complete metamorphosis!

It is no longer an open organ pipe; it is now a **closed organ pipe** of length $L' = \frac{L}{2}$.

### The Mathematical Harmony

For a closed organ pipe, the fundamental mode of vibration consists of a node at the closed end and an antinode at the open end.

This corresponds to exactly one-quarter of a wavelength:

$$L' = \frac{\lambda'}{4} \implies \lambda' = 4L'$$

Thus, the fundamental frequency $f'$ of this closed pipe is:

$$f' = \frac{v}{4L'}$$

Now, let's substitute our new active length $L' = \frac{L}{2}$ into this equation:

$$f' = \frac{v}{4 \left(\frac{L}{2}\right)}$$

Simplifying the fraction in the denominator, we get:

$$f' = \frac{v}{2L}$$

Look at this stunning result!

The new fundamental frequency $f'$ is mathematically identical to our original fundamental frequency $f$:

$$f' = f$$

Even though we cut the active length of the tube in half, the change in boundary conditions (from open-open to open-closed) perfectly compensated for the loss of length.

The fundamental frequency remains completely unchanged!

This elegant cancellation is a beautiful testament to the mathematical harmony of wave mechanics.

Animated Solution for Physics - Waves: A cylindrical tube, open at both ends, has a fundamental frequency $f$ in air. The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now

Select Answer:

Visualized Solution

Visualizing the Open Tube

Fundamental Frequency of an Open Pipe

Dipping the Tube in Water

Transition to a Closed Organ Pipe

Fundamental Frequency of a Closed Pipe

Substituting the New Length

Simplifying the Expression

Comparing the Frequencies

Exploring Further Scenarios

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Magic of Standing Waves

The Open Pipe Symphony

Dipping into the Water

The Mathematical Harmony

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