A cylinder resonance tube open at both ends has fundamental frequency f in air. Half of the length o

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

The water surface is a much denser medium than air, acting as a rigid boundary. Air molecules cannot easily displace into the water, forcing a displacement node (and a pressure antinode) to form at the boundary, just like a solid closed end.

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

No, the speed of sound $v$ depends only on the properties of the medium inside the tube (which is still air). Since the temperature and composition of the air remain the same, $v$ remains constant.

Q: What would happen if we dipped more than half of the tube?

If you dip more than half of the tube, the active air column length $L'$ becomes less than $L/2$. Since $f' = \frac{v}{4L'}$, a smaller $L'$ would result in a higher fundamental frequency ($f' > f$).

Q: Is the fundamental frequency of a closed pipe always half of an open pipe of the same length?

Yes. For a given length $L$, an open pipe has a fundamental frequency of $\frac{v}{2L}$, while a closed pipe has a fundamental frequency of $\frac{v}{4L}$, which is exactly half.

Q: Does the material of the tube affect the fundamental frequency?

No, the fundamental frequency depends only on the length of the air column and the speed of sound in the air inside it. The material of the tube walls does not affect the frequency of the air column's resonance.

Question

A cylinder resonance tube open at both ends has fundamental frequency $f$ in air. Half of the length of the tube is dipped vertically in water. The fundamental frequency of the air column now is ......

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine holding a simple cylindrical tube, completely open at both ends, and letting sound waves resonate inside it.

In its **fundamental mode of vibration**, the air column inside this open tube forms a standing wave pattern.

Because both ends are open to the atmosphere, the air molecules at the boundaries are free to move with maximum freedom. This creates **displacement antinodes** at both ends.

To satisfy this boundary condition with the longest possible wavelength, a single **displacement node** must form exactly at the center of the tube.

This wave pattern represents half of a full wave cycle, meaning the length of the tube $L$ is equal to half of the wavelength $\lambda$:

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

Using the wave equation $v = f\lambda$, where $v$ is the speed of sound in air, we can write the fundamental frequency $f$ of this open tube as:

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

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### Dipping the Tube: Changing the Boundary Conditions

Now, let us perform a fascinating experiment. We dip this tube vertically into water until exactly **half of its length** is submerged.

What happens to the physics of our system?

First, the water surface acts as a rigid, impenetrable barrier. Air molecules right at the water's surface cannot vibrate back and forth. This forces a **displacement node** to form at the water level.

Second, the top end of the tube remains open to the air, which means a **displacement antinode** must still form at the top.

By introducing water, we have transformed our open-ended tube into a **closed organ pipe** (open at one end, closed at the other).

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### Calculating the New Fundamental Frequency

Since half of the tube is submerged in water, the active air column that can resonate is now only in the upper half of the tube.

Therefore, the new effective length $L'$ of our closed air column is:

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

For a closed organ pipe of length $L'$, the fundamental mode of vibration consists of a node at the closed end and an antinode at the open end. This corresponds to a quarter of a wavelength:

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

Substituting our new length $L' = \frac{L}{2}$ into this relation, we find the new wavelength $\lambda'$:

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

Now, let us calculate the new fundamental frequency $f'$ using the wave speed relation:

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

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### The Beautiful Symmetry

Look at our two results side-by-side:

1. **Initial Open Tube:** $f = \frac{v}{2L}$
2. **Half-Dipped Closed Tube:** $f' = \frac{v}{2L}$

They are mathematically identical!

By dipping the tube halfway, two competing physical changes occurred:
* Closing one end tended to **lower** the fundamental frequency by a factor of $2$ (since a closed pipe of the same length has a frequency of $\frac{v}{4L}$).
* Halving the length of the tube tended to **raise** the fundamental frequency by a factor of $2$.

These two effects perfectly counterbalanced each other, leaving the fundamental frequency completely unchanged at **$f$**.

Animated Solution for Physics - Waves: A cylinder resonance tube open at both ends has fundamental frequency $f$ in air. Half of the length of the tube is dipped vertically in water. The fundamental frequency of the air column now is ......

Visualized Solution

Visualizing the Open Tube

Fundamental Frequency of Open Tube

Dipping the Tube in Water

Determining the New Air Column Length

Fundamental Frequency of a Closed Pipe

Substituting the New Length

Simplifying the Expression

Comparing the Frequencies

Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

Dipping the Tube

Calculating the New Fundamental Frequency

The Beautiful Symmetry

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