A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm.

Q: Why does the end correction depend on the radius of the tube?

The end correction exists because the air column's vibration doesn't stop abruptly at the open end; it spills out into the surrounding atmosphere. A wider tube (larger radius) moves a larger volume of external air, pushing the effective boundary of the antinode further out. Hence, $e$ is directly proportional to the radius $r$.

Q: Why do we use e = 0.6r instead of some other factor?

The factor $0.6$ (more precisely, about $0.61$ for an unflanged circular opening) is derived theoretically using advanced fluid dynamics and confirmed experimentally. It represents the effective mass of the air vibrating just outside the open end of a cylindrical pipe.

Q: What is the difference between the water level reading and the air column length?

The zero of the meter scale is at the top of the tube. Therefore, the reading of the water level from the top of the tube is exactly equal to the physical length of the air column $l$.

Q: Why are there only odd harmonics in a closed organ pipe?

A closed pipe must always have a node at the closed end and an antinode at the open end. This boundary condition only allows standing wave patterns where the tube length is an odd multiple of a quarter-wavelength ($\frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots$). Even multiples would result in a node or antinode at both ends, which is physically impossible for a closed-open boundary.

Q: Does the temperature of the air affect the resonance length?

Yes, absolutely! The speed of sound in air depends on temperature ($v \propto \sqrt{T}$). A higher temperature increases the speed of sound, which increases the wavelength for a fixed frequency. Consequently, the physical length $l$ required for resonance will also increase as temperature rises.

Question

A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38° C in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is

Options:

(A) 14.0 cm

(B) 15.2 cm

(C) 16.4 cm

(D) 17.6 cm

Sigma Percentile · Accepted Answer

### The Magic of Resonance: Tuning into Sound Waves

Imagine standing next to a tall, hollow tube. You strike a metal tuning fork, and it hums with a pure, clear note. As you hold it over the open mouth of the tube, nothing much happens at first. But as you slowly pour water into the tube, raising the water level, suddenly—*BOOM!*—the quiet hum erupts into a rich, booming roar.

This is not magic; it is the breathtaking phenomenon of **acoustic resonance**.

In this problem, we are invited to step into the shoes of a student performing this classic resonance column experiment. We are given a tube of diameter $4	ext{ cm}$, a tuning fork vibrating at $512	ext{ Hz}$, and the speed of sound in air at $38^\circ	ext{C}$ as $336	ext{ m/s}$. Our goal is to find the exact water level reading when the very first resonance occurs. Let's dive deep into the physics and math that govern this beautiful symphony of waves!

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### The Boundary Illusion: Why End Correction Matters

Before we scribble down any equations, let's address a subtle trap that catches many brilliant minds. When sound waves travel down our tube, they reflect off the water surface. Because water is much denser than air, it acts as a rigid boundary, forcing the air molecules right at the surface to remain completely still. This creates a **displacement node**.

At the open top of the tube, however, the air molecules are free to rush in and out. You might think that the maximum vibration—the **displacement antinode**—occurs exactly at the physical lip of the tube. But nature has a beautiful quirk!

Because the air just outside the tube's mouth also gets pushed back and forth by the vibrating column inside, the boundary of maximum vibration actually extends slightly *beyond* the physical top of the tube. This extra acoustic length is known as the **end correction**, denoted by $e$.

For a cylindrical tube of radius $r$, this end correction is given by the empirical relation:

$$e \approx 0.6r$$

Let's calculate this for our tube. We are given the diameter $D = 4	ext{ cm}$, which means the radius is:

$$r = \frac{D}{2} = 2	ext{ cm}$$

Substituting this radius into our end correction formula:

$$e = 0.6 	imes 2	ext{ cm} = 1.2	ext{ cm}$$

This means the sound wave "feels" a tube that is $1.2	ext{ cm}$ longer than it physically is!

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### The Master Equation of the Fundamental Mode

Now, let's look at the condition for the **first resonance**, also known as the **fundamental mode**.

For a tube closed at one end (by the water surface) and open at the other, the simplest standing wave pattern that can form consists of a single node at the bottom and a single antinode at the top. The distance between a consecutive node and antinode is exactly one-quarter of a wavelength ($\frac{\lambda}{4}$).

Therefore, the effective acoustic length of our air column must equal one-quarter of the wavelength:

$$l_{	ext{effective}} = l + e = \frac{\lambda}{4}$$

Where $l$ is the physical length of the air column (which corresponds to the reading on our meter scale since the zero of the scale coincides with the top of the tube).

To find the wavelength $\lambda$, we use the fundamental wave relation:

$$v = f\lambda \implies \lambda = \frac{v}{f}$$

Substituting this into our resonance condition gives us our master equation:

$$l + e = \frac{v}{4f}$$

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### The Final Calculation: Bringing It All Together

Let's substitute our known values into the master equation. We must be extremely careful with our units! The speed of sound is given in meters per second, while our tube dimensions are in centimeters. Let's convert the speed of sound to centimeters per second:

$$v = 336	ext{ m/s} = 33600	ext{ cm/s}$$

Now, let's plug in $v = 33600	ext{ cm/s}$, $f = 512	ext{ Hz}$, and $e = 1.2	ext{ cm}$:

$$l + 1.2 = \frac{33600}{4 	imes 512}$$

Let's compute the right-hand side step-by-step:

$$4 	imes 512 = 2048$$

$$\frac{33600}{2048} \approx 16.41	ext{ cm}$$

This value, $16.41	ext{ cm}$, is the total acoustic length of the air column. To find the physical length $l$, we simply subtract our end correction:

$$l = 16.41 - 1.2 = 15.21	ext{ cm}$$

Rounding to the nearest option, we get:

$$l \approx 15.2	ext{ cm}$$

This perfectly matches **Option (b)**!

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### Looking Ahead: The Symphony of Higher Harmonics

What if we kept lowering the water level? As the air column grows longer, we would eventually hit the next resonance point. For a closed pipe, only odd harmonics can exist. The second resonance (first overtone) occurs when the effective length is three-quarters of a wavelength:

$$l_2 + e = \frac{3\lambda}{4}$$

Using the same wavelength $\lambda \approx 65.6	ext{ cm}$, the next physical length would be:

$$l_2 = 3 	imes 16.41 - 1.2 = 49.23 - 1.2 = 48.03	ext{ cm}$$

By understanding these simple wave relationships, you can predict exactly where every single pocket of loud, resonant sound will occur. Physics allows us to map out the invisible architecture of sound waves in space!

Select Answer:

Visualized Solution

Visualizing the Resonance Column

Understanding End Correction $e$

Calculating the End Correction

First Resonance Condition

Relating Wavelength to Wave Speed

Substituting the Given Values

Computing the Acoustic Length $\frac{v}{4 f}$

Solving for the Physical Length $l$

The Way Forward: Exploring Higher Overtones

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Magic of Resonance

The Boundary Illusion

The Master Equation of the Fundamental Mode

The Final Calculation

Looking Ahead

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