A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm.

Q: Why do we treat the resonance column as a closed organ pipe when the top is open?

A "closed" organ pipe means it is closed at *one* end. In this experiment, the water surface acts as a rigid boundary, preventing air molecules from moving, which creates a displacement node. The open top allows maximum movement, creating an antinode. This node-antinode setup is the defining characteristic of a closed organ pipe.

Q: What exactly is end correction and why do we need it?

When sound waves reach the open end of the tube, they don't instantly stop vibrating. The air just outside the tube also participates in the vibration. This means the antinode forms slightly outside the physical tube. The end correction ($e$) accounts for this extra vibrating length.

Q: Where does the formula $e = 0.3d$ come from?

It is an empirical formula derived from experiments and theoretical acoustics for cylindrical pipes. It states that the end correction is approximately $0.6$ times the radius, or $0.3$ times the diameter of the pipe.

Q: Why is the effective length equal to $\lambda/4$ for the first resonance?

The first resonance corresponds to the fundamental mode of vibration. In a closed pipe, the simplest standing wave pattern consists of one node (at the closed end) and one antinode (at the open end). The distance between an adjacent node and antinode is always one-quarter of a wavelength ($\lambda/4$).

Q: What would happen if we performed this experiment on a hotter day?

The speed of sound in air increases with temperature. Since $v = f\lambda$ and the frequency $f$ of the tuning fork is constant, a higher speed $v$ means a longer wavelength $\lambda$. Consequently, the resonance length $L$ would also increase.

Question

A student is performing the experiment of resonance column. The diameter of the column tube is $6 	ext{ cm}$. The frequency of the tuning fork is $504 	ext{ Hz}$. Speed of the sound at the given temperature is $336 	ext{ m/s}$. The zero of the meter scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is

Options:

(A) $13 	ext{ cm}$

(B) $16.6 	ext{ cm}$

(C) $18.4 	ext{ cm}$

(D) $14.8 	ext{ cm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a physics lab, standing in front of a tall glass tube partially filled with water. You strike a tuning fork and hold it over the open top. As you slowly lower the water level, suddenly, the sound amplifies dramatically! You've just hit the **first resonance**.

This setup is a classic example of a **closed organ pipe**. The water surface acts as a rigid, impenetrable boundary for the sound waves, forcing the air molecules there to stay perfectly still. This creates a **displacement node**.

On the other hand, the open top of the tube allows the air molecules maximum freedom to vibrate, creating a **displacement antinode**.

### The Master Equation

For the first resonance (the fundamental mode), the distance between the node at the water surface and the antinode at the top is exactly one-quarter of the sound wave's wavelength, or $\frac{\lambda}{4}$.

But here is the catch—the antinode doesn't form exactly at the rim of the tube. Because the air just outside the tube also vibrates, the antinode spills over slightly. This extra distance is called the **end correction**, denoted by $e$.

Therefore, the true effective length of the vibrating air column is the physical length of the tube $L$ plus the end correction $e$.

Our master equation becomes:
$$L + e = \frac{\lambda}{4}$$

We also know the fundamental relationship between wave speed $v$, frequency $f$, and wavelength $\lambda$:
$$v = f \lambda \implies \lambda = \frac{v}{f}$$

Substituting this into our master equation gives:
$$L + e = \frac{v}{4f}$$

### Calculating the End Correction

The end correction $e$ for a cylindrical pipe depends on its diameter $d$. Empirically, it is given by:
$$e = 0.3d$$

The problem states the diameter is $6 	ext{ cm}$. Let's convert this to standard SI units (meters) to avoid any silly mistakes later:
$$d = 6 	ext{ cm} = 0.06 	ext{ m}$$

Now, we can calculate $e$:
$$e = 0.3 	imes 0.06 	ext{ m} = 0.018 	ext{ m}$$

### Final Calculation

Now, let's look at the right side of our master equation. We are given the speed of sound $v = 336 	ext{ m/s}$ and the frequency of the tuning fork $f = 504 	ext{ Hz}$.

Let's plug these values in:
$$\frac{v}{4f} = \frac{336}{4 	imes 504}$$

$$\frac{v}{4f} = \frac{336}{2016}$$

Notice how beautifully the numbers simplify. $336$ goes into $2016$ exactly $6$ times!
$$\frac{v}{4f} = \frac{1}{6} 	ext{ m} \approx 0.1667 	ext{ m}$$

Now we bring it all together to find the physical length $L$:
$$L + 0.018 = 0.1667$$

$$L = 0.1667 - 0.018 = 0.1487 	ext{ m}$$

Finally, let's convert this back to centimeters to match our options:
$$L = 14.87 	ext{ cm}$$

Rounding to one decimal place, we get **$14.8 	ext{ cm}$**. This is the reading on the meter scale where the first resonance occurs.

Select Answer:

Visualized Solution

Resonance Column Setup

First Resonance Condition

End Correction Formula

Calculating End Correction

Calculating Wavelength Term

Simplifying Wavelength Term

Finding the Water Level Reading

Second Resonance

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

Calculating the End Correction

Final Calculation

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