A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s^

Q: Why does the minimum height correspond to the fundamental mode?

Resonance occurs when a standing wave fits perfectly within the tube's boundary conditions. The fundamental mode is the simplest standing wave pattern possible (one node and one antinode), which requires the shortest column length ($l = \lambda/4$). Any higher harmonic would require a longer column.

Q: How did we get the factor of 1000 in the speed of sound formula?

The molar mass $M$ in the options is given in grams per mole. However, the standard formula for the speed of sound requires SI units (kilograms per mole). To convert grams to kilograms, we divide $M$ by $1000$ (or multiply by $10^{-3}$) in the denominator, which brings a factor of $1000$ to the numerator.

Q: Why is the water surface considered a node?

Water is much denser than gas, acting as a rigid boundary. Air molecules directly at the water surface are physically blocked from moving, meaning their displacement amplitude is zero. A point of zero displacement amplitude is, by definition, a node.

Q: What would happen if we didn't account for the experimental uncertainty?

If we only used the nominal value $l = 0.350\text{ m}$, we would get a single velocity $v = 341.6\text{ m/s}$. None of the calculated speeds for the gases ($448$, $354$, $331.8$, or $340\text{ m/s}$) match this exactly. Accounting for the uncertainty gives us a range ($336.7 - 346.5\text{ m/s}$) that clearly includes Argon ($340\text{ m/s}$).

Q: How do we know which gases are monoatomic and which are diatomic?

Neon and Argon are noble gases (Group 18), which exist as single atoms (monoatomic, $\gamma = 1.67$). Nitrogen ($N_2$) and Oxygen ($O_2$) are homonuclear diatomic molecules (diatomic, $\gamma = 1.40$).

Question

A student is performing an experiment using a resonance column and a tuning fork of frequency $244	ext{ s}^{-1}$. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350 \pm 0.005)	ext{ m}$, the gas in the tube is:

**Useful information:**
* $\sqrt{167RT} = 640	ext{ J}^{1/2}	ext{ mole}^{-1/2}$
* $\sqrt{140RT} = 590	ext{ J}^{1/2}	ext{ mole}^{-1/2}$

The molar masses $M$ in grams are given in the options. Take the value of $\sqrt{10/M}$ for each gas as given there.

Options:

(A) Neon ($M = 20, \sqrt{10/M} = 7/10$)

(B) Nitrogen ($M = 28, \sqrt{10/M} = 3/5$)

(C) Oxygen ($M = 32, \sqrt{10/M} = 9/16$)

(D) Argon ($M = 36, \sqrt{10/M} = 17/32$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a physics laboratory, holding a vibrating tuning fork over a long, vertical tube filled with water. As you slowly lower the water level, the air column inside the tube lengthens. Suddenly, at a very specific height, the sound swells into a rich, loud hum.

This is **acoustic resonance**.

In this problem, we are given a tuning fork of frequency $f = 244	ext{ s}^{-1}$. The air inside the tube has been replaced by an unknown gas. The minimum height of the air column at which resonance occurs is measured to be $l = 0.350 \pm 0.005	ext{ m}$. Our goal is to identify this mystery gas from a list of candidates: Neon, Nitrogen, Oxygen, and Argon.

Let's break this down step-by-step, combining the physics of standing waves with the kinetic theory of gases.

### The Master Equation for Resonance

The water surface acts as a rigid boundary, meaning air molecules cannot vibrate there. This creates a **displacement node** at the bottom of the air column. At the open top of the tube, air molecules are free to vibrate with maximum amplitude, forming a **displacement antinode**.

For the *minimum* height of the air column, the tube resonates in its simplest possible pattern: the **fundamental mode**.

In this mode, the length of the air column $l$ is exactly one-quarter of the wavelength $\lambda$:

$$l = \frac{\lambda}{4} \implies \lambda = 4l$$

Using the fundamental wave relation, we can express the speed of sound $v$ in the gas as:

$$v = f \lambda = 4 f l$$

Substituting our known frequency $f = 244	ext{ s}^{-1}$:

$$v = 4 	imes 244 	imes l = 976 	imes l$$

### Propagating Experimental Uncertainty

In real-world physics, measurements are never perfectly precise. Here, the length of the air column is given with an uncertainty of $\pm 0.005	ext{ m}$:

$$l_{	ext{min}} = 0.350 - 0.005 = 0.345	ext{ m}$$

$$l_{	ext{max}} = 0.350 + 0.005 = 0.355	ext{ m}$$

This uncertainty in length propagates directly into our calculated speed of sound. Let's find the lower and upper bounds for $v$:

$$v_{	ext{min}} = 976 	imes 0.345 = 336.72	ext{ m/s}$$

$$v_{	ext{max}} = 976 	imes 0.355 = 346.48	ext{ m/s}$$

Thus, the actual speed of sound in the mystery gas must lie strictly within the range:

$$\mathbf{336.7	ext{ m/s} \le v \le 346.5	ext{ m/s}}$$

### Connecting to Kinetic Theory

Now, how does the speed of sound relate to the identity of the gas? From thermodynamics, we know that the speed of sound in an ideal gas is given by:

$$v = \sqrt{\frac{\gamma RT}{M_{	ext{kg}}}}$$

where:
* $\gamma$ is the adiabatic exponent of the gas,
* $R$ is the universal gas constant,
* $T$ is the absolute temperature,
* $M_{	ext{kg}}$ is the molar mass in kilograms per mole ($M_{	ext{kg}} = M 	imes 10^{-3}	ext{ kg/mol}$, where $M$ is in grams/mole).

Substituting $M_{	ext{kg}} = M 	imes 10^{-3}$ into our velocity equation:

$$v = \sqrt{\frac{1000 \gamma RT}{M}} = \sqrt{100 \gamma RT} 	imes \sqrt{\frac{10}{M}}$$

To make our lives easier, the problem provides two highly specific approximations:
1. For a **monoatomic gas** (like Neon or Argon), $\gamma = 5/3 \approx 1.67$. This gives:
   $$\sqrt{100 \gamma RT} = \sqrt{167 RT} = 640	ext{ J}^{1/2}	ext{ mole}^{-1/2}$$
   $$\implies v_{	ext{mono}} = 640 	imes \sqrt{\frac{10}{M}}$$

2. For a **diatomic gas** (like Nitrogen or Oxygen), $\gamma = 7/5 = 1.40$. This gives:
   $$\sqrt{100 \gamma RT} = \sqrt{140 RT} = 590	ext{ J}^{1/2}	ext{ mole}^{-1/2}$$
   $$\implies v_{	ext{di}} = 590 	imes \sqrt{\frac{10}{M}}$$

### Testing the Candidates

Let's calculate the speed of sound for each of our four candidate gases using the values of $\sqrt{10/M}$ provided in the options:

1. **Neon** (Monoatomic, $M = 20$):
   $$v = 640 	imes \frac{7}{10} = 448	ext{ m/s}$$
   *(This is far too fast to fit our range of $336.7 - 346.5	ext{ m/s}$.)*

2. **Nitrogen** (Diatomic, $M = 28$):
   $$v = 590 	imes \frac{3}{5} = 354	ext{ m/s}$$
   *(This is also too fast.)*

3. **Oxygen** (Diatomic, $M = 32$):
   $$v = 590 	imes \frac{9}{16} = 331.8	ext{ m/s}$$
   *(This is slightly too slow.)*

4. **Argon** (Monoatomic, $M = 36$):
   $$v = 640 	imes \frac{17}{32} = 340	ext{ m/s}$$
   *(This is a perfect match! $340	ext{ m/s}$ lies comfortably within our experimental range of $336.7	ext{ m/s}$ to $346.5	ext{ m/s}$.)*

### Conclusion

By combining wave mechanics with thermodynamics and error propagation, we have successfully identified the mystery gas as **Argon**.

Therefore, the correct option is **(d)**.

Select Answer:

Visualized Solution

Visualizing the Resonance Column

Condition for First Resonance

Relating Wave Velocity to Length

Calculating the Range of Wave Velocity

Evaluating the Velocity Range

Speed of Sound in a Gas

Using the Given Approximations

Testing the Options

Identifying the Correct Gas

Morale & Conceptual Takeaway

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation for Resonance

Propagating Experimental Uncertainty

Connecting to Kinetic Theory

Testing the Candidates

Conclusion

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