Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled

Q: Why does Pipe B only have odd harmonics in the first part?

In a pipe closed at one end and open at the other, the closed end must be a displacement node (zero motion) and the open end must be an antinode (maximum motion). This asymmetric boundary condition only allows odd quarter-wavelengths to fit inside the pipe, meaning only odd harmonics ($1st, 3rd, 5th...$) can exist.

Q: How do we know the values of gamma for monoatomic and diatomic gases?

The adiabatic index $\gamma$ is defined as $C_p/C_v$. For a monoatomic gas, there are 3 translational degrees of freedom, giving $C_v = 3/2 R$ and $C_p = 5/2 R$, so $\gamma = 5/3$. For a diatomic gas at room temperature, there are 3 translational and 2 rotational degrees of freedom, giving $C_v = 5/2 R$ and $C_p = 7/2 R$, so $\gamma = 7/5$.

Q: Why does a pipe closed at both ends have the same fundamental frequency formula as a pipe open at both ends?

Even though one has nodes at the ends and the other has antinodes, the distance between two successive nodes (closed-closed) is exactly the same as the distance between two successive antinodes (open-open) for the fundamental mode, which is half a wavelength ($\lambda/2 = L$). Thus, both have the fundamental frequency $f = v/2L$.

Q: Does the temperature of the gas affect the ratio of the frequencies?

No, because both pipes are maintained at the same temperature. In the speed of sound formula, the temperature term $T$ cancels out completely when we take the ratio of the speeds, meaning the frequency ratio depends only on the molecular properties of the gases.

Q: What would happen to the beat frequency if we sounded both fundamental notes together in part (b)?

The fundamental frequencies are $f_A = v_A/2L$ and $f'_B = v_B/2L$. Since $f_A/f'_B = 3/4$, their frequencies are different. If sounded together, they would produce beats with a frequency equal to $|f_A - f'_B|$, which depends on the absolute length $L$ and the specific gases used.

Question

Two narrow cylindrical pipes $A$ and $B$ have the same length. Pipe $A$ is open at both ends and is filled with a monoatomic gas of molar mass $M_A$. Pipe $B$ is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass $M_B$. Both gases are at the same temperature.

(a) If the frequency to the second harmonic of pipe $A$ is equal to the frequency of the third harmonic of the fundamental mode in pipe $B$, determine the value of $M_A/M_B$.

(b) Now the open end of the pipe $B$ is closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe $A$ to that in pipe $B$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a laboratory with two organ pipes of identical length $L$.

One of them, **Pipe $A$**, is completely open at both ends, allowing air to rush freely in and out of both terminals.

This pipe is filled with a **monoatomic gas** of molar mass $M_A$.

Because it is monoatomic, its molecules behave like simple point masses, giving it an adiabatic index of:

$$\gamma_A = \frac{5}{3}$$

Right next to it sits **Pipe $B$**, which is closed at one end and open at the other.

This pipe is filled with a **diatomic gas** of molar mass $M_B$.

Its diatomic molecules can rotate, which alters its thermal properties, yielding an adiabatic index of:

$$\gamma_B = \frac{7}{5}$$

Both systems are kept at the exact same temperature $T$.

Our goal in the first part of this problem is to find the ratio of their molar masses, $M_A/M_B$, given a specific resonance condition.

---

### The Physics of Speed and Harmonics

To connect the geometry of the pipes to the properties of the gases, we must use two fundamental physical principles:

1. **The Speed of Sound in a Gas:**

$$v = \sqrt{\frac{\gamma R T}{M}}$$

This formula shows that the speed of sound depends on the stiffness-to-density ratio of the gas, which is determined by its molecular structure (via $\gamma$) and its molecular weight (via $M$).

2. **Boundary Conditions and Standing Waves:**

For **Pipe $A$** (open-open), the air molecules at both ends are free to vibrate with maximum amplitude, creating displacement antinodes at both boundaries.

The second harmonic of this pipe contains two full loops, and its frequency is:

$$f_{A,2} = 2 \left(\frac{v_A}{2L}ight) = \frac{v_A}{L}$$

For **Pipe $B$** (closed-open), the closed end forces a displacement node (zero motion), while the open end forms an antinode.

This asymmetric boundary condition only allows odd harmonics.

The third harmonic frequency is:

$$f_{B,3} = 3 \left(\frac{v_B}{4L}ight) = \frac{3v_B}{4L}$$

---

### Solving for the Molar Mass Ratio

We are given that these two frequencies are equal:

$$f_{A,2} = f_{B,3}$$

Substituting our expressions for the frequencies:

$$\frac{v_A}{L} = \frac{3v_B}{4L}$$

Cancelling the common length $L$ from both sides gives the ratio of the speeds of sound in the two pipes:

$$\frac{v_A}{v_B} = \frac{3}{4}$$

Now, we substitute the thermodynamic speed of sound formula into this ratio:

$$\frac{\sqrt{\frac{\gamma_A R T}{M_A}}}{\sqrt{\frac{\gamma_B R T}{M_B}}} = \frac{3}{4}$$

Since both gases are at the same temperature $T$, the constants $R$ and $T$ cancel out, leaving:

$$\sqrt{\frac{\gamma_A M_B}{\gamma_B M_A}} = \frac{3}{4}$$

Squaring both sides to eliminate the radical:

$$\frac{\gamma_A M_B}{\gamma_B M_A} = \frac{9}{16}$$

Rearranging this equation to isolate the ratio of molar masses $M_A/M_B$:

$$\frac{M_A}{M_B} = \frac{\gamma_A}{\gamma_B} 	imes \frac{16}{9}$$

Now, we substitute the values of $\gamma$ for monoatomic and diatomic gases:

$$\frac{\gamma_A}{\gamma_B} = \frac{5/3}{7/5} = \frac{25}{21}$$

Substituting this back into our ratio:

$$\frac{M_A}{M_B} = \frac{25}{21} 	imes \frac{16}{9} = \mathbf{\frac{400}{189}}$$

This is the elegant result for Part (a).

---

### Part (b): Changing the Boundary Conditions

Now, let's modify the experiment.

We seal the open end of **Pipe $B$**, making it closed at both ends.

A pipe closed at both ends has nodes at both boundaries, which is mathematically identical to a pipe open at both ends in terms of its allowed wavelengths.

Its fundamental frequency is now:

$$f'_B = \frac{v_B}{2L}$$

Meanwhile, **Pipe $A$** remains open at both ends, with a fundamental frequency of:

$$f_A = \frac{v_A}{2L}$$

We want to find the ratio of these fundamental frequencies:

$$\frac{f_A}{f'_B} = \frac{\frac{v_A}{2L}}{\frac{v_B}{2L}} = \frac{v_A}{v_B}$$

Notice how the geometric factor $2L$ cancels out completely, leaving only the ratio of the speeds of sound!

From our calculations in Part (a), we already know that:

$$\frac{v_A}{v_B} = \frac{3}{4}$$

Therefore, the ratio of their fundamental frequencies is simply:

$$\frac{f_A}{f'_B} = \mathbf{\frac{3}{4}}$$

This beautifully simple result shows how the fundamental properties of the gases dictate the resonance of the system, independent of the complex molecular mass ratio we calculated earlier!

Visualized Solution

Visualizing the Setup

The Speed of Sound Formula

Second Harmonic of Pipe $A$

Third Harmonic of Pipe $B$

Equating the Frequencies

Finding the Ratio of Speeds

Substituting the Speed Formula

Squaring Both Sides

Substituting Gamma Values

Final Calculation for Part (a)

Part (b): Closing Pipe $B$

Fundamental Frequency of Pipe $A$

Ratio of Fundamental Frequencies

Final Answer for Part (b)

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Physics of Speed and Harmonics

Solving for the Molar Mass Ratio

Part (b)

Similar Questions

A closed organ pipe of length $L$ and an open organ pipe contain gases of densities $ρ_{1}$ and $ρ_{2}$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is $300 m/s$ , the frequency difference between the fundamental and second harmonic of this pipe is ...... Hz. [2020, 8 Jan Shift-I]

An organ pipe $P_{1}$ closed at one end vibrating in its first harmonic and another pipe $P_{2}$ open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of $P_{1}$ and $P_{2}$ is

Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes $A$ and $B$ is

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by $100 Hz$ than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

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