Velocity of sound in air is 320 m/s. A pipe closed at one end has a length of 1 m. Neglecting end co

Q: Why can't even harmonics exist in a closed organ pipe?

Even harmonics would require an even number of quarter-wavelengths to fit inside the pipe. This would result in either having nodes at both ends or antinodes at both ends. Since one end of the pipe is closed (forcing a node) and the other is open (forcing an antinode), this boundary condition can only be satisfied by an odd number of quarter-wavelengths.

Q: The answer key in my book says (a, c, d). Is that wrong?

Yes, that is a well-known typographical error in some older textbook answer keys. A frequency of $320\text{ Hz}$ is exactly $4$ times the fundamental frequency of $80\text{ Hz}$. Since $4$ is an even number, it represents the $4^{\text{th}}$ harmonic, which cannot exist in a closed pipe. The correct answer is definitely (a, b, d).

Q: What is the difference between a displacement node and a pressure node?

They are completely opposite! A displacement node is a point where air molecules do not move at all (like at the closed end of a pipe). Because the molecules are compressed against this wall, the pressure variation is at its maximum, making it a pressure antinode. Conversely, a displacement antinode (open end) is a pressure node.

Q: Does neglecting end correction affect our final answer?

Neglecting end correction simplifies the math by assuming the displacement antinode occurs exactly at the open end of the pipe. In reality, the antinode occurs slightly outside the open end by a distance of about $0.6r$ (where $r$ is the radius of the pipe). Since we are told to neglect it, we assume the effective length is exactly equal to the physical length of $1\text{ m}$.

Q: How would the allowed frequencies change if the pipe were open at both ends?

If the pipe were open at both ends, it would have displacement antinodes at both ends. This allows both even and odd harmonics to exist, with the fundamental frequency being $f_1 = v/(2L) = 320/(2 \times 1) = 160\text{ Hz}$. The allowed frequencies would then be $160\text{ Hz}, 320\text{ Hz}, 480\text{ Hz}, \dots$

Question

Velocity of sound in air is $320	ext{ m/s}$. A pipe closed at one end has a length of $1	ext{ m}$. Neglecting end corrections, the air column in the pipe can resonate for sound of frequency

Options:

(A) $80	ext{ Hz}$

(B) $240	ext{ Hz}$

(C) $320	ext{ Hz}$

(D) $400	ext{ Hz}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing next to a cylindrical organ pipe that is **closed at one end** and open at the other.

This physical boundary condition is incredibly beautiful because it forces the air molecules inside to behave in a very specific way.

At the closed end, the air molecules are blocked by a solid wall, meaning they cannot vibrate back and forth. This creates a **displacement node**.

At the open end, however, the air molecules are completely free to rush in and out of the pipe, vibrating with maximum amplitude. This creates a **displacement antinode**.

### The Master Equation

Because the pipe must always have a node at one end and an antinode at the other, the length of the pipe $L$ must always be an **odd multiple of a quarter wavelength**:

$$L = (2n-1)\frac{\lambda}{4}$$

Since the speed of sound is related to frequency and wavelength by $v = f\lambda$, we can substitute $\lambda = \frac{v}{f}$ into our boundary condition to find the allowed resonant frequencies:

$$f_n = (2n-1)\frac{v}{4L}$$

where $n = 1, 2, 3, \dots$ represents the harmonic mode index.

This formula tells us a profound truth: **closed organ pipes only support odd harmonics** ($1^{	ext{st}}, 3^{	ext{rd}}, 5^{	ext{th}}, \dots$). Even harmonics are physically forbidden because they would require either a node at both ends or an antinode at both ends, which violates our boundary conditions.

### Step-by-Step Calculation

Let's substitute our given values into this master equation. We are given:
- Speed of sound, $v = 320	ext{ m/s}$
- Length of the pipe, $L = 1	ext{ m}$

#### 1. The Fundamental Frequency ($n=1$)

For the first harmonic, we set $n=1$:

$$f_1 = (2(1)-1)\frac{320}{4(1)} = 1 	imes 80	ext{ Hz} = 80	ext{ Hz}$$

This matches **Option (a)** perfectly.

#### 2. The Third Harmonic ($n=2$)

For the next allowed mode, we set $n=2$:

$$f_3 = (2(2)-1)\frac{320}{4(1)} = 3 	imes 80	ext{ Hz} = 240	ext{ Hz}$$

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

#### 3. The Fifth Harmonic ($n=3$)

For the third allowed mode, we set $n=3$:

$$f_5 = (2(3)-1)\frac{320}{4(1)} = 5 	imes 80	ext{ Hz} = 400	ext{ Hz}$$

This matches **Option (d)** perfectly.

### Resolving the Textbook Typo

If you look at some older answer keys for this classic 1989 question, you might notice they list the correct options as **(a), (c), and (d)**.

However, let's look at the physics. For a frequency of $320	ext{ Hz}$ (Option c) to resonate, it would have to be an even multiple of our fundamental frequency:

$$\frac{320	ext{ Hz}}{80	ext{ Hz}} = 4$$

Since $4$ is an even integer, this corresponds to the $4^{	ext{th}}$ harmonic, which is **physically impossible** in a closed organ pipe.

Therefore, the correct options are mathematically and physically proven to be **(a), (b), and (d)**. Always trust the fundamental laws of physics over a printing press typo!

Animated Solution for Physics - Waves: Velocity of sound in air is $320 m/s$ . A pipe closed at one end has a length of $1 m$ . Neglecting end corrections, the air column in the pipe can resonate for sound of frequency

Select Answer:

Visualized Solution

Understanding the Closed Organ Pipe Setup

The Resonance Condition Formula

Calculating the Fundamental Frequency ( $n = 1$ )

Evaluating $f_{1}$

Setting up the Third Harmonic ( $n = 2$ )

Evaluating $f_{3}$

Setting up the Fifth Harmonic ( $n = 3$ )

Evaluating $f_{5}$

Checking Option (c) ( $320 Hz$ )

Concluding the Correct Options

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

Step-by-Step Calculation

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Resolving the Textbook Typo

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