A vibrating string of certain length l under a tension T resonates with a mode corresponding to the

Q: Why is the first overtone of a closed organ pipe the third harmonic and not the second?

A closed organ pipe only supports odd harmonics because of its boundary conditions (a node at the closed end and an antinode at the open end). Therefore, the first allowed frequency above the fundamental (the first overtone) is the third harmonic.

Q: How does increasing the tension of the string affect its frequency?

The frequency of a stretched string is directly proportional to the square root of its tension ($f \propto \sqrt{T}$). Therefore, increasing the tension always increases the frequency of the string.

Q: What is a beat frequency?

Beat frequency is the absolute difference between the frequencies of two superposing waves ($f_b = |f_1 - f_2|$). It represents the periodic variation in intensity heard when two sound waves of slightly different frequencies interfere.

Q: Why did the beat frequency decrease when the tension was increased?

Since the tuning fork's frequency ($344\text{ Hz}$) was higher than the string's initial frequency ($340\text{ Hz}$), increasing the string's frequency brought it closer to the tuning fork's frequency, thereby reducing the difference (the beat frequency) from $4\text{ Hz}$ to $2\text{ Hz}$.

Q: What would happen if the tube was open at both ends?

If the tube were open at both ends, it would support all harmonics (both even and odd). The first overtone would then be the second harmonic, given by $f = 2(v/2L) = v/L$.

Question

A vibrating string of certain length $l$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75	ext{ cm}$ inside a tube closed at one end. The string also generates $4	ext{ beats/s}$ when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to $2$ per second. Assuming the velocity of sound in air to be $340	ext{ m/s}$, the frequency $n$ of the tuning fork in Hz is

Options:

(A) 344

(B) 336

(C) 117.3

(D) 109.3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a physics laboratory, surrounded by the rich, warm hum of acoustic instruments.

On one side, you have a cylindrical tube closed at one end—an organ pipe.

On the other side, a tightly stretched string is clamped at both ends, ready to vibrate.

These two systems, though physically distinct, are linked by a beautiful phenomenon: **resonance**.

Our first task is to determine the frequency of the sound wave inside the closed organ pipe.

### The Physics of Closed Pipes

For a pipe closed at one end, the air molecules at the closed boundary are completely restricted from moving.

This physical constraint forces a **displacement node** to form at the closed end.

Conversely, at the open end, the air molecules are free to rush in and out, creating a **displacement antinode**.

These boundary conditions dictate that only odd harmonics can exist inside the tube.

The mathematical expression for these allowed frequencies is:

$$f_m = (2m-1)\frac{v}{4L}$$

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

### Calculating the First Overtone

The fundamental mode ($m=1$) is the first harmonic.

The next possible mode of vibration ($m=2$) is the **first overtone**, which corresponds to the **third harmonic**.

Let's write down the formula for this specific frequency:

$$f_p = 3\left(\frac{v}{4L}ight)$$

We are given the speed of sound in air as $v = 340	ext{ m/s}$ and the length of the tube as $L = 75	ext{ cm} = 0.75	ext{ m}$.

Substituting these values into our equation:

$$f_p = 3\left(\frac{340}{4 	imes 0.75}ight)$$

Notice how beautifully the numbers simplify:

$$4 	imes 0.75 = 3.0$$

This simplifies our expression to:

$$f_p = 3\left(\frac{340}{3}ight) = 340	ext{ Hz}$$

The acoustic frequency of the first overtone is exactly $340	ext{ Hz}$.

### Coupling with the Vibrating String

The problem states that the vibrating string resonates with this mode of the air column.

This means the initial frequency of the string, $f_s$, must be exactly equal to the frequency of the pipe:

$$f_s = 340	ext{ Hz}$$

Now, we introduce a tuning fork of unknown frequency $n$.

When sounded together with the string, they produce a beat frequency of $4	ext{ Hz}$:

$$|f_s - n| = 4	ext{ Hz}$$

This absolute difference yields two possible values for the tuning fork's frequency:

$$n = 340 + 4 = 344	ext{ Hz}$$

$$	ext{or}$$

$$n = 340 - 4 = 336	ext{ Hz}$$

### Resolving the Ambiguity via Tension

To determine which of these two frequencies is correct, we must analyze the effect of increasing the string's tension.

The fundamental frequency of a stretched string is given by:

$$f_s = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$$

This formula shows that the frequency is directly proportional to the square root of the tension ($f_s \propto \sqrt{T}$).

When the tension $T$ is slightly increased, the frequency of the string $f_s$ must also increase ($f_s > 340	ext{ Hz}$).

Let's test both candidate frequencies under this condition:

* **Case 1: $n = 344	ext{ Hz}$**
  Initially, $f_s = 340	ext{ Hz}$, giving $|340 - 344| = 4	ext{ beats/s}$.
  As tension increases, $f_s$ rises (for example, to $342	ext{ Hz}$).
  The new beat frequency becomes $|342 - 344| = 2	ext{ beats/s}$.
  This matches the problem's condition perfectly!

* **Case 2: $n = 336	ext{ Hz}$**
  Initially, $f_s = 340	ext{ Hz}$, giving $|340 - 336| = 4	ext{ beats/s}$.
  As tension increases, $f_s$ rises (for example, to $342	ext{ Hz}$).
  The new beat frequency becomes $|342 - 336| = 6	ext{ beats/s}$.
  This contradicts the problem, which states that the beat frequency reduces.

### Final Conclusion

Thus, the frequency of the tuning fork must be **$344	ext{ Hz}$**.

This corresponds to option **(a)**.

Select Answer:

Visualized Solution

Visualizing the Coupled Systems

Acoustic Modes of a Closed Pipe

Identifying the First Overtone

Evaluating $f_{p}$

Simplifying the Expression

Initial Frequency of the String

Understanding Beats with the Tuning Fork

Tension and Frequency Relationship

Testing $n = 344 Hz$

Testing $n = 336 Hz$

Final Answer

Exploring Further Variations

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Physics of Closed Pipes

Calculating the First Overtone

Coupling with the Vibrating String

Resolving the Ambiguity via Tension

Final Conclusion

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