A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm l

Q: Why is the first overtone of the string considered the second harmonic, while for the closed pipe, the first overtone is the third harmonic?

A string fixed at both ends can support all integer harmonics ($n = 1, 2, 3, \dots$), so its first overtone (the next mode after the fundamental) is the second harmonic ($n=2$). A pipe closed at one end, however, only supports odd harmonics ($n = 1, 3, 5, \dots$) due to its boundary conditions. Therefore, its first overtone is the third harmonic ($n=3$).

Q: How did we know that the string's frequency must be greater than the pipe's frequency ($f_s > f_p$)?

We know that frequency is proportional to the square root of tension ($f_s \propto \sqrt{T}$). Decreasing tension decreases $f_s$. If $f_s$ were less than $f_p$, decreasing $f_s$ would move it further away from $f_p$, increasing the beat frequency $|f_s - f_p|$. Since the problem states that decreasing tension decreases the beat frequency, $f_s$ must be greater than $f_p$ so that decreasing $f_s$ brings it closer to $f_p$.

Q: What is linear mass density, and why did we need to calculate it?

Linear mass density ($\mu$) is the mass per unit length of the string ($\mu = m/l$). It is a fundamental property of the medium that determines how fast a wave can travel through it under a given tension, according to the formula $v = \sqrt{T/\mu}$.

Q: What would happen to the beat frequency if the temperature of the air inside the pipe increased?

An increase in temperature increases the speed of sound in air ($v_{\text{sound}} \propto \sqrt{T_{\text{temp}}}$). This would increase the fundamental frequency of the pipe ($f_p$). Depending on whether $f_s$ is greater or less than the new $f_p$, the beat frequency $|f_s - f_p|$ would change accordingly.

Q: Can we produce beats with any two frequencies, or do they have to be close to each other?

Beats are physically produced by any two overlapping waves, but the human ear can only distinctly perceive and count beats if the frequency difference is small (typically less than $10$ to $15\text{ Hz}$). If the difference is larger, we perceive it as a complex chord or two separate tones rather than a periodic throbbing.

Question

A string $25	ext{ cm}$ long and having a mass of $2.5	ext{ g}$ is under tension. A pipe closed at one end is $40	ext{ cm}$ long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, $8	ext{ beats/s}$ are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is $320	ext{ m/s}$, find the tension in the string.

Sigma Percentile · Accepted Answer

### Introduction

Imagine standing in a physics laboratory, surrounded by the gentle hum of vibrating strings and the resonant echoes of organ pipes.

This problem brings together two of the most beautiful phenomena in wave mechanics: **standing waves** and **acoustic beats**.

We are presented with a stretched string and a closed organ pipe sounding together.

By analyzing how their frequencies interact to produce beats, and how a change in tension shifts this interaction, we can unlock the exact tension holding the string taut.

Let's embark on this step-by-step journey to solve this classic JEE problem.

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### Analyzing the Closed Organ Pipe

First, let's focus on the closed organ pipe.

A pipe closed at one end of length $l_2 = 40	ext{ cm} = 0.4	ext{ m}$ supports standing air columns.

The boundary conditions dictate that there must be a displacement node at the closed end and a displacement antinode at the open end.

For the **fundamental mode** of vibration, the length of the pipe corresponds to exactly one-quarter of a wavelength:

$$l_2 = \frac{\lambda_p}{4} \implies \lambda_p = 4l_2$$

Using the wave equation $v = f \lambda$, we can write the fundamental frequency of the pipe $f_p$ as:

$$f_p = \frac{v_{	ext{sound}}}{4l_2}$$

Given that the speed of sound in air is $v_{	ext{sound}} = 320	ext{ m/s}$, let's substitute the values:

$$f_p = \frac{320}{4 	imes 0.4} = \frac{320}{1.6} = 200	ext{ Hz}$$

Thus, our closed organ pipe is humming at a steady frequency of **$200	ext{ Hz}$**.

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### Analyzing the Vibrating String

Now, let's turn our attention to the stretched string of length $l_1 = 25	ext{ cm} = 0.25	ext{ m}$ and mass $m = 2.5	ext{ g} = 2.5 	imes 10^{-3}	ext{ kg}$.

The string is clamped at both ends, meaning it must have displacement nodes at both boundaries.

The problem states that the string is vibrating in its **first overtone**.

For a string fixed at both ends, the fundamental mode (first harmonic) has one loop.

The first overtone corresponds to the **second harmonic**, which consists of two complete loops.

Therefore, the frequency of the first overtone $f_s$ is given by:

$$f_s = 2 \left(\frac{v_1}{2l_1}ight) = \frac{v_1}{l_1}$$

where $v_1$ is the velocity of the transverse wave on the string.

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### Resolving the Beat Frequency Ambiguity

When the string and the pipe vibrate simultaneously, they produce a beat frequency of $8	ext{ beats/s}$.

Beats occur due to the superposition of two waves of slightly different frequencies, and the beat frequency is simply the absolute difference between them:

$$f_{	ext{beat}} = |f_s - f_p| = 8	ext{ Hz}$$

This mathematical absolute value yields two possible frequencies for the string:

1. $f_s = f_p + 8 = 208	ext{ Hz}$
2. $f_s = f_p - 8 = 192	ext{ Hz}$

To determine which of these is the physical reality, we must use the crucial clue: **"decreasing the tension in the string decreases the beat frequency."**

Let's analyze the physics of tension and frequency.

The wave velocity on a stretched string is given by:

$$v_1 = \sqrt{\frac{T}{\mu}}$$

where $T$ is the tension and $\mu$ is the linear mass density.

Since $f_s = \frac{v_1}{l_1}$, we have:

$$f_s = \frac{1}{l_1} \sqrt{\frac{T}{\mu}} \implies f_s \propto \sqrt{T}$$

If we **decrease the tension $T$**, the frequency of the string $f_s$ must **decrease**.

Now, let's look at how a decrease in $f_s$ affects the beat frequency $f_{	ext{beat}} = |f_s - f_p|$:

* If $f_s$ was initially $192	ext{ Hz}$ (which is less than $f_p = 200	ext{ Hz}$), then decreasing $f_s$ further (say, to $190	ext{ Hz}$) would increase the gap between $f_s$ and $f_p$, thereby **increasing** the beat frequency (to $10	ext{ Hz}$).
* If $f_s$ was initially $208	ext{ Hz}$ (which is greater than $f_p = 200	ext{ Hz}$), then decreasing $f_s$ (say, to $206	ext{ Hz}$) would bring it closer to $f_p$, thereby **decreasing** the beat frequency (to $6	ext{ Hz}$).

Since the problem explicitly states that decreasing the tension *decreases* the beat frequency, we must conclude that:

$$f_s > f_p \implies f_s = 208	ext{ Hz}$$

This is a beautiful piece of logical deduction!

---

### Calculating Tension in the String

Now that we have the exact frequency of the string, $f_s = 208	ext{ Hz}$, we can calculate the wave velocity $v_1$:

$$v_1 = f_s 	imes l_1 = 208 	imes 0.25 = 52	ext{ m/s}$$

Next, let's calculate the linear mass density $\mu$ of the string:

$$\mu = \frac{m}{l_1} = \frac{2.5 	imes 10^{-3}	ext{ kg}}{0.25	ext{ m}} = 0.01	ext{ kg/m}$$

Finally, using the relation $v_1 = \sqrt{\frac{T}{\mu}}$, we solve for the tension $T$:

$$T = \mu v_1^2$$

$$T = 0.01 	imes (52)^2 = 0.01 	imes 2704 = 27.04	ext{ N}$$

### Conclusion

The tension in the string is exactly **$27.04	ext{ N}$**.

By systematically breaking down the standing wave equations for both the organ pipe and the string, and carefully applying the physical constraint of tension variation, we arrived at a precise and elegant solution.

This problem highlights how mathematical equations and physical intuition work hand-in-hand to describe the harmony of sound!

Enter Numerical Value:

Visualized Solution

Visualizing the Standing Waves

Calculating the Pipe's Fundamental Frequency

Evaluating $f_{p}$

Expressing the String's First Overtone Frequency

Analyzing the Beat Frequency Condition

Applying the Tension Constraint

Calculating String Wave Velocity $v_{1}$

Computing $v_{1}$

Finding Linear Mass Density $μ$

Solving for Tension $T$

Exploring Further Variations

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

Analyzing the Closed Organ Pipe

Analyzing the Vibrating String

Resolving the Beat Frequency Ambiguity

Calculating Tension in the String

Conclusion

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