A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is v

Q: Why is the fundamental frequency of a closed pipe v/4L and not v/2L?

A pipe closed at one end must have a displacement node at the closed end and an antinode at the open end. The simplest wave pattern that fits this boundary condition is a quarter of a wavelength ($L = \lambda/4$), which leads to $f = v/4L$. An open pipe has antinodes at both ends, fitting a half wavelength ($L = \lambda/2$), leading to $f = v/2L$.

Q: How did the 2nd harmonic frequency of the string simplify to just v/L?

For a string fixed at both ends, the fundamental frequency is $v/2L$. The $n$-th harmonic frequency is $n(v/2L)$. For the second harmonic ($n=2$), the frequency is $2(v/2L) = v/L$. The two loops formed on the string make its length exactly equal to one wavelength.

Q: Why does resonance mean the frequencies are equal? Can they be harmonics of each other?

Resonance occurs when one vibrating system drives another system to oscillate with greater amplitude at a specific frequency. For maximum energy transfer, the driving frequency (the string's vibration) must match one of the natural frequencies of the driven system (the pipe's air column). In this problem, we are explicitly told the second harmonic of the string resonates with the fundamental frequency of the pipe, meaning these two specific frequencies are equal.

Q: What is the difference between linear mass density and total mass?

Linear mass density ($\mu$) is the mass per unit length of the string (measured in $\text{kg/m}$), representing how heavy each meter of the string is. The total mass ($m$) is the actual mass of the entire string, obtained by multiplying the linear mass density by the total length of the string ($m = \mu \times L$).

Q: Does the speed of sound in air affect the speed of the wave on the string?

No, they are physically independent. The speed of sound in air depends on the temperature and properties of air, while the wave speed on the string depends on the tension and mass of the string. However, because the two systems are in resonance, their frequencies are coupled, allowing us to mathematically relate the two speeds.

Question

A hollow pipe of length $0.8	ext{ m}$ is closed at one end. At its open end a $0.5	ext{ m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50	ext{ N}$ and the speed of sound is $320	ext{ ms}^{-1}$, the mass of the string is

Options:

(A) $5	ext{ g}$

(B) $10	ext{ g}$

(C) $20	ext{ g}$

(D) $40	ext{ g}$

Sigma Percentile · Accepted Answer

### Introduction to Wave Resonance

Imagine two completely different physical systems—a column of air trapped inside a hollow tube and a tightly stretched metal string.

At first glance, they seem to have nothing in common.

One is a fluid medium supporting longitudinal pressure waves, while the other is a solid medium supporting transverse displacement waves.

Yet, through the magic of **resonance**, they can be perfectly synchronized to vibrate at the exact same frequency.

This problem explores this beautiful coupling, requiring us to bridge the physics of organ pipes and stretched strings to find the mass of the vibrating string.

---

### Step 1: Analyzing the Closed Organ Pipe

Let's first look at the hollow pipe.

We are told it has a length of $l_2 = 0.8	ext{ m}$ and is closed at one end.

When air inside a closed pipe is excited, it forms standing wave patterns.

Because one end is closed, the air molecules there are restricted from moving, creating a **displacement node**.

At the open end, the air molecules can vibrate freely, creating a **displacement antinode**.

For the fundamental mode (the simplest possible standing wave), the distance from a node to an adjacent antinode is exactly a quarter of a wavelength:

$$l_2 = \frac{\lambda_{	ext{pipe}}}{4} \implies \lambda_{	ext{pipe}} = 4 l_2$$

Now, we can express the fundamental frequency of this pipe using the wave relation $v = f \lambda$:

$$f_{	ext{pipe}} = \frac{v_{	ext{sound}}}{4 l_2}$$

Substituting the given values ($v_{	ext{sound}} = 320	ext{ ms}^{-1}$ and $l_2 = 0.8	ext{ m}$):

$$f_{	ext{pipe}} = \frac{320}{4 	imes 0.8} = \frac{320}{3.2} = 100	ext{ Hz}$$

This means the air column is vibrating back and forth exactly $100$ times every second!

---

### Step 2: Analyzing the Vibrating String

Now, let's shift our focus to the stretched string of length $l_1 = 0.5	ext{ m}$.

Since it is clamped at both ends, the ends must be displacement nodes.

When vibrating in its second harmonic, the string forms two complete loops, which means its length is exactly equal to one full wavelength:

$$l_1 = \lambda_{	ext{string}}$$

The frequency of the second harmonic is given by:

$$f_{	ext{string}} = \frac{v_{	ext{string}}}{\lambda_{	ext{string}}} = \frac{v_{	ext{string}}}{l_1}$$

Since the string resonates with the pipe, their frequencies must be identical:

$$f_{	ext{string}} = f_{	ext{pipe}} = 100	ext{ Hz}$$

Substituting this back into our frequency equation for the string:

$$\frac{v_{	ext{string}}}{0.5} = 100 \implies v_{	ext{string}} = 50	ext{ ms}^{-1}$$

We have successfully determined that the transverse wave travels along the string at a speed of $50	ext{ ms}^{-1}$.

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### Step 3: Connecting Wave Speed to Mass

How do we find the mass of the string from its wave speed?

We use the classic formula for the speed of a transverse wave on a stretched string:

$$v_{	ext{string}} = \sqrt{\frac{T}{\mu}}$$

Where $T$ is the tension in the string ($50	ext{ N}$) and $\mu$ is the linear mass density (mass per unit length, $m/l_1$).

Squaring both sides to solve for $\mu$:

$$v_{	ext{string}}^2 = \frac{T}{\mu} \implies \mu = \frac{T}{v_{	ext{string}}^2}$$

Substituting our known values:

$$\mu = \frac{50}{50^2} = \frac{50}{2500} = 0.02	ext{ kg m}^{-1}$$

Finally, the total mass $m$ of the string is the linear mass density multiplied by the total length of the string:

$$m = \mu 	imes l_1 = 0.02	ext{ kg m}^{-1} 	imes 0.5	ext{ m} = 0.01	ext{ kg} = 10	ext{ g}$$

Thus, the mass of the string is **$10	ext{ g}$**, which corresponds to **Option (b)**.

Select Answer:

Visualized Solution

Visualizing the Coupled System

Fundamental Frequency of the Closed Pipe

Computing $f_{pipe}$

Second Harmonic of the Stretched String

Applying the Resonance Condition

Finding $v_{string}$

Wave Speed Formula

Calculating $μ$

Calculating Total Mass $m$

Exploring Variations

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Wave Resonance

Step 1

Step 2

Step 3

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