/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 An organ pipe is open at both en... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 43

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the length of the pipe?

Short Answer

Expert verified
The length of the pipe is approximately 1.964 meters.

Step by step solution

01

Understand the Problem

We need to find the length of an organ pipe open at both ends, producing sound at its third harmonic. We know the frequency of the third harmonic is 262 Hz, and the speed of sound is 343 m/s.
02

Relationship in an Open Pipe

For a pipe open at both ends, harmonics are produced at frequencies given by \( f_n = n\frac{v}{2L} \), where \( v \) is the speed of sound, \( L \) is the length of the pipe, and \( n \) is the harmonic number. In this case, since we are dealing with the third harmonic, \( n = 3 \).
03

Solve for Length of Pipe using Third Harmonic Equation

Substitute the given values into the formula for the third harmonic: \( 262 = 3\frac{343}{2L} \). Rearrange to find \( L \):\[ L = \frac{3 \times 343}{2 \times 262} \]
04

Calculate the Length

Calculate the value of \( L \) using the formula from Step 3:\[ L = \frac{1029}{524} \approx 1.964 \text{ m} \]
05

Final Check

Verify the calculation by plugging the length back into the formula and checking if it gives the correct frequency for the third harmonic: \( 262 \approx 3\frac{343}{2 \times 1.964} \), which is approximately correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Harmonics
In the world of sound waves, harmonics are fundamental. They are the distinct frequencies at which sound resonates in musical instruments. For an organ pipe open at both ends, the harmonics are very special because they include all integer multiples of the fundamental frequency.

An open pipe supports harmonics at frequencies given by the formula:
  • \( f_n = n\frac{v}{2L} \)
Where:
  • \( f_n \) is the frequency of the nth harmonic,
  • \( n \) is the harmonic number (1 for the first harmonic, 2 for the second, and so on),
  • \( v \) is the speed of sound,
  • \( L \) is the length of the pipe.

For open pipes, harmonics are particularly rich and predictable. This is because the pipe supports standing waves at multiples of half wavelengths. These harmonics create overtones that enrich musical tones, contributing to the instrument's sound quality.
Speed of Sound
The speed of sound is a key factor in how we perceive and compute sound in various environments. At sea level and room temperature (around 20掳C or 68掳F), the speed of sound in air is approximately 343 meters per second (m/s). This value can vary slightly depending on factors such as air temperature, humidity, and pressure.

Why does this matter in organ pipes or other musical instruments? The speed of sound helps determine the frequency of sound waves, allowing us to calculate harmonics accurately. In our exercise, the speed of sound is essential in calculating the pipe length that results in a specific harmonic frequency. By knowing the speed at which sound travels through air, we can predict and tune musical notes, combining the physical properties of the pipe with the behavior of sound waves.
Frequency Calculation
Calculating frequency in an open pipe involves understanding how the speed of sound, harmonic number, and pipe length interplay. Using the formula \( f_n = n\frac{v}{2L} \), we can solve for any variable involved, whether it's frequency, pipe length, or speed of sound.

In this exercise, we know the frequency of the third harmonic (262 Hz) and the speed of sound (343 m/s). By substituting these values into the equation, we can isolate \( L \), the length of the pipe:
  • Rearrange the equation to solve for \( L \): \( L = \frac{n\cdot v}{2\cdot f_n} \)
  • Substitute the known values: \( L = \frac{3\cdot 343}{2\cdot 262} \)
  • This gives \( L \approx 1.964 \) meters.

This calculation showcases how interconnected these concepts are, allowing for precise control in creating musical sounds. Calculating the correct frequency ensures that each note played on an instrument is harmonious and pleasant to the ear.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two waves are traveling in opposite directions on the same string. The displacements caused by the individual waves are given by \(y_{1}=(24.0 \mathrm{mm}) \sin (9.00 \pi t-1.25 \pi x)\) and \(y_{2}=(35.0 \mathrm{mm}) \sin (2.88 \pi t+\) \(0.400 \pi x\) ). Note that the phase angles \((9.00 \pi t-1.25 \pi x)\) and \((2.88 \pi t+\) \(0.400 \pi x\) ) are in radians, \(t\) is in seconds, and \(x\) is in meters. At \(t=4.00 \mathrm{s}\) what is the net displacement (in \(\mathrm{mm}\) ) of the string at (a) \(x=2.16 \mathrm{m}\) and (b) \(x=2.56 \mathrm{m} ?\) Be sure to include the algebraic sign \((+\) or \(-)\) with your answers.

A thin \(1.2-\mathrm{m}\) aluminum rod sustains a longitudinal standing wave with vibration antinodes at each end of the rod. There are no other antinodes. The density and Young's modulus of aluminum are, respectively, \(2700 \mathrm{kg} / \mathrm{m}^{3}\) and \(6.9 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) What is the frequency of the rod's vibration?

Two speakers, one directly behind the other, are each generating a 245-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343 \mathrm{m} / \mathrm{s}\).

A string that is fixed at both ends has a length of \(2.50 \mathrm{m}\). When the string vibrates at a frequency of \(85.0 \mathrm{Hz}\), a standing wave with five loops is formed. (a) What is the wavelength of the waves that travel on the string? (b) What is the speed of the waves? (c) What is the fundamental frequency of the string?

A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \times 10^{5} \mathrm{Pa}\). The tube has a length of \(0.75 \mathrm{m} .\) Mercury (mass density \(\left.=13600 \mathrm{kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Quantum Physics

Read Explanation

Force

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.