/*! 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 26 The end of a stopped pipe is to ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length \(L\). what fraction of \(L\) should be cut off so that the fundamental mode of the resulting open pipe has the same frequency as the fifth harmonic \((n=5)\) of the original stopped pipe?

Short Answer

Expert verified
Cut \(\frac{3}{5}\) of the pipe's length.

Step by step solution

01

Understanding the Problem

We need to find a fraction of the total length \( L \) of a stopped pipe to cut off so that the frequency of the fundamental mode of an open pipe equals the frequency of the 5th harmonic of the original stopped pipe.
02

Formulas and Initial Setup

For a stopped pipe, the fundamental frequency is \( f_n = \frac{nv}{4L} \), where \( n \) is the harmonic. The 5th harmonic has \( n=5 \), so its frequency is \( f_5 = \frac{5v}{4L} \). For an open pipe, the fundamental frequency is \( f_{ ext{open}} = \frac{v}{2L_\text{open}} \). We want these frequencies to be equal.
03

Equate Frequencies

Set the frequency of the open pipe to the frequency of the 5th harmonic of the original pipe: \( \frac{v}{2L_\text{open}} = \frac{5v}{4L} \). Simplifying, we get \( 2L_\text{open} = \frac{4L}{5} \) or \( L_\text{open} = \frac{2L}{5} \).
04

Calculate Length to Cut

Determine the length of the pipe that should be cut off: \( L - L_\text{open} = L - \frac{2L}{5} = \frac{3L}{5} \). This means \( \frac{3}{5} \) of the total length should be cut.

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 91Ó°ÊÓ!

Key Concepts

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

Stopped Pipe
In acoustics, a stopped pipe is a tube with one closed end. The sound waves created within a stopped pipe form a specific set of harmonics, where only odd harmonics are present. This occurs because the closed end acts as a node where no movement occurs, while the open end acts as an antinode where maximum displacement and sound pressure occur.
The fundamental frequency of a stopped pipe is given by the formula:
  • \( f_1 = \frac{v}{4L} \)
where:
  • \( f_1 \) is the fundamental frequency,
  • \( v \) is the speed of sound in air,
  • \( L \) is the length of the stopped pipe.
For the harmonics in a stopped pipe, the frequencies follow the pattern \( f_n = \frac{nv}{4L} \), where \( n \) takes odd integer values (1, 3, 5,...). Understanding these principles is crucial when modifying a pipe to alter its acoustic properties.
Open Pipe
An open pipe, unlike a stopped pipe, is a tube with both ends open. This type of pipe supports both odd and even harmonics because both ends can serve as antinodes. Thus, sound waves in an open pipe can achieve a wider range of harmonics compared to a stopped pipe.
The fundamental frequency of an open pipe is calculated differently from that of a stopped pipe. The equation used is:
  • \( f_{\text{open}} = \frac{v}{2L} \)
where:
  • \( f_{\text{open}} \) is the fundamental frequency of the open pipe,
  • \( v \) is the speed of sound,
  • \( L \) is the length of the open pipe.
This equation indicates that the fundamental frequency of an open pipe is directly proportional to the speed of sound and inversely proportional to twice the length of the pipe. This contrasts the stopped pipe formula, making modifications between the two fascinating for altering sound characteristics.
Sound Waves
Sound waves in pipes, whether stopped or open, are longitudinal waves that travel through air. These waves can be visualized as compressions and rarefactions moving through the air inside the pipe. The behavior of sound waves in pipes forms the basis for musical instruments and various acoustic devices.
While sound waves propagate, they set up standing wave patterns in the pipe, generating specific frequencies. In a stopped pipe, these standing waves only permit odd harmonics due to the boundary conditions at the closed end. Conversely, in an open pipe, the conditions at both ends of the pipe allow for a broader harmonic series, including both odd and even harmonics.
These patterns and harmonics help in determining the sound qualities and are essential for calculating resonant frequencies, guiding modifications, and shaping understanding of acoustic behaviors.
Pipe Length Calculation
Pipe length plays a crucial role in determining the frequency of sound waves produced. Calculating the required length for a specific frequency or modifying a pipe's length to achieve a desired frequency involves applying fundamental principles of harmonics and wave equations.
When modifying a stopped pipe into an open pipe, a certain amount of its length must be removed to achieve specific frequency goals. For example, in the given exercise, the goal was to make the frequency of the fundamental mode of an open pipe equal the fifth harmonic of the original stopped pipe. This involved finding the fraction of the pipe's length to cut off.
The calculation was conducted by solving the equation where the frequency of the now-open pipe equates to the fifth harmonic frequency of the stopped pipe. The result, \( L_{\text{open}} = \frac{2L}{5} \), indicates the new length of the pipe, and the remaining \( \frac{3L}{5} \) portion represents the length needing removal. Such calculations emphasize the relationship between pipe length and harmonic frequencies.

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

For cranial ultrasound, why is it advantageous to use frequencies in the kHZ range rather than in the MHz range? A. The antinodes of the standing waves will be closer together at the lower frequencies than at the higher frequencies. B. There will be no standing waves at the lower frequencies. C. Cranial bones will attenuate the ultrasound more at the lower frequencies than at the higher frequencies. D. Cranial bones will attenuate the ultrasound less at the lower frequencies than at the higher frequencies.

You are driving down the road at \(15.6 \mathrm{~m} / \mathrm{s}(35 \mathrm{mph}\) ) when an ambulance passes you with its siren blaring. The ambulance siren produces a frequency of \(700 \mathrm{~Hz}\). As the ambulance approaches you from behind, you hear a frequency of \(740 \mathrm{~Hz}\). (a) What is the speed of the ambulance? (b) What frequency do you hear after the ambulance passes?

A wire with mass \(40.0 \mathrm{~g}\) is stretched so that its ends are tied down at points \(80.0 \mathrm{~cm}\) apart. The wire vibrates in its fundamental mode with frequency \(60.0 \mathrm{~Hz}\) and with an amplitude of \(0.300 \mathrm{~cm}\) at the antinodes. (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire.

Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the speed of light. (a) FM radio. Find the wavelength of an FM radio station signal broadcasting at a frequency of \(104.5 \mathrm{MHz}\). (b) X-rays. X-rays have a wavelength of about 0.10 nm. What is their frequency? (c) The Big Bang. Microwaves with a wavelength of \(1.1 \mathrm{~mm}\), left over from soon after the Big Bang. have been detected. What is their frequency? (d) Sunburn. Sunburn (and skin cancer) is caused by ultraviolet light waves having a frequency of around \(10^{16} \mathrm{~Hz}\). What is their wavelength? (e) SETI. It has been suggested that extraterrestrial civilizations (if they exist) might try to communicate by using electromagnetic waves having the same frequency as that given off by the spin flip of the electron in hydrogen, which is \(1.43 \mathrm{GHz}\). To what wavelength should we tune our telescopes in order to search for such signals? (f) Microwave ovens. Microwave ovens cook food with electromagnetic waves of frequency around \(2.45 \mathrm{GHz}\). What wavelength do these waves have?

The longest pipe found in most medium-sized pipe organs is \(4.88 \mathrm{~m}\) (16 ft). What is the wavelength of the note corresponding to the fundamental mode if the pipe is (a) open at both ends and (b) open at one end and closed at the other?

See all solutions

Recommended explanations on Physics Textbooks

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.