/*! 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 24 An organ pipe is made to play a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An organ pipe is made to play a low note at \(27.5 \mathrm{Hz}\), the same as the lowest note on a piano. Assuming a sound speed of \(343 \mathrm{m} / \mathrm{s},\) what length open-open pipe is needed? What length open-closed pipe would suffice?

Short Answer

Expert verified
The length of the open-open pipe needed to play a low note at 27.5 Hz is 6.24 meters, while the length of the open-closed pipe would be 3.12 meters.

Step by step solution

01

- Finding the length of the open-open pipe

The fundamental frequency (also called the first harmonic) of an open-open pipe is given by the formula \(f = v / (2L)\) where \(f\) is the frequency, \(v\) is the speed of sound, and \(L\) is the length of the pipe. The frequency \(f\) is given as \(27.5 Hz\) and the speed of sound \(v\) as \(343 m/s\). The length \(L\) of the pipe can be rearranged as \(L = v / (2f)\). Substituting the given values into this formula, the correct pipe length is \(L = 343 / (2*27.5) m\).
02

- Calculation for open-open pipe

Now, calculate the length of the open-open pipe by plugging the values into the formula. This gives us \(L = 343 / (2*27.5) m = 6.24 m\). Therefore, the length of the open-open pipe needed to play a note at 27.5 Hz is 6.24 meters.
03

- Finding the length of the open-closed pipe

For an open-closed pipe, the fundamental frequency is given by the formula \(f = v / (4L)\), where \(f\) is the frequency, \(v\) is the speed of sound, and \(L\) is the length of the pipe. Again, the frequency \(f\) is given as \(27.5 Hz\) and the speed of sound \(v\) as \(343 m/s\). The length \(L\) of the pipe can be found by rearranging as \(L = v / (4f)\). Substituting the given values into this formula gives the length of the open-closed pipe.
04

- Calculation for open-closed pipe

The length of the open-closed pipe is found by substituting the values into the rearranged formula, giving \(L = 343 / (4*27.5) m\). Calculating this gives \(L = 343 / (4*27.5) m = 3.12 m\). Therefore, the length of the open-closed pipe needed to play a note at 27.5 Hz is 3.12 meters.

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.

Fundamental Frequency of Pipes
Understanding the fundamental frequency of pipes is essential for anyone delving into organ pipe physics. It’s the lowest frequency at which a pipe can resonate or produce a musical tone. Specifically, an open-open pipe (also known as an open-ended pipe) has both ends open to the air, which allows the air particles at each end to move freely. On the other hand, an open-closed pipe has one end sealed off, affecting the resonance conditions dramatically.

For open-open pipes, the fundamental frequency formula is given by \( f = \frac{v}{2L} \), where \( f \) is the frequency, \( v \) represents the speed of sound, and \( L \) is the length of the pipe. Since air can move at both ends, the pipe generates a standing wave with an antinode (point of maximum oscillation) at each end.

In contrast, the formula for an open-closed pipe's fundamental frequency is \( f = \frac{v}{4L} \). The closed end of the pipe forces a node (point of zero oscillation), while the open end has an antinode. The fundamental frequency is lower compared to an open-open pipe of the same length because only a quarter wavelength (\( \lambda/4 \) ) fits into an open-closed pipe to meet these boundary conditions.
Sound Speed and Wavelength Relationship
The relationship between sound speed, wavelength, and frequency is of paramount importance in the realm of acoustics. This core concept can be articulated through the equation \( v = f \lambda \), where \( v \) is the speed of sound in air, \( f \) is the frequency, and \( \lambda \) is the wavelength of the sound wave.

The speed of sound in air (\( v \) ) is approximately \( 343 \text{m/s} \) at room temperature, though it varies with temperature and, to a lesser degree, humidity. Wavelength (\( \lambda \) ) can be perceived as the physical distance between two points in phase on consecutive waves, like the distance from crest to crest. The higher the frequency of the sound wave, the shorter its wavelength will be, and vice versa. It's essential always to remember that sound speed remains constant in a given medium under the same conditions, which helps us predict and calculate wavelengths and frequencies for musical instruments like organ pipes.
Open-Open and Open-Closed Pipe Harmonics
Diving deeper into the physics of organ pipes, one encounters harmonics or overtones, which are an integral part of the rich sound we hear from these instruments. Harmonics are multiples of the fundamental frequency, and the conditions in which they form differ between open-open and open-closed pipes.

An open-open pipe supports harmonics that are whole number multiples of the fundamental frequency (\( f, 2f, 3f, \dots \)). These result from the standing wave patterns where the length of the pipe is equal to an integer multiple of half wavelengths (\( L = n\frac{\lambda}{2} \)) for the nth harmonic.

For open-closed pipes, only odd-numbered harmonics are present (\( f, 3f, 5f, \dots \)). This limitation is due to the need for a node at the closed end and an antinode at the open end, so only an odd multiple of quarter wavelengths (\( L = \frac{(2n-1)\lambda}{4} \)) fits the given pipe length. Consequently, the harmonic series in a pipe with one closed end skips even harmonics, resulting in a distinct sound spectrum compared to an open-open pipe.

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

In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube, which can lead to standing waves. Suppose \(0.70 \mathrm{MHz}\) ultrasound is directed through a layer of tissue with a bone \(0.55 \mathrm{cm}\) below the surface. Will standing waves be created? Explain.

Musicians can use beats to tune their instruments. One flute is properly tuned and plays the musical note A at exactly \(440 \mathrm{Hz}\). A second player sounds the same note and hears that her instrument is slightly "flat" (that is, at too low a frequency). Playing at the same time as the first flute, she hears two loud-soft-loud beats per second. What is the frequency of her instrument?

Although the vocal tract is quite complicated, we can make a simple model of it as an open-closed tube extending from the opening of the mouth to the diaphragm, the large muscle separating the abdomen and the chest cavity. What is the length of this tube if its fundamental frequency equals a typical speech frequency of \(200 \mathrm{Hz}\) ? Assume a sound speed of \(350 \mathrm{m} / \mathrm{s}\). Does this result for the tube length seem reasonable, based on observations on your own body?

In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the position of one worker. The details of practical systems are complex, but we can present a simple example that gives you the idea. Suppose a machine \(5.0 \mathrm{m}\) away from a worker emits a persistent \(80 \mathrm{Hz}\) hum. To cancel the sound at the worker's location with a speaker that exactly duplicates the machine's hum, how far from the worker should the speaker be placed? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\).

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, \(0.75 \mathrm{m}\) wide and \(1.5 \mathrm{m}\) long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of \(343 \mathrm{m} / \mathrm{s}\).

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.