/*! 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 23 Standing sound waves are produce... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Standing sound waves are produced in a pipe that is 1.20 \(\mathrm{m}\) long. For the fundamental frequency and the first two over-tones, determine the locations along the pipe (measured from the left end) of the displacement nodes if (a) the pipe is open atboth ends; (b) the pipe is closed at the left end and open at the right end.

Short Answer

Expert verified
Open pipe: 0.6m, 0.3/0.6/0.9m, 0.2/0.4/0.6/0.8/1.0m; Closed pipe: 0.3m, 0.9m.

Step by step solution

01

Identify the Conditions for an Open Pipe

For a pipe open at both ends, there will be displacement antinodes at both ends. Nodes will occur at distances where the displacement changes from positive to negative or vice versa. The fundamental frequency will have a node at the middle as it is half a wavelength long.
02

Calculate Node Positions for an Open Pipe

For the fundamental frequency of a pipe open at both ends, nodes occur at \(\frac{L}{2}\). For overtones, they occur at fractions like \(\frac{L}{n}\) for the 1st overtone (second harmonic), the pattern is repeated. Since the pipe's length \(L = 1.20 \; \text{m}\), nodes are at 0.6 m for the fundamental, 0.3 m, 0.6 m, and 0.9 m for the first overtone, and 0.2 m, 0.4 m, 0.6 m, 0.8 m, and 1.0 m for the second overtone.
03

Identify the Conditions for a Closed Pipe

For a pipe closed at one end and open at the other, there is a node at the closed end and an antinode at the open end. The fundamental frequency is a quarter wavelength, and each subsequent overtone adds half a wavelength.
04

Calculate Node Positions for a Closed Pipe

For a closed pipe, nodes occur at \(\frac{L}{4}\) for the fundamental frequency, \(\frac{3L}{4}\) for the first overtone, and \(\frac{5L}{4}\) for the second overtone. For the given length \(L = 1.20 \; \text{m}\), nodes are at 0.30 m for the fundamental, 0.90 m for the first overtone, and beyond the 1.20 m physical length for the second overtone.
05

Summarize the Node Locations

For an open pipe, nodes are at 0.6 m (fundamental), 0.3, 0.6, 0.9 m (first overtone), and 0.2, 0.4, 0.6, 0.8, 1.0 m (second overtone). For a closed pipe, nodes are at 0.30 m (fundamental) and 0.90 m (first overtone); the second overtone extends beyond the pipe's length.

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
The fundamental frequency of a vibrating system, like a pipe, is the lowest possible frequency at which the system can resonate. It is often termed as the first harmonic. For a pipe, this frequency is determined by its length and whether it is open or closed at its ends.
The fundamental frequency sets the primary mode of vibration, which means it has the least number of waves or vibrations possible between the pipe's ends.
In an open pipe, both ends act as anti-nodes where the air displacement is at its maximum. The distance between these points is one-half of a wavelength, represented as \( \frac{\lambda}{2} = L \).
Conversely, for a closed pipe, one end is a node (minimum displacement) and the other an anti-node (maximum displacement). The fundamental has a wavelength of four times the pipe's length, given by \( \frac{\lambda}{4} = L \).
Understanding the fundamental frequency is crucial for predicting the behavior of overtones, helping to explain why musical instruments produce multiple tones at different frequencies.
Open and Closed Pipes
Open and closed pipes differ in how they support standing waves due to how their ends interact with air displacement.
  • **Open Pipes:** Both ends open, allowing air to move freely. Displacement nodes, points of minimal air movement, occur centrally when considering the fundamental frequency.
  • **Closed Pipes:** One end sealed, imposing a node with no air movement. Displacement nodes, under these conditions, occur at specific intervals dictated by the pipe's length and the harmonic number.
These properties impact the wavelengths and consequently the frequencies measurable within the pipe.
For an open pipe, the fundamental frequency corresponds to half a wavelength fitting into the pipe length. The pattern multiples to accommodate different harmonics moving through, resulting in additional nodes and anti-nodes.
For a closed pipe, the longest wave fitting is a quarter wavelength, establishing the fundamental. This methodarily changes for successive overtones, with each overtone prominently altering the length needed for the next harmonic in odd integers.
Displacement Nodes
Displacement nodes in the context of standing waves are regions within a resonating medium where there is minimal or no movement of particles.
This concept can be visualized by thinking of two fixed points with a string tied around, where it barely vibrates at these points — these are your nodes.
In open pipes, displacement nodes appear at even intervals along the pipe. The fundamental frequency has its node exactly in the center. For instance, in a pipe 1.20 m long, the node for this primary mode will be precisely at 0.60 m.
In closed pipes, nodes are compulsory at the closed end. With the pipe closed on one side and open at another, nodes occur one-fourth into the length for the fundamental frequency. Using our example, a node would present itself at 0.30 m in a 1.20 m long pipe.
Nodes are key in understanding the sound quality and pitch, revealing where maximum constructive or destructive interference occurs.

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

\(\bullet\) (a) If the amplitude in a sound wave is doubled, by what factor does the intensity of the wave increase? (b) By what fac- tor must the amplitude of a sound wave be increased in order to increase the intensity by a factor of 9\(?\)

\(\cdot\) Mapping the ocean floor. The ocean floor is mapped by sending sound waves (sonar) downward and measuring the time it takes for their echo to return. From this information, the ocean depth can be calculated if one knows that sound travels at 1531 \(\mathrm{m} / \mathrm{s}\) in seawater. If a ship sends out sonar pulses and records their echo 3.27 s later, how deep is the ocean floor at that point, assuming that the speed of sound is the same at all depths?

\(\bullet\) The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

\(\bullet\) How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 10.0\(\%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

Tuning a cello. A cellist tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must she stretch that portion of the string? (b) What percentage increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

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.