/*! 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 77 \(\bullet\) Two organ pipes, ope... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 77

\(\bullet\) Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 \(\mathrm{cm} .\) Find the frequency of the beat they produce when playing together in their fundamental.

Short Answer

Expert verified
The beat frequency is approximately 1.30 Hz.

Step by step solution

01

Understand the Problem

The problem involves two organ pipes that are closed at one end and open at the other. These types of pipes have odd harmonics and their fundamental frequency can be calculated using the formula for pipes closed at one end.
02

Fundamental Frequency of a Closed Pipe

The fundamental frequency of a pipe closed at one end is given by \( f = \frac{v}{4L} \), where \(v\) is the speed of sound in air (approximately 343 m/s) and \(L\) is the length of the pipe. First, calculate the fundamental frequency for the original pipe with \(L = 1.14\) m.
03

Calculate Frequency for the Original Pipe

Substitute \(L = 1.14\) m into the formula: \[ f_1 = \frac{343}{4 \times 1.14} \approx 75.22 \, \text{Hz} \]
04

Adjust Length for the Extended Pipe

The second pipe is lengthened by 2.00 cm, which is 0.02 m. Thus, its new length is \(L = 1.14 + 0.02 = 1.16\) m.
05

Calculate Frequency for the Lengthened Pipe

Substitute \(L = 1.16\) m into the formula: \[ f_2 = \frac{343}{4 \times 1.16} \approx 73.92 \, \text{Hz} \]
06

Calculate the Beat Frequency

The beat frequency is the absolute difference between the frequencies of the two pipes: \[ f_{beat} = |f_1 - f_2| \approx |75.22 - 73.92| = 1.30 \, \text{Hz} \]

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.

Organ Pipes
Organ pipes are fascinating instruments. They can be open at both ends or open at one end and closed at the other. In this exercise, we're considering pipes that are open at one end and closed at the other, known as closed pipes.

These pipes produce sound through the vibration of air inside, and their length, as well as the speed of sound in the air, determines the frequencies they produce. A key characteristic of closed pipes is that they produce only odd harmonics.
  • Closed pipes produce sound at frequencies of the odd-numbered harmonics (1st, 3rd, 5th, etc.).
  • This means the fundamental frequency is the lowest, followed by the third harmonic, and so on.
  • The formula used for calculating the fundamental frequency assumes the pipe is closed at one end, creating nodes and antinodes in the air column.
Fundamental Frequency
Understanding the fundamental frequency is crucial for musical instruments like organ pipes. The fundamental frequency is the first harmonic and is the lowest frequency produced by the oscillating air column in a pipe.

For closed organ pipes, the fundamental frequency can be computed using the formula \( f = \frac{v}{4L} \), where \(v\) is the speed of sound and \(L\) is the length of the pipe. Using this formula:
  • The speed of sound in air is typically about 343 meters per second (m/s).
  • Substitute the length of the pipe into the formula to find the fundamental frequency.
  • The calculated frequency tells us how fast the air column inside the pipe vibrates.
Beat Frequency
When two sounds of nearly identical frequency are played together, they produce a phenomenon known as "beats." This is an interference pattern between two waves of slightly different frequencies.

The beat frequency is the rate at which the volume of the combined sound rises and falls:
  • It is found by taking the absolute difference between the two frequencies.
  • If two organ pipes are played together with slightly different lengths, they create a beat frequency.
  • This is heard as a periodic variation in volume, which is useful in tuning instruments.
In the exercise, the beat frequency is calculated using \(|f_1 - f_2|\), showing how the slight difference in pipe lengths creates a perceptible change in sound intensity.
Speed of Sound
The speed of sound is a critical factor when calculating sound frequencies in organ pipes. It is generally accepted that the speed of sound in air at room temperature is approximately 343 meters per second. However, this can vary slightly based on temperature and humidity.

Factors affecting the speed of sound include:
  • Temperature: Warmer air makes sound travel faster due to increased energy and movement of molecules.
  • Humidity: More humid air also increases the speed since moist air is less dense.
  • Medium: Sound travels at different speeds in different media (faster in solid materials than in gases).
Understanding these factors helps you appreciate how sound production in organ pipes can be influenced by environmental conditions.

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

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 \(\mathrm{D} ?\)

Transverse waves on a string have wave speed \(8.00 \mathrm{m} / \mathrm{s},\) amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m} .\) These waves travel in the \(x\) direction, and at \(t=0\) the \(x=0\) end of the string is at \(y=0\) and moving downward. (a) Find the frequency, period, and wave number of these waves. (b) Write the equation for \(y(x, t)\) describing these waves. (c) Find the transverse displacement of a point on the string at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) .

\(\bullet\) Transverse waves are traveling on a long string that is under a tension of 4.00 \(\mathrm{N}\) . The equation describing these waves is $$y(x, t)=(1.25 \mathrm{cm}) \sin \left[\left(415 \mathrm{s}^{-1}\right) t-\left(44.9 \mathrm{m}^{-1}\right) x\right]$$ Find the linear mass density of this string.

\(\bullet\) A certain transverse wave is described by the equation $$y(x, t)=(6.50 \mathrm{mm}) \sin 2 \pi\left(\frac{t}{0.0360 \mathrm{s}}-\frac{x}{0.280 \mathrm{m}}\right)$$ Determine this wave's (a) amplitude, (b) wavelength, (c) frequency, (d) speed of propagation, and (e) direction of propagation.

\(\cdot\) 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?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Force

Read Explanation

Electricity

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Work Energy and Power

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.