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Two organ pipes, open at ond 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 1.29 Hz.

Step by step solution

01

Calculate the Length of Each Pipe

The original length of each pipe is 1.14 m. After lengthening one pipe by 2.00 cm, the new length of that pipe becomes 1.14 + 0.02 = 1.16 m.
02

Determine Fundamental Frequency of the Original Pipe

The fundamental frequency of a pipe closed at one end is determined by the formula \( f = \frac{v}{4L} \), where \( v \) is the speed of sound in air (approximately 343 m/s) and \( L \) is the length of the pipe. For the original pipe, \( L = 1.14 \, \text{m} \). Hence, \( f_1 = \frac{343}{4 \times 1.14} \approx 75.22 \, \text{Hz} \).
03

Calculate Fundamental Frequency of the Lengthened Pipe

Applying the same formula for the lengthened pipe: \( L = 1.16 \, \text{m} \), \( f_2 = \frac{343}{4 \times 1.16} \approx 73.93 \, \text{Hz} \).
04

Determine Beat Frequency

The beat frequency is calculated by taking the absolute difference between the two fundamental frequencies of the pipes: \( f_{\text{beat}} = |f_1 - f_2| = |75.22 - 73.93| \approx 1.29 \, \text{Hz} \).

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Key Concepts

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

Organ Pipes
Organ pipes come in various designs, but the specific type we're discussing here is one that is open at one end and closed at the other. This configuration is known as a closed-end pipe.

An organ pipe of this type supports standing waves, which are crucial for creating sound. When air is blown into the pipe, these standing waves form, leading to resonant frequencies.

In a closed-end pipe, the closed end acts as a node, where there is no air movement, and the open end acts as an antinode, where the air moves the most.

This characteristic determines the harmonic series a pipe can produce. Unlike open pipes, closed-end pipes only support odd harmonics, which can affect the types of sound they produce significantly. Understanding these principles is essential when considering musical instruments like organs, where the quality and type of sound are critical.

The pipe length is critical; in this exercise, we see a change in one pipe's length influences the sound and causes a beat frequency when both pipes are played together.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system like an organ pipe resonates. It is the most critical frequency and forms the basis for all other harmonics and overtones produced.

For a pipe that is closed at one end, the fundamental frequency, denoted by \( f \), is calculated using the formula \( f = \frac{v}{4L} \) where \( v \) is the speed of sound in air, and \( L \) is the length of the pipe.

This formula reflects the specific standing wave pattern permissible in a closed-end pipe, where only one-quarter wavelength fits into the pipe's length.

In the given problem, different lengths cause variations in fundamental frequency, leading to the production of beats when sounds from two pipes interact. Understanding this principle explains why the smallest changes in length or environment can drastically affect the sound quality of a pipe organ or any resonant system.
Speed of Sound
The speed of sound in air is a vital parameter in acoustics and plays a crucial role in the resonant frequencies of musical instruments like organ pipes.

Typically, the speed of sound in air at room temperature (20°C) is approximately 343 meters per second. This value can fluctuate depending on temperature, humidity, and atmospheric pressure, but for simplicity and most basic calculations, we use this standard figure.

Knowing the speed of sound allows us to link the physical length of a resonant column with the sound frequencies it produces.
  • For open-end organ pipes, the speed determines how wavelengths are structured inside the pipe.
  • For closed-end pipes, it helps in determining the fundamental frequency using the formula \( f = \frac{v}{4L} \).


Understanding the speed of sound helps us explore how two pipes with slightly different lengths can create a beat frequency due to variations in fundamental frequencies, such as in our exercise.

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Most popular questions from this chapter

CP A person is playing a small flute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 Hz. If the speed of sound is \(344.0 \mathrm{m} / \mathrm{s},\) for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

A jet plane flies overhead at Mach 1.70 and at a constant altitude of 950 \(\mathrm{m}\) . (a) What is the angle \(\alpha\) of the shock-wave cone? (b) How much time after the plane passes directly overhead do you hear the sonic boom? Neglect the variation of the speed of sound with altitude.

Tuning a Violin. A violinist is tuning her instrument to concert \(\mathrm{A}(440 \mathrm{Hz}) .\) She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 3 \(\mathrm{Hz}\) , which increases to 4 \(\mathrm{Hz}\) when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3 -Hz beat? (b) To get her violin perfectly tuned to concert \(\Lambda,\) should she tighten or loosen her string from what it was when she heard the 3 -Hz beat?

Wagnerian Opera. A man marries a great Wagnerian soprano but, alas, he discovers he cannot stand Wagnerian opera. In order to save his eardrums, the unhappy man decides he must silence his larklike wife for good. His plan is to tie her to the front of his car and send car and soprano speeding toward a brick wall. This soprano is quite shrewd, however, having studied physics in her student days at the music conservatory. She realizes that this wall has a resonant frequency of 600 \(\mathrm{Hz}\) , which means that if a continuous sound wave of this frequency hits the wall, it will fall down, and she will be saved to sing more Isoldes. The car is heading toward the wall at a high speed of 30 \(\mathrm{m} / \mathrm{s}\) . (a) At what frequency must the soprano sing so that the will will crumble? (b) What frequency will the soprano hear reflected from the wall just before it crumbles?

CP You have a stopped pipe of adjustable length close to a taut \(85.0-\mathrm{cm}, 7.25-\mathrm{g}\) wire under a tension of 4110 \(\mathrm{N}\) . You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be?

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