/*! 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 68 The frequency of the note \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

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The frequency of the note \(\mathrm{F}_{4}\) is 349 \(\mathrm{Hz} .\) (a) If an organ pipe is open at one end and closed at the other, what length must it have for its fundamental mode to produce this note at \(20.0^{\circ} \mathrm{C}\) (b) At what air temperature will the frequency be 370 \(\mathrm{Hz}\) , corresponding to a rise in pitch from F to F-sharp? (Ignore the change in length of the pipe due to the temperature change.)

Short Answer

Expert verified
(a) The length must be 0.246 meters. (b) The frequency will be 370 Hz at approximately 54.47°C.

Step by step solution

01

Equation for Frequency of Pipe

An open-closed pipe produces a fundamental frequency according to the formula: \( f = \frac{v}{4L} \), where \( f \) is the frequency, \( v \) is the speed of sound in air, and \( L \) is the length of the pipe.
02

Calculate Speed of Sound at 20°C

The speed of sound in air as a function of temperature can be calculated using: \( v = 331.4 + 0.6 \times T \) m/s, where \( T \) is the temperature in Celsius. At \( 20.0^{\circ} \mathrm{C} \), \( v = 331.4 + 0.6 \times 20 = 343.4 \) m/s.
03

Find Length for Frequency 349 Hz

Rearrange the formula to solve for \( L \): \( L = \frac{v}{4f} \). Substitute \( v = 343.4 \) m/s and \( f = 349 \) Hz: \( L = \frac{343.4}{4 \times 349} = 0.246 \) meters.
04

Calculate New Speed of Sound for 370 Hz

Rearrange the formula to solve for the new speed of sound \( v' \) when \( f = 370 \) Hz: \( v' = 4Lf = 4 \times 0.246 \times 370 = 364.08 \) m/s.
05

Find New Temperature for 370 Hz

The new speed of sound equation is \( v' = 331.4 + 0.6 \times T' \). Set \( v' = 364.08 \) m/s and solve for \( T' \): \( 364.08 = 331.4 + 0.6 \times T' \). Rearranging gives: \( T' = \frac{364.08 - 331.4}{0.6} \approx 54.47 \)°C.

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Key Concepts

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

Frequency of sound
The frequency of sound refers to the number of vibrations or cycles per second, measured in Hertz (Hz). It's like the beat in a song; more beats mean a higher frequency and pitch. Different musical notes have specific frequencies. For example, the note F4 has a frequency of 349 Hz.
This means if you hear the note F4, the air is vibrating at 349 cycles per second. Understanding frequency helps in various applications, from music to acoustics. In our exercise, we need to find an organ pipe's length that produces this note. Frequency is crucial because it defines the sound you hear.
Speed of sound
The speed of sound is how fast sound waves travel through a medium, like air. It's affected by several factors, including temperature. At room temperature, which is around 20°C, the speed of sound in air is about 343.4 meters per second.
The formula to calculate the speed of sound in air is: \[ v = 331.4 + 0.6 imes T \] where \( T \) is the temperature in Celsius.
  • As temperature increases, sound travels faster.
  • At 20°C, we calculate it to be 343.4 m/s.
This concept helps when considering how sound will behave in different environments, such as calculating how a pipe length needs to change with temperature.
Fundamental frequency
The fundamental frequency is the lowest frequency produced by any vibrating object, including musical instruments. For pipes, this is the simplest form of vibration. Pipes can be open at both ends or one end closed. Our exercise uses a pipe closed at one end to produce a fundamental frequency of 349 Hz.
In acoustics for such a pipe, the formula for the fundamental frequency is: \[ f = \frac{v}{4L} \]
  • \( f \) is the frequency.
  • \( v \) is the speed of sound.
  • \( L \) is the length of the pipe.
Using this equation, we can find the appropriate length of the pipe to achieve the desired frequency, as done in our exercise.
Organ pipe acoustics
Organ pipes work on the principle of acoustics where sound waves resonate within a cylindrical shape, like the pipe. There are two types of pipes: open-open and open-closed. In open-closed pipes, the closed end reflects sound waves, allowing a standing wave to form, which gives rise to the fundamental frequency.
In our exercise, the pipe is designed to produce a note at 349 Hz. To calculate its length when open at one end, we use the equation \[ L = \frac{v}{4f} \]Here, you need the speed of sound and your desired frequency. We used a calculated speed of sound at 20°C to determine the correct pipe length for this note to resonate perfectly.
This principle is crucial for tuning musical instruments, ensuring they produce the correct pitches when played.

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Most popular questions from this chapter

The Sacramento City Council adopted a law to reduce the allowed sound intensity level of the much-despised leaf blowers from their current level of about 95 \(\mathrm{dB}\) to 70 \(\mathrm{dB}\) . With the new law, what is the ratio of the new allowed intensity to the previously allowed intensity?

Ep A New Musical Instrument. You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length \(L\) and diameter \(L / 10 .\) You have stretched a string of mass per unit length \(\mu\) across the open end of the tube. The other end of the tube is closed. To produce the musical effect you're looking for, you want the frequency of the third-harmonic standing wave on the string to be the same as the fundamental frequency for sound waves in the air column in the tube. The speed of sound waves in this air column is \(v_{\mathrm{s}}\) . (a) What must be the tension of the string to produce the desired effect? (b) What happens to the sound produced by the instrument if the tension is changed to twice the value calculated in part (a)? (c) For the tension calculated in part (a), what other harmonics of the string, if any, are in resonance with standing waves in the air column?

BIO The Human Voice. The human vocal tract is a pipe that extends about 17 \(\mathrm{cm}\) from the lips to the vocal folds (also called "vocal cords") near the middle of your throat. The vocal folds behave rather like the reed of a clarinet, and the vocal tract acts like a stopped pipe. Estimate the first three standing-wave frequencies of the vocal tract. Use \(v=344 \mathrm{m} / \mathrm{s} .\) The answers are only an estimate, since the position of lips and tongue affects the motion of air in the vocal tract.)

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