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The frequency of the note F\(_4\) is 349 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\(^\circ\)C? (b) At what air temperature will the frequency be 370 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
The pipe length is 0.246 m. The air temperature for 370 Hz is 55.13°C.

Step by step solution

01

Understand the Organ Pipe Frequency

An organ pipe open at one end and closed at the other supports standing waves where the fundamental frequency is given by \(f = \frac{v}{4L}\). Here, \(v\) is the speed of sound in air and \(L\) is the length of the pipe.
02

Calculate Speed of Sound in Air at 20°C

The speed of sound in air at \(20\,^{\circ}\mathrm{C}\) is calculated using \(v = 331 + 0.6 \times \text{temperature in °C}\). For \(20\,^{\circ}\mathrm{C},\) \(v = 343\, \text{m/s}\).
03

Calculate Length of Pipe for F\(_4\)

Using the formula \(f = \frac{v}{4L}\), with \(f = 349\, \text{Hz}\) and \(v = 343\, \text{m/s}\), rearrange to find \(L\): \[ L = \frac{v}{4f} = \frac{343\, \text{m/s}}{4 \times 349\, \text{Hz}} = 0.246\, \text{m} \].
04

Calculate Required Speed of Sound for 370 Hz

To find the speed of sound that results in a frequency of 370 Hz using the same pipe length, use \(f = \frac{v}{4L}\). Solving for \(v\), we find: \[ v = 4Lf = 4 \times 0.246\, \text{m} \times 370\, \text{Hz} = 364.08\, \text{m/s} \].
05

Calculate Temperature for Desired Speed of Sound

With the speed of sound formula \(v = 331 + 0.6T\), equate and solve for temperature \(T\): \[ 364.08 = 331 + 0.6T \] \[ T = \frac{364.08 - 331}{0.6} = 55.13\, ^{\circ}\mathrm{C} \].

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Key Concepts

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

Standing Waves in Organ Pipes
Standing waves are a fascinating concept in physics, especially when it comes to musical instruments like organ pipes. In these pipes, sound waves reflect back and forth between the open and closed ends.
This creates nodes and antinodes, forming standing waves. For a pipe that is open at one end and closed at the other, the fundamental mode of vibration is a quarter of a wavelength.
This means that the length of the pipe (\(L\)) is one-fourth of the wavelength of the sound. Consequently, the fundamental frequency is expressed as:
  • \(f = \frac{v}{4L}\)
Where \(f\) is the frequency, \(v\) is the speed of sound, and \(L\) is the length of the pipe.
Understanding standing waves helps us to calculate the frequencies produced by the organ pipes and tailor them to produce specific notes.
Fundamental Frequency Calculation
Calculating the fundamental frequency of an organ pipe is quite simple once you understand the formula. For any pipe, when you know the speed of sound and the desired frequency, you can find the required length.
Using the formula \(f = \frac{v}{4L}\), you can rearrange it to find the length:
  • \(L = \frac{v}{4f}\)
Plug in the known values of \(v\) and \(f\) to get the length. For instance, if the speed of sound at 20°C is 343 m/s and you need to produce a 349 Hz frequency, the length of the pipe would be:
  • \(L = \frac{343}{4 \times 349} = 0.246\) m
This length will give you the desired note when the pipe is played.
Speed of Sound in Air
The speed of sound in air is an essential factor in acoustics. It is not a constant and changes with temperature. At a baseline of 0°C, the speed of sound in air is approximately 331 m/s.
To calculate the speed of sound at different temperatures, you can use the formula:
  • \(v = 331 + 0.6 \times \text{temperature in °C}\)
For example, at 20°C, the speed of sound is:
  • \(v = 331 + 0.6 \times 20 = 343\, \text{m/s}\)
This relationship allows us to predict how sound frequencies will change with varying air conditions, assisting in making performance adjustments.
Temperature Effect on Sound Frequency
Temperature affects the frequency of the sound produced by an organ pipe because it changes the speed of sound. Higher temperatures increase the speed of sound, leading to higher frequencies.
This means that an organ pipe will produce a higher pitch on hot days than on cold ones.
To find the temperature at which a certain frequency is achieved, you'd adjust the speed of sound. If you need the frequency to be 370 Hz, first calculate the needed speed of sound.
Using the rearranged equation \(v = 4Lf\), with \(L\) being 0.246 m:
  • \(v = 4 \times 0.246 \times 370 = 364.08\, \text{m/s}\)
Then, you solve for temperature with:
  • \(v = 331 + 0.6T\)
By solving for \(T\), you find the air temperature required:
  • \(T = \frac{364.08 - 331}{0.6} = 55.13\, °C\)
This helps in understanding how to compensate for temperature changes in tuning musical instruments.

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Most popular questions from this chapter

A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 m/s, and the crests of the waves ahead of the duck are spaced 0.12 m apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?

The longest pipe found in most medium-size pipe organs is 4.88 m (16 ft) long. What is the frequency of the note corresponding to the fundamental mode if the pipe is (a) open at both ends, (b) open at one end and closed at the other?

A car alarm is emitting sound waves of frequency 520 Hz. You are on a motorcycle, traveling directly away from the parked car. How fast must you be traveling if you detect a frequency of 490 Hz?

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At point \(A, 3.0 \mathrm{~m}\) from a small source of sound that is emitting uniformly in all directions, the sound intensity level is \(53 \mathrm{~dB}\). (a) What is the intensity of the sound at \(A ?\) (b) How far from the source must you go so that the intensity is one-fourth of what it was at \(A\) ? (c) How far must you go so that the sound intensity level is one-fourth of what it was at \(A ?\) (d) Does intensity obey the inverse-square law? What about sound intensity level?

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