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A closed organ pipe has a fundamental frequency of \(528 \mathrm{~Hz}(\mathrm{a} \mathrm{C}\) note \()\) at \(20{ }^{\circ} \mathrm{C}\). What is the fundamental frequency of the pipe when the temperature is \(0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The fundamental frequency at 0°C is 509 Hz.

Step by step solution

01

Understanding the Relationship between Frequency and Temperature

The frequency of sound in a medium changes with temperature because the speed of sound in air changes with temperature. The formula for the speed of sound \( v \) in air as a function of temperature \( T \) in Celsius is \( v = 331 + 0.6T \).
02

Calculation of Speed of Sound at 20°C

First, calculate the speed of sound in air at 20°C. Using the formula for speed of sound: \[ v_{20} = 331 + 0.6 imes 20 = 343 \text{ m/s} \]
03

Calculation of Speed of Sound at 0°C

Now, calculate the speed of sound in air at 0°C. Using the same formula: \[ v_{0} = 331 + 0.6 imes 0 = 331 \text{ m/s} \]
04

Relating Frequency to Speed of Sound

The frequency of the fundamental note in a closed pipe is directly proportional to the speed of sound in air. Thus, \( \frac{f_{0}}{f_{20}} = \frac{v_{0}}{v_{20}} \), where \( f_{0} \) is the frequency at 0°C and \( f_{20} \) is the frequency at 20°C (528 Hz).
05

Solving for Fundamental Frequency at 0°C

Substitute the known values into the proportionality relation: \[ \frac{f_{0}}{528} = \frac{331}{343} \]Solving for \( f_{0} \), we get: \[ f_{0} = 528 \times \frac{331}{343} = 509 \text{ Hz} \]

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Key Concepts

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

Sound Frequency
Sound frequency refers to the number of vibrations or cycles a sound wave completes in one second, measured in Hertz (Hz). It determines the pitch of the sound we hear. The more cycles per second, the higher the frequency, and thus, the higher the pitch.

For instance, a sound frequency of 528 Hz means that the sound wave vibrates 528 times in one second. Frequencies are a crucial aspect of acoustics as they define the unique tones and notes we hear, ranging from low bass sounds to high treble notes.

Understanding sound frequency is essential when dealing with musical instruments, such as an organ pipe, where different frequencies correspond to different musical notes. The frequency can be influenced by various factors including the speed of sound, which is affected by temperature changes, as we'll explore further.
Closed Pipe Resonance
Closed pipe resonance involves vibrations within a pipe that is closed at one end and open at the other. Such pipes can produce resonance at specific frequencies, known as the fundamental frequency and its harmonics. These are the frequencies at which the natural standing wave patterns fit perfectly within the length of the pipe.

For a closed organ pipe, the fundamental frequency represents the lowest frequency at which the pipe resonates. It forms a quarter-wave of the sound inside the pipe, with a node at the closed end and an antinode at the open end.
  • The pipe length is roughly one fourth of the wavelength of the fundamental frequency.
  • This is why these pipes have a deeper tone compared to open pipes of the same length.
Each change in environmental conditions, like temperature, can affect the resonance by changing the speed of sound and thereby altering the frequency at which the pipe resonates.
Temperature Effects on Sound Speed
Temperature significantly influences the speed of sound. As the temperature increases, the speed of sound in air also increases. This is because air molecules move more rapidly at higher temperatures, facilitating faster transmission of sound waves.

The relationship is approximately linear and can be defined by the formula: \[ v = 331 + 0.6T \]where \( v \) is the speed of sound in meters per second and \( T \) is the temperature in degrees Celsius.

For instance, at 20°C, the speed of sound is calculated to be 343 m/s, but at 0°C, it reduces to 331 m/s. Thus, warmer air results in higher sound speed, while colder air results in slower sound speed. This change directly affects the frequency in closed pipes, resulting in lower frequencies in colder air.
Fundamental Frequency Calculation
Calculating the fundamental frequency in a closed pipe involves understanding the direct proportionality between frequency and the speed of sound. In this context:\[ \frac{f_0}{f_{20}} = \frac{v_0}{v_{20}} \]where \( f_0 \) and \( f_{20} \) are the frequencies at 0°C and 20°C respectively, and \( v_0 \) and \( v_{20} \) are the corresponding speeds of sound at these temperatures.

To find the fundamental frequency at 0°C, if the known frequency at 20°C is 528 Hz:
  • First, determine the speed of sound at both temperatures using the formula \( v = 331 + 0.6T \).
  • Then, substitute these values into the proportionality equation to solve for \( f_0 \).
  • The solution yields a new frequency of approximately 509 Hz at 0°C when the speed of sound in air is slower than at 20°C.
This calculation ensures instruments are tuned appropriately in varying temperatures to maintain consistent sound quality.

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Most popular questions from this chapter

An open organ pipe with a length of \(50.0 \mathrm{~cm}\) is oscillating in its second-overtone or third-harmonic mode. Assume the air to be at room temperature and the pipe to be at rest in still air. A person moves toward this pipe at \(2.00 \mathrm{~m} / \mathrm{s}\) and, at the same time, away from a highly reflective wall. (a) Will the observer hear beats: (1) yes, (2) no, or (3) can't tell from the data given? (b) Calculate the frequency of sound emitted. (c) Calculate the beat frequency the observer would hear. [Hint: There are two frequencies, one directly from the pipe and one from the wall.]

A rock band (with loud speakers) has an average intensity level of \(110 \mathrm{~dB}\) at a distance of \(15 \mathrm{~m}\) from the band. Assuming the sound is radiated equally over a hemisphere in front of the band, what is the total power output?

A person holds a rifle horizontally and fires at a target. The bullet has a muzzle speed of \(200 \mathrm{~m} / \mathrm{s},\) and the person hears the bullet strike the target \(1.00 \mathrm{~s}\) after firing it. The air temperature is \(72^{\circ} \mathrm{F}\). What is the distance to the target?

The first three natural frequencies of an organ pipe are \(126 \mathrm{~Hz}, 378 \mathrm{~Hz},\) and \(630 \mathrm{~Hz}\). (a) Is the pipe an open or a closed pipe? (b) Taking the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\), find the length of the pipe.

Bats typically give off an ultrahigh-frequency sound at about \(50000 \mathrm{~Hz}\). If a bat is approaching a stationary object at \(18.0 \mathrm{~m} / \mathrm{s},\) what will be the reflected frequency it detects? Assume the air in the cave is at \(5^{\circ} \mathrm{C} .\)

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