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An organ pipe that is closed at one end is filled with helium. The pipe has a fundamental frequency of \(660 \mathrm{~Hz}\) in air at \(20^{\circ} \mathrm{C}\). What is the pipe's fundamental frequency with the helium in it?

Short Answer

Expert verified
The fundamental frequency in helium is approximately 1937 Hz.

Step by step solution

01

Understand the Concept

An organ pipe closed at one end operates on the principle of resonance. For a closed pipe, the fundamental frequency is determined by the speed of sound in the medium filling the pipe.
02

Know the Relationship Between Speed of Sound and Frequency

The fundamental frequency of a pipe can be determined by the formula \( f = \frac{v}{4L} \) where \( v \) is the speed of sound in the medium and \( L \) is the length of the pipe. Thus, \( f \propto v \). Since the pipe's physical properties remain constant, the change in frequency is directly related to the change in the speed of sound.
03

Compute the Speed of Sound in Air and Helium

At \( 20^{\circ} \, \text{C} \), the speed of sound in air, \( v_{\text{air}} \), is approximately \( 343 \, \text{m/s} \). The speed of sound in helium, \( v_{\text{helium}} \), is approximately \( 1007 \, \text{m/s} \).
04

Calculate the New Frequency

Since the frequency is directly proportional to the speed of sound in the medium, the frequency in helium can be calculated by the ratio of the speeds of sound. Thus, the new frequency with helium, \( f_{\text{helium}} \), is \( f_{\text{helium}} = f_{\text{air}} \times \frac{v_{\text{helium}}}{v_{\text{air}}} \).
05

Solve for Helium Frequency

Given \( f_{\text{air}} = 660 \, \text{Hz} \), compute \( f_{\text{helium}} = 660 \, \text{Hz} \times \frac{1007 \, \text{m/s}}{343 \, \text{m/s}} \approx 1936.92 \, \text{Hz} \).

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Key Concepts

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

Organ Pipe Physics
Organ pipe physics involves understanding how sound waves behave in tube-like structures. Organs often use pipes to produce sound by pushing air through them, causing the air to resonate. Each pipe creates a different sound depending on its length and whether it is open at both ends, or closed at one end. A closed pipe, which is closed at one end and open at the other, produces sound based on specific resonant frequencies.
This type of pipe emphasizes odd harmonics, meaning only the odd-numbered resonant frequencies can sound prominently. The air column in the pipe vibrates when the length matches the wavelength of these odd harmonics. The lowest resonant frequency is known as the fundamental frequency, and it is what primarily determines the pitch we hear from the pipe.
  • Closed pipes have a unique sound due to resonance principles.
  • Fundamental frequency depends on pipe length and medium's speed of sound.
  • Odd harmonics are emphasized in closed pipe systems.
Speed of Sound
The speed of sound is a key factor in determining the frequency of sound that a pipe can produce. It varies depending on the medium through which sound waves travel. In the context of gases, temperature, and the type of gas significantly affect the speed of sound. For example, at room temperature around 20°C, the speed of sound in air is approximately 343 meters per second (m/s).
Helium, on the other hand, allows sound to travel much faster, at about 1007 m/s under similar conditions. This difference is due to helium having a lower density and a different molecular structure compared to air. As a result, sound waves can pass through helium more quickly, leading to higher frequencies for the same pipe length.
  • The medium's properties significantly affect the speed of sound.
  • With helium, sound travels faster than in air.
  • Speed of sound impacts the fundamental frequency in organ pipes.
Closed Pipe Resonance
Closed pipe resonance is an interesting phenomenon where resonance occurs in a pipe that is closed at one end. This arrangement means the pipe supports standing waves that form at only odd multiples of the fundamental frequency. The closed end serves as a node (a point of zero amplitude) while the open end remains an antinode (a point of maximum amplitude).
The fundamental frequency, the lowest frequency of vibration, is important because it establishes the baseline pitch of the sound produced. The length of the pipe and the speed of sound in the medium determine this frequency, highlighting the directly proportional relationship between speed of sound and frequency. If the speed of sound increases, so does the fundamental frequency.
  • Closed pipes resonate at odd harmonic frequencies.
  • Fundamental frequency sets the tone's pitch.
  • Speed of sound and pipe length determine resonance characteristics.
Wave Speed Ratio
The wave speed ratio measures the change in sound frequency when the speed of sound in the medium changes. Since the fundamental frequency is proportional to the speed of sound, altering the medium changes this frequency based on the ratio of the wave speeds in the different media.
In our exercise, replacing air with helium in a closed pipe dramatically increases the frequency due to helium's much faster speed of sound. By calculating the wave speed ratio, \(\frac{1007 \, \text{m/s}}{343 \, \text{m/s}}\), \ we can determine how the fundamental frequency scales up. This ratio is used to easily compute how much faster the frequency becomes in a new medium without changing the pipe's dimensions.
  • The wave speed ratio helps calculate frequency changes between media.
  • It shows how fundamental frequency scales with speed of sound.
  • Critical for transitioning sound characteristics between different gases.

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Most popular questions from this chapter

During practice in a huddle, a quarterback shouts the play in anticipation of crowd noise during the actual game. To a receiver \(0.750 \mathrm{~m}\) away from the quarterback in the huddle, it seems as loud as the noise from a screaming child. When they get into practice formation, the quarterback yells at twice the output power, yet the instructions seem only about as loud as normal conversation. Use typical values in Table 14.2 to estimate how far from the quarterback the receiver is in the formation.

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?

If the distance to a sound source is halved, (a) will the sound intensity level change by a factor of (1) 2 (2) \(1 / 2,\) (3) \(4,\) (4) \(1 / 4\), or (5) none of the preceding? Why? (b) What is the change in the sound intensity level?

What is the intensity level of a \(23-\mathrm{dB}\) sound after being amplified (a) ten thousand times, (b) a million times, (c) a billion times?

The speed of sound in human tissue is on the order of \(1500 \mathrm{~m} / \mathrm{s}\). A 3.50 -MHz probe is used for an ultrasonic procedure. (a) If the effective physical depth of the ultrasound is 250 wavelengths, what is the physical depth in meters? (b) What is the time lapse for the ultrasound to make a round trip if reflected from an object at the effective depth? (c) The smallest detail capable of being detected is on the order of one wavelength of the ultrasound. What would this be?

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