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Chapter 16: Problem 9

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 \(\mathrm{m} / \mathrm{s}\) when the gas temperature is \(22.0^{\circ} \mathrm{C}\) . For a certain experiment, you need to have the same uscillator produce sound of waveleugth 28.5 \(\mathrm{cm}\) in this gas. What should the gas temperature be to achieve this wavelength?

Short Answer

Expert verified
The gas temperature should be approximately 74.29掳C.

Step by step solution

01

Understand the Relationship Between Wavelength, Frequency, and Speed

The fundamental relationship between the speed of sound \( v \), frequency \( f \), and wavelength \( \lambda \) is given by the equation \( v = f \cdot \lambda \). This equation will be our key tool in solving the problem.
02

Identify the Known Values

We have the initial conditions given as:- Frequency \( f = 1250 \) Hz, - Initial speed of sound \( v_1 = 325 \) m/s, and - Initial gas temperature \( T_1 = 22.0^{\circ} \mathrm{C} \) which is \( 22.0 + 273.15 = 295.15 \) K.We need to find the new temperature \( T_2 \) where the wavelength \( \lambda_2 \) is 0.285 m.
03

Relate Speed with Temperature

The speed of sound in a gas is proportional to the square root of its absolute temperature, given by the formula:\[ v = \sqrt{\frac{k T}{m}} \]where \( k \) is a constant and \( m \) is the molecular mass. From this, \( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \). This equation will relate the change in speed to the change in temperature.
04

Calculate the New Speed of Sound

Using the new wavelength \( \lambda_2 = 0.285 \) m and the same frequency \( f = 1250 \) Hz, we can calculate the new speed of sound:\[ v_2 = f \cdot \lambda_2 \]\[ v_2 = 1250 \cdot 0.285 = 356.25 \text{ m/s} \]
05

Use the Proportion to Find New Temperature

Substitute the values into the proportion:\[ \frac{356.25}{325} = \sqrt{\frac{T_2}{295.15}} \]Squaring both sides, we have:\[ \left( \frac{356.25}{325} \right)^2 = \frac{T_2}{295.15} \]Calculating further:\[ T_2 = \left( \frac{356.25}{325} \right)^2 \times 295.15 \approx 347.44 \text{ K} \]
06

Convert Temperature to Celsius

Finally, we convert the new temperature from Kelvin back to Celsius:\[ T_2 = 347.44 - 273.15 = 74.29^{\circ} \text{C} \]

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Key Concepts

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

Sound Wave Propagation
Sound wave propagation is the journey of sound waves through different mediums. These waves are longitudinal, meaning they move in the same direction as the energy that is being transmitted. Sound waves consist of compressions and rarefactions, which are areas where the molecules in the medium are bunched together and spread apart, respectively. This is how sound creates the vibrations we hear and experience.

In an ideal gas, which is a hypothetical gas used to simplify calculations, sound travels depending on the arrangement and movement of molecules. The speed of sound in a gas is determined by the medium's properties, including density and elasticity. The denser the medium, the slower the sound waves move, and vice versa. Hence, understanding wave propagation is essential in predicting how sound will travel in various mediums and under different conditions.
Ideal Gas Behavior
Ideal gas behavior refers to the theoretical behavior of gases as explained by the ideal gas law. This simplifies real-life gas behavior through assumptions like no molecular interaction and negligible volume of gas particles.

For sound wave calculations, we use the properties of ideal gases to predict wave speed. The speed of sound in an ideal gas \( v \) is related to the temperature and molecular mass of the gas as \( v = \sqrt{\frac{kT}{m}} \), where \( k \) is a constant. This formula indicates that gas temperature affects the speed of sound, demonstrating the link between gas behavior and sound propagation. Understanding this helps us predict how changes in conditions, such as temperature, will affect sound travel, which is essential for experiments and real-world applications.
Temperature Effects on Sound
The temperature of a gas significantly affects the speed of sound. As the temperature increases, the molecules in the gas gain more energy, move faster, and bounce off each other with more force. This increases the speed at which sound waves travel through the gas.

For example, in the original problem, the sound wave needs to travel faster in order to achieve a certain wavelength with the same oscillator frequency. By increasing the gas temperature, we increase the energy of the molecules, thereby increasing the speed of sound. This method showcases the direct relationship between temperature and sound velocity, where higher temperatures lead to faster sound propagation. This concept is vital in fields like meteorology and acoustics, where understanding sound behavior under varying temperature conditions is critical.

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Most popular questions from this chapter

Two loudspeakers, \(A\) and \(B(\mathrm{Fig} .16 .40),\) are driven by the same amplifier and emit sinusoidal waves in phase. Speaker \(B\) is 2.00 \(\mathrm{m}\) to the right of speaker \(A\) . Consider point \(Q\) along the exten- sion of the line connecting the speakers, 1.00 \(\mathrm{m}\) to the right of speaker \(B .\) Both speakers emit sound waves that travel directly from the speaker to point \(Q .\) (a) What is the lowest frequency for which constructive interference occurs at point \(Q ?\) (b) What is the lowest frequency for which destructive interference occurs at point \(Q ?\)

A soprano and a bass are singing a duet. While the soprano \(\operatorname{sings}\) an \(A^{\prime \prime}\) at 932 . Hz, the bass sings an \(A^{\prime \prime}\) but three octaves lower. In this concert hall, the density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its bulk modulus is \(1.42 \times 10^{5} \mathrm{Pa}\) . In order for their notes to have the same Rnund intensity level, what must he (a) the ratio of the pressur amplitude of the bass to that of the soprano, and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in \(\mathrm{m}\) and \(\mathrm{nm}\) ) does the soprano produce to sing her \(A^{\prime \prime}\) at 72.0 \(\mathrm{dB} ?\)

An organ pipe has two successive harmonics with frequencies 1372 and 1764 \(\mathrm{Hz}\) (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

The fundamental frequency of a pipe that is open at both ends is 594 Hz (a) How long is this pipe? If one end is now closed, find (b) the wavelength and (c) the frequency of the new fundamental.

On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz}\) . If the stationary female hears a tone of 1240 \(\mathrm{Hz}\) , what is the speed of sound in the atmosphere of Arrakis?
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