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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 oscillator produce sound of wavelength 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 66.0掳C.

Step by step solution

01

Determine the relationship between wavelength, frequency, and velocity

The relationship between the velocity of a wave \( v \), the frequency \( f \), and the wavelength \( \lambda \) is given by the equation: \[ v = f \times \lambda \] This equation will help us relate these three variables.
02

Calculate current wavelength at 22掳C

First, let's calculate the wavelength at which the wave currently travels at 22掳C. We are given the velocity \( v = 325 \, \mathrm{m/s} \) and the frequency \( f = 1250 \, \mathrm{Hz} \). Using \( v = f \times \lambda \), we have: \[ \lambda = \frac{v}{f} = \frac{325}{1250} \, \mathrm{m} = 0.26 \, \mathrm{m} \] Hence, the current wavelength is 0.26 meters.
03

Relate wavelength and temperature change

Using the fact that the velocity of sound in a gas is related to the temperature \( T \) by the equation: \[ v \propto \sqrt{T} \] Since \( v = f \times \lambda \), we can use \[ f \times \lambda \propto \sqrt{T} \]This means the ratio of frequencies to the square root of their respective temperatures is constant.
04

Calculate new temperature for required wavelength

We need the wavelength to be 0.285 meters. Let the temperature at this condition be \( T' \). Using the proportionality from step 3: \[ \frac{f \times \lambda'}{\sqrt{T'}} = \frac{f \times \lambda}{\sqrt{T}} \] Where \( \lambda = 0.26 \mathrm{m} \), \( \lambda' = 0.285 \mathrm{m} \), and \( T = 295.15 \mathrm{K} \). Now, solve for \( T' \):\[ \frac{0.285}{\sqrt{T'}} = \frac{0.26}{\sqrt{295.15}} \]Squaring both sides:\[ \frac{0.285^2}{T'} = \frac{0.26^2}{295.15} \]\[ \Rightarrow T' = 295.15 \times \left( \frac{0.285}{0.26} \right)^2 \approx 339.1 \mathrm{K} \]Convert to Celsius: \[ T' - 273.15 \approx 66.0 ^{\circ} \mathrm{C} \].Hence, the gas temperature should be approximately \( 66.0^{\circ} \mathrm{C} \).

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Key Concepts

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

Oscillator
An oscillator is an essential component in physics, particularly when studying sound waves. It can be thought of as a device or system that produces rhythmic vibrations or oscillations. These oscillations can create sound waves at a specific frequency.
An everyday example of an oscillator is a tuning fork. When struck, it vibrates at a steady frequency. This vibration generates sound waves that we can hear.
  • Oscillators are used in various devices like clocks, radios, and even musical instruments.
  • They help in controlling the pitch of the sound produced in these devices.
By understanding oscillators, we can predict and manipulate the sound they produce, such as altering the frequency to achieve a desired pitch.
Frequency
Frequency refers to the number of waves that pass a certain point every second. It's usually measured in Hertz (Hz).
In our exercise, the oscillator has a frequency of 1250 Hz. This means that it causes the sound waves to complete 1250 cycles per second.
  • Frequency dictates the pitch of the sound. Higher frequencies are perceived as higher-pitched sounds.
  • Frequency is a key component when calculating other aspects of sound waves, such as wavelength and velocity.
Understanding frequency allows us to adjust sounds for different environments and applications, being a critical factor in audio technologies and acoustics.
Wavelength
Wavelength is the distance between consecutive peaks鈥攐r troughs鈥攐f a wave. It's usually measured in meters or centimeters.
In sound waves, wavelength is directly related to the speed of sound and frequency. Based on the formula \[ v = f \times \lambda \],where 'v' is velocity, 'f' is frequency, and '\( \lambda \)' is wavelength, we see that wavelength can be calculated for a given speed and frequency.
  • Understanding the relationship of these elements allows us to manipulate sound propagation in various conditions.
  • For the oscillator at 1250 Hz and speed 325 m/s, the initial wavelength is 0.26 meters.
Managing wavelength is crucial in fields like music production and architectural acoustics, where sound behavior in spaces is significant.
Temperature Effect on Sound
Temperature can significantly affect the speed at which sound travels through a medium. As temperature increases, the speed of sound in air generally increases due to increased energy within the molecules.
This is because sound waves are mechanical waves that transfer energy through vibrations. When a gas, like air, is warmer, its molecules move more rapidly and transfer sound waves more efficiently.
  • The speed of sound is approximately proportional to the square root of the temperature (in Kelvins).
  • A change in temperature, therefore, affects the wavelength when the frequency remains constant.
In our exercise, adjusting the temperature allowed us to achieve a desired wavelength of 0.285 meters, showing how environmental adjustments can tailor sound properties for specific needs.

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Most popular questions from this chapter

CP The sound from a trumpet radiates uniformly in all directions in \(20^{\circ} \mathrm{C}\) air. At a distance of 5.00 \(\mathrm{m}\) from the trumpet the sound intensity level is 52.0 \(\mathrm{dB}\) . The frequency is 587 \(\mathrm{Hz}\) . (a) What is the pressure amplitude at this distance? (b) What is the displacement amplitude? (c) At what distance is the sound intensity level 30.0 \(\mathrm{dB}\) ?

A railroad train is traveling at 30.0 \(\mathrm{m} / \mathrm{s}\) in still air. The frequency of the note emitted by the train whistle is 262 \(\mathrm{Hz}\) . What frequency is heard by a passenger on a train moving in the opposite direction to the tirst at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first and (b) receding from the first?

Tuning a Violin. A violinist is tuning her instrument to concert \(\mathrm{A}(440 \mathrm{Hz}) .\) She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 3 \(\mathrm{Hz}\) , which increases to 4 \(\mathrm{Hz}\) when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3 -Hz beat? (b) To get her violin perfectly tuned to concert \(\Lambda,\) should she tighten or loosen her string from what it was when she heard the 3 -Hz beat?

A long tube contains air at a pressure of 1.00 atm and a temperature of \(77.0^{\circ} \mathrm{C}\) . The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 \(\mathrm{Hz}\) . Resonance is produced when the piston is at distances \(18.0,55.5,\) and 93.0 \(\mathrm{cm}\) from the open end. (a) From these measurements, what is the speed of sound in air at \(77.0^{\circ} \mathrm{C} ?\) (b) From the result of part (a), what is the value of \(\gamma ?\) (c) These data show that a displacement antinode is slightly outside of the open end of the tube. How far outside is it?

BIO The Vocal Tract. Many opera singers (and some pop singers) have a range of about 2\(\frac{1}{2}\) octaves or even greater. Suppose a soprano's range extends from A below middle \(\mathrm{C}\) (frequency 220 \(\mathrm{Hz}\) ) up to \(\mathrm{E}^{\mathrm{b}}\) -flat above high \(\mathrm{C}\) (frequency 1244 \(\mathrm{Hz}\) . Although the vocal tract is quite complicated, we can model it as a resonating air column, like an organ pipe, that is open at the top and closed at the bottom. The column extends from the mouth down to the diaphragm in the chest cavity, and we can also assume that the lowest note is the fundamental. How long is this column of air if \(v=354 \mathrm{m} / \mathrm{s} ?\) Does your result seem reasonable, on the basis of observations of your own body?
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