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The distance between a loudspeaker and the left ear of a listener is \(2.70 \mathrm{~m}\). (a) Calculate the time required for sound to travel this distance if the air temperature is \(20^{\circ} \mathrm{C}\). (b) Assuming that the sound frequency is \(523 \mathrm{~Hz}\), how many wavelengths of sound are contained in this distance?

Short Answer

Expert verified
(a) Time: 0.00787 s; (b) Wavelengths: 4.12

Step by step solution

01

Determine the speed of sound in air at 20°C

The speed of sound in air can be calculated using the formula \( v = 331.4 + 0.6 imes T \) where \( T \) is the temperature in degrees Celsius. For \( T = 20^{\circ} \text{C} \): \[ v = 331.4 + 0.6 \times 20 = 343.4 \, \text{m/s} \].
02

Calculate the time required for sound to travel 2.70 m

Use the formula \( t = \frac{d}{v} \), where \( d \) is the distance and \( v \) is the speed of sound. Here, \( d = 2.70 \, \text{m} \) and \( v = 343.4 \, \text{m/s} \): \[ t = \frac{2.70}{343.4} \approx 0.00787 \, \text{seconds} \].
03

Calculate the wavelength of the sound

The wavelength \( \lambda \) can be calculated with \( \lambda = \frac{v}{f} \), where \( f \) is the frequency of the sound. Here, \( v = 343.4 \, \text{m/s} \) and \( f = 523 \, \text{Hz} \): \[ \lambda = \frac{343.4}{523} \approx 0.656 \, \text{m} \].
04

Find the number of wavelengths in 2.70 m

Divide the total distance \( 2.70 \) by the wavelength \( 0.656 \): \[ N = \frac{2.70}{0.656} \approx 4.12 \]. This means there are approximately 4.12 wavelengths in this distance.

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Key Concepts

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

Understanding the Speed of Sound
The speed of sound is how fast sound waves travel through a medium. It's important to know that sound travels by compressing and expanding the particles it moves through. Whether it's air, water, or metal, each medium affects the speed of sound differently.

On a normal day at 20°C, the speed of sound in air is about 343.4 meters per second. This is because the vibrations move faster in warmer air due to the energized molecules.

To find this, you can use the formula: \[ v = 331.4 + 0.6 \times T \]where \( T \) is the temperature in Celsius. This formula helps us understand how temperature affects the sound speed: higher temperatures mean faster sound!

Choosing the right medium and considering temperature are vital factors in acoustic physics.
Measuring Sound Wavelength
Wavelength is like measuring the length of one complete cycle of a sound wave. You can think of it as the distance between two peaks in a wave.

Wavelength is connected to frequency and speed of sound. To calculate it, use: \[ \lambda = \frac{v}{f} \]where \( \lambda \) is wavelength, \( v \) is speed of sound, and \( f \) is frequency.

In this case, with a frequency of 523 Hz, you get a wavelength of approximately 0.656 meters.

Knowing the wavelength helps us determine how sound behaves in different environments and is crucial for designing acoustic spaces.
Exploring the Frequency of Sound
Sound frequency is all about how often the sound waves hit your ear per second. It's measured in cycles per second, or Hertz (Hz).

Frequency determines the pitch of the sound.
  • High frequency means a high pitch (like a whistle).
  • Low frequency means a low pitch (like a drum).
For example, a frequency of 523 Hz gives a clear, musical note.

This is essential in music and communications. Understanding sound frequency helps us tune instruments and decide the right sound for events.
Calculating Distance via Sound
Distance calculation using sound is an interesting application of physics. It's about finding out how far the sound has traveled or will travel.

When you know the speed of sound and how long it takes for the sound to reach you, you can calculate the distance traveled by the sound wave. The formula used is:\[ t = \frac{d}{v} \]where \( t \) is time, \( d \) is distance, and \( v \) is speed of sound.

In our example, the sound takes approximately 0.00787 seconds to travel 2.70 meters.

This calculation can be used in sonar technology, where sound waves are used to measure distances underwater.
The Fascination of Acoustic Physics
Acoustic physics deals with the study of sound, its production, transmission, and effects. It's a wide field that turns simple sound into complex science.

Acoustic physics helps us in many areas:
  • Designing concert halls for the best sound experience by manipulating echoes and reverberations.
  • Improving communication technology through better microphones and speakers.
  • Understanding natural phenomena like how animals use sound waves for communication.
The integration of speed, wavelength, frequency, and distance calculations forms the backbone of acoustic physics, demonstrating its relevance in our daily lives. Embracing this concept can enrich our understanding of the world of sound.

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Most popular questions from this chapter

Review Interactive Solution \(\underline{16.55}\) at for one approach to this problem. A dish of lasagna is being heated in a microwave oven. The effective area of the lasagna that is exposed to the microwaves is \(2.2 \times 10^{-2} \mathrm{~m}^{2}\). The mass of the lasagna is \(0.35 \mathrm{~kg},\) and its specific heat capacity is \(3200 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). The temperature rises by \(72 \mathrm{C}^{\circ}\) in 8.0 minutes. What is the intensity of the microwaves in the oven?

A steel cable of cross-sectional area \(2.83 \times 10^{-3} \mathrm{~m}^{2}\) is kept under a tension of \(1.00 \times 10^{4} \mathrm{~N}\). The density of steel is \(7860 \mathrm{~kg} / \mathrm{m}^{3}\) (this is not the linear density). At what speed does a transverse wave move along the cable?

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{~m}^{2},\) has a linear density of \(7.0 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of \(46 \mathrm{~m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). What will be the speed of the wave when the temperature is lowered by \(14 \mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{~m},\) a speed of \(5.2 \mathrm{~m} /\) \(\mathrm{s},\) and a frequency of \(14 \mathrm{~Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

Concept Questions Using \(1 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\) as a reference intensity, the threshold of hearing for an average young person is \(0 \mathrm{~dB}\). Two individuals who are not average have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{~dB}\) and \(\beta_{2}=+12.0 \mathrm{~dB} .(\mathrm{a})\) Is the sound intensity at a level of \(-8.00 \mathrm{~dB}\) greater than, or less than, that at \(0 \mathrm{~dB} ?\) (b) Which individual has the better hearing? Why? Problem How many times greater (or less) is the sound intensity when person 1 \(\left(\beta_{1}=-8.00 \mathrm{~dB}\right)\) hears it at the threshold of hearing compared to when person 2 \(\left(\beta_{2}=+12.0 \mathrm{~dB}\right)\) hears it at the threshold of hearing? Check to see that your answer is consistent with your answers to the Concept Questions.

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