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Chapter 16: Problem 98

A bat emits a sound whose frequency is \(91 \mathrm{kHz}\). The speed of sound in air at \(20.0^{\circ} \mathrm{C}\) is \(343 \mathrm{m} / \mathrm{s} .\) However, the air temperature is \(35^{\circ} \mathrm{C},\) so the speed of sound is not \(343 \mathrm{m} / \mathrm{s}\). Assume that air behaves like an ideal gas, and find the wavelength of the sound.

Short Answer

Expert verified
The wavelength of the sound is approximately 0.00388 m.

Step by step solution

01

Determine the Speed of Sound at 35掳C

The speed of sound in air as a function of temperature (in Celsius) is given by the formula \( v = 331.4 + 0.6T \), where \( T \) is the temperature in degrees Celsius. Here, \( T = 35 \). Plugging this into the formula gives: \[ v = 331.4 + 0.6 \times 35 = 352.9 \, \mathrm{m/s} \]
02

Use the Wave Equation to Find the Wavelength

The wavelength \( \lambda \) of a wave is given by the equation \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound you just calculated, and \( f \) is the frequency of the sound. Here, \( f = 91 \, \mathrm{kHz} = 91,000 \, \mathrm{Hz} \). Substitute the values into the equation:\[ \lambda = \frac{352.9}{91,000} \approx 0.003877 \, \mathrm{m} \]

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Key Concepts

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

Speed of Sound
The speed of sound is an important concept in acoustics, referring to how fast sound waves travel through a medium, like air. At a standard temperature of 20掳C, sound travels at 343 meters per second.
However, the speed of sound is not constant and varies with temperature. It's crucial to consider this variation when dealing with real-world applications of sound.
  • Temperature Dependence: The speed of sound increases with an increase in air temperature. This happens because molecules move more quickly in warmer air, allowing sound waves to travel faster.
  • Calculation Formula: To find the speed of sound at any given temperature in Celsius, use the formula: \[ v = 331.4 + 0.6T \]where \( v \) is the speed of sound in "m/s" and \( T \) is the temperature in degrees Celsius.
Applying this, at 35掳C, the speed is 352.9 m/s, which is faster than at 20掳C.
Ideal Gas Law
The ideal gas law is a fundamental equation in physics and chemistry that relates the pressure, volume, temperature, and number of moles of a gas. Although not directly used to calculate the speed of sound in the provided example, it underpins why temperature influences sound speed.
  • Formula: The ideal gas law is expressed as:\[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
  • Relation to Sound: In acoustics, the behavior of sound waves in gases can be predicted using the properties of gases as described by this law. It implies the effect of temperature on gas density and pressure, affecting sound speed.
The connection lies in how the speed of sound changes with temperature, due to changes in air pressure and density as described by this law.
Wavelength Calculation
Understanding wavelength involves relating the distance a wave covers over one complete cycle to its speed and frequency. Using the wave equation, wavelength is determined by dividing the speed of sound, \( v \), by frequency, \( f \).
  • Wavelength Formula: \[ \lambda = \frac{v}{f} \]where \( \lambda \) is the wavelength in meters.
  • Example Calculation: For a bat emitting sound at a frequency of 91 kHz at a speed of 352.9 m/s:\[ \lambda = \frac{352.9}{91,000} \approx 0.003877 \text{ meters} \]
This calculation gives you the length of one wave in meters, crucial in understanding the acoustics involved in real-world scenarios.
Frequency
Frequency is central to analyzing sound. It is the number of cycles a sound wave completes in one second, measured in Hertz (Hz).
In the context of the problem, the frequency of the bat's sound is 91 kHz, or 91,000 Hz.
  • Relation to Pitch: Higher frequency sounds are perceived as higher pitches. Conversely, lower frequency sounds appear as lower-pitched notes.
  • Impact on Wavelength: Frequency inversely affects wavelength; as frequency increases, wavelength decreases for a constant speed.
Understanding frequency enhances comprehension of sound characteristics, especially when exploring natural phenomena like echolocation in bats.

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Most popular questions from this chapter

A middle-aged man typically has poorer hearing than a middle-aged woman. In one case a woman can just begin to hear a musical tone, while a man can just begin to hear the tone only when its intensity level is increased by \(7.8 \mathrm{dB}\) relative to the just-audible intensity level for the woman. What is the ratio of the sound intensity just detected by the man to the sound intensity just detected by the woman?

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{m} / \mathrm{s},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C},\) how many times greater is the speed of sound in the rail than in the air?

(a) A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed \(v\) of the wave on the rope in terms of the distance \(y\) above the bottom end of the rope and the magnitude \(g\) of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{m}\) and \(2.0 \mathrm{m}\) above the bottom end of the rope.

Using an intensity of \(1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\) as a reference, the threshold of hearing for an average young person is 0 dB. Person 1 and person 2 , who are not average, have thresholds of hearing that are \(\beta_{1}=-8.00 \mathrm{dB}\) and \(\beta_{2}=+12.0 \mathrm{dB}\). What is the ratio \(I_{1} / I_{2}\) of the sound intensity \(I_{1}\) when person 1 hears the sound at his own threshold of hearing compared to the sound intensity \(I_{2}\) when person 2 hears the sound at his own threshold of hearing?
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