/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 A bat emits a sound whose freque... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(343 \mathrm{~m} / \mathrm{s} .\) Find the wavelength of the sound.

Short Answer

Expert verified
The wavelength of the sound is approximately \(3.84 \mathrm{~mm}\).

Step by step solution

01

Determine the Speed of Sound

The speed of sound in air changes with temperature and can be calculated using the formula:\[ v = v_0 \sqrt{\frac{T}{T_0}} \]where \(v_0 = 343 \mathrm{~m/s}\) is the speed of sound at \(20^{\circ} \mathrm{C}\), \(T\) is the absolute temperature in Kelvin at the desired condition (\(35^{\circ} \mathrm{C}\)), and \(T_0\) is the absolute temperature in Kelvin at \(20^{\circ} \mathrm{C}\). First, convert the temperatures to Kelvin: \[ T = 35^{\circ} \mathrm{C} + 273.15 = 308.15 \mathrm{~K} \] \[ T_0 = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{~K} \]Now substitute these into the formula:\[ v = 343 \sqrt{\frac{308.15}{293.15}} \approx 349.2 \mathrm{~m/s} \]
02

Apply the Wave Equation Formula

We use the wave equation to find the wavelength, which is:\[ \lambda = \frac{v}{f} \]where \(v\) is the speed of sound calculated in Step 1, and \(f = 91000 \mathrm{Hz}\) is the frequency of the sound emitted by the bat.Substituting the known values:\[ \lambda = \frac{349.2 \mathrm{~m/s}}{91000 \mathrm{~Hz}} \approx 0.00384 \mathrm{~m} \]
03

Finalize the Result

The calculations show that the wavelength of the sound emitted by the bat at \(35^{\circ} \mathrm{C}\) is approximately \(0.00384 \mathrm{~m}\) or \(3.84 \mathrm{~mm}\). This reflects the relationship between speed, frequency, and wavelength.

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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 when dealing with wave propagation through different media. It significantly differs depending on the medium in which the sound is traveling, such as air, water, or solids. In air, the speed of sound is influenced by the temperature. This is because sound travels by compressing and expanding air molecules, and these molecules move more quickly at higher temperatures.
For a standard reference, the speed of sound in air at a temperature of 20°C is approximately 343 meters per second (m/s). However, as temperature varies, so does the speed. The formula to calculate this speed at different temperatures is given by:
  • \[ v = v_0 \sqrt{\frac{T}{T_0}} \]
Where:
  • \( v_0 \) is the standard speed of sound at 20°C, which is 343 m/s.
  • \( T \) is the absolute temperature at the desired condition, measured in Kelvin.
  • \( T_0 \) is the absolute temperature at 20°C, also in Kelvin.
By using this formula, we can accurately determine the speed of sound at any given temperature in air, thereby aiding in many practical applications such as acoustics and sonar.
Wave Equation
The wave equation is fundamental for understanding sound waves and their behavior. It expresses the relationship between the speed of the wave, its frequency, and its wavelength. This relationship is vital for calculations involving sound waves emitted by sources such as musical instruments or animals like bats.
The standard wave equation is:
  • \[ \lambda = \frac{v}{f} \]
Where:
  • \( \lambda \) represents the wavelength of the wave.
  • \( v \) is the speed of the wave.
  • \( f \) is the frequency of the wave.
By understanding this equation, we can calculate any one of these three variables if the other two are known. This is particularly useful in physics and engineering, where understanding the behavior of waves is crucial.
Temperature Conversion
Temperature conversion is a necessary step when dealing with formulas that use Kelvin, as many foundational physics formulas are based on absolute temperatures. Celsius, Fahrenheit, and Kelvin are the three main scales.
To convert a temperature from Celsius to Kelvin, which is often necessary in physics problems, use the following formula:
  • \[ T(K) = T(^{\circ}C) + 273.15 \]
This conversion is essential because Kelvin is the SI unit for temperature and absolute zero, the point where no heat energy remains in a substance, is 0 Kelvin. Understanding these conversions allows for accurate calculation of properties like the speed of sound at any given temperature.
Wavelength Calculation
Calculating the wavelength of a sound wave requires understanding both the speed of the sound within its medium and its frequency. The wavelength is the distance between successive crests of the wave, and it gives insight into how sound propagates over distances.
Given the speed of sound (\( v \)) and frequency (\( f \)), the formula for calculating wavelength \( \lambda \) is:
  • \[ \lambda = \frac{v}{f} \]
In practical applications, such as determining the wavelength of a bat's echolocation signal, knowing these two parameters can paint a detailed picture of how effectively the sound waves will travel and reflect off objects. The outcome is beneficial for understanding navigation and communication in both natural and engineered systems.

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Most popular questions from this chapter

Humans can detect a difference in sound intensity levels as small as \(1.0 \mathrm{~dB}\). What is the ratio of the sound intensities?

Concept Questions Example 4 in the text discusses an ultrasonic ruler that displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose you use this ruler underwater, rather than in air. (a) Is the speed of sound in water greater than, less than, or equal to the speed of sound in air? (b) Is the reading on the ruler greater than, less than, or equal to the actual distance? Provide reasons for your answers. Problem The actual distance from the ultrasonic ruler to an object is \(25.0 \mathrm{~m} .\) The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\) and \(\rho=1025\) \(\mathrm{kg} / \mathrm{m}^{3},\) respectively. Assume that the ruler uses a preset value of \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air, and determine the distance reading on its display. Verify that your answer is consistent with your answers to the Concept Questions.

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?

The middle \(C\) string on a piano is under a tension of \(944 \mathrm{~N}\). The period and wavelength of a wave on this string are \(3.82 \mathrm{~ms}\) and \(1.26 \mathrm{~m}\), respectively. Find the linear density of the string.

Refer to at for one approach to this type of problem. Two trucks travel at the same speed. They are far apart on adjacent lanes and approach each other essentially head-on. One driver hears the horn of the other truck at a frequency that is \(1.14\) times the frequency he hears when the trucks are stationary. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). At what speed is each truck moving?

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