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Chapter 23: Problem 23

What is the speed of sound in air when the air temperature is 31 \({ }^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The speed of sound in air at 31掳C is 349.9 m/s.

Step by step solution

01

Understanding the Formula for Speed of Sound

The speed of sound in air can be approximated using the formula \( v = 331.3 + 0.6T \), where \( v \) is the speed of sound in meters per second, and \( T \) is the temperature in degrees Celsius.
02

Substituting Temperature into the Formula

Substitute the given temperature, 31 \( ^{ ext{掳C}} \), into the formula: \( v = 331.3 + 0.6 imes 31 \).
03

Calculating the Product

Calculate the product of 0.6 and 31: \( 0.6 imes 31 = 18.6 \).
04

Adding to the Base Value

Add the result from the previous step to the base speed of 331.3: \( 331.3 + 18.6 = 349.9 \).
05

Final Answer

Conclude that the speed of sound in air at 31掳C is \( 349.9 ext{ m/s} \).

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Key Concepts

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

Physics Formula
The behavior of sound as it travels through the air is described by a physics formula that helps us calculate the speed of sound based on temperature. This important formula is expressed as \( v = 331.3 + 0.6T \), where \( v \) represents the speed of sound in meters per second and \( T \) is the air temperature in degrees Celsius. This formula provides a mathematical approach to understanding how temperature affects the speed at which sound waves move. The base speed of 331.3 m/s reflects the speed of sound at 0 degrees Celsius. For every degree Celsius increase in temperature, the speed of sound increases by 0.6 meters per second. This straightforward equation allows us to quickly compute sound speed in varying thermal conditions, facilitating scientific and practical applications alike.
Temperature Effect on Sound
Temperature plays a significant role in determining how quickly sound waves can travel through the air. As temperature increases, so does the energy of air particles. This results in faster movement and collisions between particles, which in turn allows sound waves to propagate more rapidly. This is why the speed of sound increases with rising temperatures.
In contrast, when the temperature drops, air particles move more slowly and have less energy, causing sound waves to travel more slowly. For instance, on a hot day at 31掳C, sound travels faster compared to a cold one.
  • Warm air speeds up sound.
  • Cool air slows down sound.
By understanding the connection between temperature and sound speed, we can predict and adjust for the changes in sound transmission in different weather conditions.
Sound Speed in Air
The speed of sound in air is an essential concept in both physics and engineering. It's vital for a wide variety of applications, from music to meteorology. The speed is not constant and depends mainly on the temperature of the air through which it is traveling.
At 31掳C, using our formula, the speed of sound is approximately 349.9 m/s. This indicates that sound travels quite quickly under such conditions, due to the high activity and energy in air molecules.
  • At 0掳C, sound speed is about 331.3 m/s.
  • Each degree rise in temperature increases speed by 0.6 m/s.
Knowing the speed of sound for various temperatures is crucial when predicting how sound will behave, making it useful for designing acoustic equipment, predicting weather, and even in GPS technology that relies on sound waves for positioning. Understanding sound speed builds the bridge to more advanced studies in acoustics and wave physics.

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Most popular questions from this chapter

A car moving at \(20 \mathrm{~m} / \mathrm{s}\) with its horn blowing \((f=1200 \mathrm{~Hz})\) is chasing another car going at \(15 \mathrm{~m} / \mathrm{s}\) in the same direction. What is the apparent frequency of the horn as heard by the driver being chased? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). This is a Doppler problem. Draw the observer-to-source arrow; that's the positive direction. Both the source and the observer are moving in the negative direction. Hence, we use \(-\mathrm{u}_{o}\) and \(-\mathrm{v}_{\mathrm{s}}\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(1200 \mathrm{~Hz}) \frac{340-15}{340-20}=1.22 \mathrm{kHz} $$ Because the source is approaching the observer, the latter will measure an increase in frequency.

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\). Using Eq. (23.2) $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter \(20 .\)

An autofocusing camera sends out a pulse of ultrasound and determines distance from the time of return. Roughly what is the operative speed of sound? How far is the subject from the camera if the time interval between launch and return is \(8.00 \mathrm{~ms}\) when the temperature of the air is \(23.50^{\circ} \mathrm{C}\) ?

If the intensity of sound changes, the sound level will change. Suppose then that the sound level goes from \(\beta_{i}\) to \(\beta_{f}\) such that \(\beta_{f}-\) \(\beta_{i}=\Delta \beta .\) Show that $$ \Delta \beta=10 \log _{10}\left(I_{f} / I_{i}\right) $$ [ Hint: Remember that \(\log _{10} \mathrm{~A}-\log _{10} \mathrm{~B}=\log _{10} \mathrm{~A} / \mathrm{B} .\) ]

A person riding a power mower may be subjected to a sound of intensity \(2.00 \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2}\). What is the intensity level to which the person is subjected?
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