How long would it take a sound wave to travel 150m

AI Thread Summary
The discussion revolves around calculating the time it takes for a sound wave to travel 150 meters in air, considering a temperature increase from 5 to 15 degrees Celsius. The speed of sound in air is given as 331 m/s at 0 degrees Celsius, increasing by 0.590 m/s for each degree rise in temperature. The calculations involve integrating the velocity over the distance, resulting in a time of approximately 0.67 seconds, while the book states the answer is 0.445 seconds. A participant clarifies the initial temperature should be 5 degrees, leading to a recalculation that aligns with the expected answer. The excitement of applying physics to real-world scenarios, such as estimating the distance of a thunderstorm, is also expressed.
rpthomps
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Homework Statement

The sound speed in air at 0 degrees Celsius is 331 m/s, and for temperatures within a few tens of degrees of 0 degrees celsius it increases at the rate 0.590 m/s for every degree celsius increase in temperature. How long would it take a sound wave to travel 150 m over a path where the temperature rises linearly from 5 degrees to 15 degrees at the other end?

Homework Equations

d= 150 mv=331+0.59T##T=(\frac{1}{15})x+5\\dT=\frac{1}{15}dx##

The Attempt at a Solution



##dt=\frac { dx }{ v }\\T_{ total }=\int _{ 0 }^{ 150 }{ \frac { dx }{ v } }\\=\int _{ 0 }^{ 150 }{ \frac { 15dT }{ 331+0.59T } }\\=\frac { 15 }{ 0.59 } ln(331+0.59T){ | }_{ 0 }^{ 15 }=0.67##Back of the book says .445 s
 
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rpthomps said:

Homework Statement

The sound speed in air at 0 degrees Celsius is 331 m/s, and for temperatures within a few tens of degrees of 0 degrees celsius it increases at the rate 0.590 m/s for every degree celsius increase in temperature. How long would it take a sound wave to travel 150 m over a path where the temperature rises linearly from 5 degrees to 15 degrees at the other end?

Homework Equations

d= 150 mv=331+0.59T##T=(\frac{1}{15})x+5\\dT=\frac{1}{15}dx##

The Attempt at a Solution



##dt=\frac { dx }{ v }\\T_{ total }=\int _{ 0 }^{ 150 }{ \frac { dx }{ v } }\\=\int _{ 0 }^{ 150 }{ \frac { 15dT }{ 331+0.59T } }\\=\frac { 15 }{ 0.59 } ln(331+0.59T){ | }_{ 0 }^{ 15 }=0.67##Back of the book says .445 s
Is the initial temperature 0° ?
 
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Yup, that's it. the lower boundary should be 5 degrees. When I recalculate, I get the correct answer. Thank you sir!
 
how thrilling is it that using a simple formula v = d/t, you can calculate how far a thunderstorm is from you! So exciting! I love Physics!
 
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