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

In summary, using the given information, the time it would take a sound wave to travel 150 m over a path where the temperature rises linearly from 5 degrees to 15 degrees is 0.445 seconds. This is calculated by integrating the equation dt = dx/v and substituting the given values, resulting in a total time of 0.67 seconds. However, after correcting the lower boundary to 5 degrees, the correct answer is obtained. Additionally, the use of the formula v = d/t can also be used to calculate the distance of a thunderstorm from an observer, making Physics an exciting subject.
  • #1
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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 m


v=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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  • #2
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 m


v=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° ?
 
  • #3
Yup, that's it. the lower boundary should be 5 degrees. When I recalculate, I get the correct answer. Thank you sir!
 
  • #4
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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