Calculating the Width of a Valley Using Echo Time Delay

  • Thread starter Thread starter tdk
  • Start date Start date
  • Tags Tags
    Properties Sound
Click For Summary
SUMMARY

The problem involves calculating the width of a valley based on the time delay of echoes from vertical walls. The rifle shot is heard as an echo 2.0 seconds after firing, and the second echo is heard 2.0 seconds after the first. Given the speed of sound at 340 meters per second, the total time for the sound to travel to both walls and back is 4.0 seconds. Therefore, the width of the valley is calculated as 340 meters/second multiplied by 2 seconds, resulting in a width of 340 meters.

PREREQUISITES
  • Understanding of sound wave propagation and speed
  • Basic knowledge of time-distance calculations
  • Familiarity with echo phenomena
  • Ability to interpret and create diagrams for physics problems
NEXT STEPS
  • Study the principles of sound wave reflection and echo
  • Learn about the speed of sound in different mediums
  • Explore time-distance problems in physics
  • Practice drawing diagrams to visualize physics problems
USEFUL FOR

Students studying physics, educators teaching sound wave concepts, and anyone interested in practical applications of echo and sound propagation in real-world scenarios.

tdk
Messages
13
Reaction score
0

Homework Statement



"A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 2.0 s after the rifle was fired. The echo from the other wall is heard 2.0 s after the first echo. How wide is the valley?

Homework Equations



I don't know any relevant equations. v=wavelength x frequency? Do I have to convert something into decibels?

The Attempt at a Solution



time=2 s
distance=?
.
 
Last edited by a moderator:
Physics news on Phys.org
No nothing that cpplex--just that the speed of sound is 340meters/sec. Hint: to avoid simple errors, draw a picture of the reflecting sound waves so as to include all the distances.
 
But I don't understand how you apply the speed of sound to the problem. Do you multiply it by 2?.
 
Last edited by a moderator:
know leave the speed alone and as i suggested draw a diagram. In this case, try to isolate the time requires for sound to span the valley. Hint its 1/2 the sum...
 
alright pugfug, I told you I unclicked the "Automatically parse links in text" and that's why the link wasnt clickable anymore. And now everytime I post I unclick it so it doesn't come up as a link. Do you understand that? I don't get why I can't post here if the link isn't even clickable, what's the harm?

Thanks Denverdoc I'll try that.
 
Last edited by a moderator:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
10
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K