Depth of a well (using speed of sound)

Click For Summary

Homework Help Overview

The problem involves determining the depth of a well based on the time it takes for a stone to fall and the sound of the splash to return. The context includes the speed of sound and its relation to the temperature of the air.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the role of the speed of sound and question the assumptions made about the time taken for the stone to fall versus the time for the sound to travel back. There is an exploration of the equations involved and the need for a proper understanding of the motion of the stone.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions, and exploring different interpretations of the time taken for each part of the scenario. Some guidance has been offered regarding the need to separate the time for the stone's fall and the sound's travel.

Contextual Notes

There is uncertainty regarding the correct substitution into the equations and the interpretation of the total time of 2 seconds, as well as the implications of the stone's motion and the speed of sound.

halo168
Messages
12
Reaction score
0

Homework Statement


A stone is dropped from rest into a well. Th sound of the splash is heard exactly 2.00 s later. Find the depth of the well if the air temperature is 10.0 degrees Celsius.

Homework Equations


  • How does the speed of sound play a role in this?
  • How can I find the depth?

The Attempt at a Solution


I used v= [(331)*sqrt(1 + T/273)]. If v = (lambda)/t, then won't lambda equal [(331)*sqrt(1 + T/273)] * t?
I assumed that t would equal 1 s because it takes 1 s for the stone to hit the bottom and 1 s for the echo to be heard (2s total). Is that a correct assumption? As a result, my answer was that lambda = 337.01 m but the correct well depth were supposed to be 18.5 m...

What did I do wrong?
 
Physics news on Phys.org
The stone accelerates as it falls, starting from zero speed. It takes some time to fall to the bottom of the well. The sound it makes when it hits the water travels back up the well at the speed of sound, which is a constant speed and much faster than the stone ever was moving. So the time for the two paths (stone falling, sound rising) is not the same.
 
  • Like
Likes   Reactions: halo168
If it took the same time for the stone to hit the water as it took for the sound to reach you, that means that the stone broke the speed of sound!

You need an equation to account for the position of the stone as a function of time.
 
  • Like
Likes   Reactions: halo168
The main problem is that I am not sure why the depth is 18.5 m while I had solved for lambda and got 337.01 m. What did I do wrong in terms of substitution into the equation?
 
You don't know how long it took for the sound travel up the well. You have to find at what time the stone hit the bottom.
 
halo168 said:
The main problem is that I am not sure why the depth is 18.5 m while I had solved for lambda and got 337.01 m. What did I do wrong in terms of substitution into the equation?
You right if you're using echo as distance measurement.
 
Dr Claude has your answer. It is a two part problem. You must solve part 1 first.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
13K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
3
Views
4K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
3
Views
2K