Sound Wave Problem -- time of flight in air versus in water

AI Thread Summary
The discussion revolves around understanding the propagation of sound waves in different media, specifically air and water. Participants clarify that sound travels as a longitudinal wave and that the time of flight for sound reaching both a diver underwater and a person on land can be analyzed using the equation t = d/v. It is emphasized that the velocities of sound differ in air and water, necessitating separate calculations for each medium. The conversation includes the importance of visualizing the wavefront and recognizing the total distance sound travels, which is not simply additive due to the differing paths taken in each medium. Ultimately, the key takeaway is that the time taken for sound to reach both observers can be equated to solve for unknowns in the problem.
Selfless_Gene
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Homework Statement


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Homework Equations



v = d/t
Solve for t. t = d/v

The Attempt at a Solution


In my General Physics 2 course we are doing sound waves I have the answer to the problem which is 90.8m I am trying to understand the concepts of sound wave. So please correct me if I am wrong,

1. The sound from the horn is propagating as a longitudinal wave in all directions.
2. The Diver and and the person on ground hears the sound at the same time. Does this mean we can use this equation t=d/v because the time is the same for both the diver and the person on the ground?

3. I am unsure of the math behind this problem.
 

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Selfless_Gene said:
Does this mean we can use this equation t=d/v
It depends how you are using that. Note that the sound to the diver travels through two different media and the velocities are different. You need to look up what those velocities are.
You are not told the height above water of the ears of the person on dry land, so take the 22m as their distance from the horn.
 
Is it just me, or is that diver way out of her depth?

(I got an answer even deeper than the OPs.)
 
haruspex said:
It depends how you are using that. Note that the sound to the diver travels through two different media and the velocities are different. You need to look up what those velocities are.
You are not told the height above water of the ears of the person on dry land, so take the 22m as their distance from the horn.
So when I subtract the time t2 which is the time the sound of the horn hits the surface of the water by t1 which the time that the sound of the horn reaches the person on the ground. This can tell me the time the sound travels in the water? I don't understand this portion of the problem its in the picture below listed as t1-t2 time for which the sound travels through water. I don't understand the relationship and how can subtracting t1-t2 tell us the time the sound travels through the water.

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Selfless_Gene said:
how can subtracting t1-t2 tell us the time the sound travels through the water
When the wave front that heads down from the horn towards the diver reaches the water, how far has the wave front traveling horizontally to the land gone?
(And how did 1.6m become 2m?)
DaveC426913 said:
is that diver way out of her depth?
Rather deep for normal recreational diving, but I believe the record is over 300m, and that was in seawater.
 
haruspex said:
When the wave front that heads down from the horn towards the diver reaches the water, how far has the wave front traveling horizontally to the land gone?
(And how did 1.6m become 2m?)

Rather deep for normal recreational diving, but I believe the record is over 300m, and that was in seawater.
This is the same problem just with different values. I don't understand what you mean by the traveling wave front. The sound wave is propagating in all directions. The wave travels from the horn to the surface of the water which is 1.6 m and horizonally the same wave front also travels 22m to the horizontally.
 
Selfless_Gene said:
This is the same problem just with different values. I don't understand what you mean by the traveling wave front. The sound wave is propagating in all directions. The wave travels from the horn to the surface of the water which is 1.6 m and horizonally the same wave front also travels 22m to the horizontally.
You can think of it either way. The question is the same. When it reaches the surface of the water, how far has it still to go to reach the person on land?
 
haruspex said:
You can think of it either way. The question is the same. When it reaches the surface of the water, how far has it still to go to reach the person on land?
the distance it has to travel is 22 m since the person on land is is 22 m away.
 
Selfless_Gene said:
the distance it has to travel is 22 m since the person on land is is 22 m away.
That's the total distance from the horn, but that is not what I asked.
Draw a diagram, showing the whole wavefront when it first touches the water. How much further does it have to go to reach the landlubber?
 
  • #10
haruspex said:
That's the total distance from the horn, but that is not what I asked.
Draw a diagram, showing the whole wavefront when it first touches the water. How much further does it have to go to reach the landlubber?

I drew a diagram. Any did some analysis on this what i observed was the following The wave front traveling from the surface of the water has to move up 1.6m and horizontally 22m to reach the landlubber. So the distance from the wave front touching the water to the person is 23.6m
 
  • #11
Selfless_Gene said:
I drew a diagram. Any did some analysis on this what i observed was the following The wave front traveling from the surface of the water has to move up 1.6m and horizontally 22m to reach the landlubber. So the distance from the wave front touching the water to the person is 23.6m
Why would the sound go from the horn straight down to the water then bounce up and across to reach the land? What is wrong with the straight line from source to receiver?
As I wrote, you can treat it as two separate wavefronts leaving at the same time and place but heading in different directions, or as a single wave front spreading out. If you take the second view, you should have drawn concentric circles around the horn.
 
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  • #12
haruspex said:
Why would the sound go from the horn straight down to the water then bounce up and across to reach the land? What is wrong with the straight line from source to receiver?
As I wrote, you can treat it as two separate wavefronts leaving at the same time and place but heading in different directions, or as a single wave front spreading out. If you take the second view, you should have drawn concentric circles around the horn.
I think I am understanding better because you mention concentric circles the waves will go directly to the person the ground from the horn. Correct me if I am wrong but the total distance the wave traveling in the air is 23.6 m, then the wave touches the water after this distance. I don't understand why we have to do t1-t2 = the time wave travels in water. Wouldn't it be t1+t2?
 
  • #13
Selfless_Gene said:
the waves will go directly to the person the ground from the horn.
Right.
Selfless_Gene said:
the total distance the wave traveling in the air is 23.6 m
No.
As I wrote in post #2, we do not know the height of the person's ears relative to the horn, so you will have to assume they are at the same height. The sound that goes from the horn to those ears will take a straight line, so it is 22m.
Draw a horizontal line from horn to ears and label it 22m.

That sound leaves the horn at the same instant that it also sets out straight down towards the diver. At time t2 the sound reaches the surface of the water. To represent the wavefront at this instant draw a circle centred on the horn and touching the surface of the water. What is the radius of this circle?

As the part of the wave entering the water continues down to the diver, the part moving horizontally continues towards the land. How much further does it have to travel to land after time t2?
 
  • #14
One way to do this is to set it up as two equations with equivalence.

Both observers hear the sound at the same time.
Therefore, the mtine to Landlubber equals the time to Diver.
TL = TD

TL is easily calculated.
TD is the sum of the time in air (easily calced) and the time in water (has one unknown).
Solve for the unknown.
 
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