Calculating distance using the speed of sound through two different mediums

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Homework Help Overview

The problem involves calculating the distance from David to a beaver based on the time it takes for sound to travel through two different mediums: water and air. The speeds of sound in these mediums are given, along with a time difference for the sound's arrival.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between distance, speed, and time, with some attempting to express the travel times for sound in both mediums using a variable for distance. There are questions about how to set up the equations correctly and what the implications of the time difference are.

Discussion Status

Participants have made several attempts to formulate equations based on the problem's parameters, with some expressing uncertainty about the next steps. There is ongoing exploration of the relationship between the two travel times and how to isolate the distance variable.

Contextual Notes

There is a noted confusion regarding the interpretation of the time difference and how it relates to the distances traveled in each medium. Participants are working under the assumption that the total distance is the same for both paths taken by the sound.

Joe Schmoe
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Homework Statement



David is swimming when he hears a beaver sap the water with its tail, first through the water (v= 1400 meters/second) and then 0.95 seconds later (after the sound reaches him through the water) through the air (v= 340 meters/second).


Homework Equations


Find the distance from David to the beaver.


The Attempt at a Solution


This one stumped me. I thought maybe I 'd have to compare the two velocities and do some cross multiplication but I really don't know.
 
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Joe Schmoe said:

Homework Statement



David is swimming when he hears a beaver sap the water with its tail, first through the water (v= 1400 meters/second) and then 0.95 seconds later (after the sound reaches him through the water) through the air (v= 340 meters/second).


Homework Equations


Find the distance from David to the beaver.


The Attempt at a Solution


This one stumped me. I thought maybe I 'd have to compare the two velocities and do some cross multiplication but I really don't know.

What equations do you think apply here? If you happened to know the distance and velocity, how would you find the time? Write the expressions for the time of travel for both modes of transport (air, water) using a variable to represent the unknown distance.
 
Well, I obviously know that d=t×v. This would mean that in the air the sound travels 323 m (d= 0.95×340). The problem is that isn't the total distance, there remains an additional distance that I can't figure out. If I sub in the time value into an equation with the sound velocity in water (d=0.95×1400) then the sound will have traveled 1330 m. But I am sure that neither of these two values are the actual distance. There remains another step but I am unsure what to do.
 
The distance traveled in both mediums is a constant. 0.95 seconds is a relation between the time taken by the sound wave in both mediums to cover this distance.
 
Last edited:
Joe Schmoe said:
Well, I obviously know that d=t×v. This would mean that in the air the sound travels 323 m (d= 0.95×340).

The problem is that isn't the total distance, there remains an additional distance that I can't figure out. If I sub in the time value into an equation with the sound velocity in water (d=0.95×1400) then the sound will have traveled 1330 m. But I am sure that neither of these two values are the actual distance. There remains another step but I am unsure what to do.
No, the 0.95 seconds is the difference in time between the arrival of the sound by the two paths.

There is only one unknown distance, the distance between the beaver and the person. Call that distance d. Write the expressions for the two arrival times of the sound assuming distance is d; so t1 = ?, t2 = ?.
 
OK, so t1 (water)= d/1400
t2 (air)= d/340
 
Joe Schmoe said:
OK, so t1 (water)= d/1400
t2 (air)= d/340

Okay. Now, what is the relationship between t1 and t2? (This is where the .95 seconds comes into play).
 
Then, t2=t1+0.95
so: d/1400=d/340+0.95
 
[STRIKE]Yes. Now write another equation using speed=distance/time.[/STRIKE]
 
Last edited:
  • #10
no, sorry, I mean: d/340=d/1400+0.95
 
  • #11
Joe Schmoe said:
no, sorry, I mean: d/340=d/1400+0.95

You can edit your posts. :smile:
 
  • #12
You mean like this: 340=d/(d/1400)+0.95
 
  • #13
Joe Schmoe said:
You mean like this: 340=d/(d/1400)+0.95

Using your equation d/340=d/1400+0.95, can you not solve for d?
 
  • #14
So would the answer be 690m?
 
  • #16
OK, my bad. I am just unsure at how to rearrange d/340=d/1400+0.95 to isolate d. Any ideas?
 

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