# Calculating distance using the speed of sound through two different mediums

• Joe Schmoe
In summary, David hears a beaver sap the water with its tail at a velocity of 1400 meters/second and then hears the sound again 0.95 seconds later at a velocity of 340 meters/second. The distance between David and the beaver can be found by solving the equation d/340=d/1400+0.95, where d represents the unknown distance.
Joe Schmoe

## 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.

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:
no, sorry, I mean: d/340=d/1400+0.95

Joe Schmoe said:
no, sorry, I mean: d/340=d/1400+0.95

You can edit your posts.

You mean like this: 340=d/(d/1400)+0.95

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?

So would the answer be 690m?

OK, my bad. I am just unsure at how to rearrange d/340=d/1400+0.95 to isolate d. Any ideas?

## 1. How is the speed of sound through two different mediums calculated?

The speed of sound through two different mediums can be calculated by dividing the distance traveled by the time it takes for the sound to travel through the medium.

## 2. What factors affect the speed of sound through a medium?

The speed of sound through a medium is affected by the temperature, density, and elasticity of the medium. Higher temperatures, lower densities, and higher elasticity will result in a faster speed of sound.

## 3. How does the speed of sound through air compare to the speed of sound through water?

The speed of sound through air is significantly slower than the speed of sound through water. This is because air is less dense and less elastic than water, resulting in a slower propagation of sound waves.

## 4. Can the speed of sound through a medium change?

Yes, the speed of sound through a medium can change depending on the conditions of the medium. For example, the speed of sound through air will increase with higher temperatures, as the air molecules will be moving faster and therefore, sound waves can travel faster.

## 5. How is the distance between two points calculated using the speed of sound through two different mediums?

The distance between two points can be calculated by multiplying the speed of sound through the first medium by the time it takes for sound to travel through that medium, and then multiplying the speed of sound through the second medium by the time it takes for sound to travel through that medium. The sum of these two distances will give the total distance between the two points.

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