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Quick question about sound

  • Thread starter Zondrina
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Zondrina

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1. The problem statement, all variables and given/known data

A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water and the air (340 m/s) simultaneously.

The echo in the water takes 0.4s to return. How much additional time will it take for the echo in the air to return?

2. Relevant equations

##v = \frac{Δd}{Δt}##

3. The attempt at a solution

So my take is that ##Δt = 0.4s## for the water echo and a quick look on wiki told me the speed of sound under salt water is ##v = 1560 m/s##.

This yields ##Δd = vΔt = 624m##.

So the salt water echo takes 0.4 seconds to travel a total distance of 624 m at a rate of 1560 m/s.

Now I'm asked to find how much additional time will it take for the echo in the air to return? So I'm being asked to find ##Δt## for the sound in the air.

I'm given that ##v = 340 m/s## in the air ( By the question ) so the pieces look like they're going to fall together.

I just need to confirm one thing, would i still use ##Δd = 624m## in this next calculation? If so why? If not, why not?

EDIT : I forgot to mention the reason I'm doing this is because I know when I subtract the time it took for the air echo to come back by the time it took for the salt water echo to come back I will get the difference the question is asking for.

EDIT 2 : Nevermind I was reading this question incorrectly. I got my answer.
 
Last edited:

Redbelly98

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I just need to confirm one thing, would i still use ##Δd = 624m## in this next calculation? If so why? If not, why not?
Yes. This is simply the distance that the boat is from the cliff.
 

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