Sounds propagation in different mediums

  • Thread starter mgiddy911
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In summary, the speed of sound in steel is about 4.2 kilometers per second. A steel rail is struck with a hammer, and there is an observer some distance away with one ear to the rail. The observer measures 1.60 seconds between hearing the sound through the rail and hearing the sound through the air. If the air temperature is 20 degrees Celsius and there is no wind, how far away is the observer?
  • #1
mgiddy911
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I am utterly confused with this problem, I am not quite sure how to formulate an equation to solve for the answer

:
The speed of sound in steel is about 4.2 kilometers per second. A steel rail is struck with a hammer, and there is an observer some distance away with one ear to the rail. The observer measures 1.60 seconds between hearing the sound through the rail and hearing the sound through the air. If the air temperature is 20 degrees Celsius and there is no wind, how far away is the observer?
Just looking for someone to show me where to begin
any help appreciated?
 
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  • #2
So far I know you use 343 m/s for speed of sound in air at 20 degrees celcius, and you use 4.2km/s (4200m/s) as speed of sound in the steel, for both you can use the equation v=d/t but both d and t are unknown, also you know t(air)-t(steel) = 1.6 seconds, from there i am not sure how to proceed, i have been trying to come up with a way to combine all of those equations but am comming up short...
 
  • #3
Yes, both d and t are unknown- that means you need two equations to solve for the two unknown. Those two equations are the motion through the air and through the rail. The distance sound has to travel in either case is d but there are two different times: call them tair and tsteel.

The speed of sound through the air is 343 m/s so d/tair= 343 which is the same as tair= d/343.

The speed of sound through steel is 4200 m/s so d/tsteel= 4200 which is the same as tsteel= d/4200.

What you DO know is the difference in the times: t1- t2= 1.6 s.

Okay subtract one equation from another:
tair- tsteel= 1.6 s= d/343- d/4200. Now you have a very simple equation to solve for d.
 
  • #4
Thanks!

Thank you, however right before I read this post i stumbled upon a way i think will work, it may be the same as what u said however, i just thought tat u know the dstances n ar and steel are equal, so d=vt; vt(air)=vt(steel) ; then vt(air)=v(t-1.6) and then solve for t of air, then plug that into the equation v=d/t to solve for d in air. does this work also?
 

1. How does sound travel through air?

Sound travels through air through a series of compressions and rarefactions of air molecules. When a sound is produced, it creates a vibration that travels through the air, causing the air molecules to bump into each other and create a sound wave. This wave then travels through the air until it reaches our ears.

2. How does sound travel through water?

Sound travels through water in a similar way as it does through air. However, water is denser than air, so sound waves travel at a much faster speed. The molecules in water are also closer together, allowing the sound wave to travel further and with less loss of energy.

3. How does sound travel through solids?

Sound travels through solids by causing vibrations in the molecules of the solid material. These vibrations then travel through the solid, similar to how a wave travels through water. The denser the material, the faster the sound will travel.

4. How does temperature affect sound propagation?

Temperature affects sound propagation by altering the speed at which sound waves travel through a medium. Warmer air and liquids have faster molecular movement, allowing sound waves to travel faster. In solids, higher temperatures can also cause molecules to vibrate more, increasing the speed of sound.

5. How does elevation affect sound propagation?

Elevation affects sound propagation in a similar way as temperature. At higher elevations, the air is less dense, causing sound waves to travel slower. This means that sounds may appear quieter at higher elevations due to the decrease in air molecules for the sound wave to travel through.

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