Physics 20 Sound and Sound Waves (13 of 49) Sound Intensity at a Distance
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- čas přidán 28. 08. 2024
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In this video I will show you how to calculate I=? at distances of 10m, 1000m, and 10,000m.
Getting enlightened. Thanks
Thank you so much for the precise explanations. It helps a lot !!
Ehhhhh not really it not broken down enough then throws w in there without an explanation as to what it is….
The way that you explain concepts is just amazing, last semester I had Thermodynamics which I ended up getting an A with your help! Thank you so much. Now I'm taking Physics, and I started to watch the sound waves videos from the first one. But I'm just wondering if we always have to use (w/m^2) unit for this formula I2= (R1/R2)^2 I1. Since I wasn't able to get the right answer when I used dB! I'd really appreciate it if you could help me with this. Thank you
+Niloufar Ghazavi
Yes, with intensity you always have to use W/m^2.
When adding intensities you have to convert from dB to W/m^2.
There are some example videos.
that was great, but what happens during an echo ? Why don't sound waves cancel each other before they make it back to me ?
There are a number of potential situations and conditions that could be considered, but those are not part of this analysis.
Great lesson!
Thank you. 🙂
where did he get 10^-4 from? I thought intensity was at 80 dB
We have videos in that playlist that explain how to convert from sound in dB to sound levels in W/m^2
what if the intensity is lower than 10m, for example, 5m. What is the computation for that? thanks
Just replace the 10 m with 5 m in the ratio.
how does this work with negative dB? and what's the name of the formula to find the dB at 10 meters, or is it an estimate?
The equation is on the next video. (No negative dB for sound intensity). czcams.com/video/shnaAp498OU/video.html
sir
your answer by default match. your method is wrong. First you convert intensity into watt per m2. Otherwise wrong many answer.
How does A=4*PI*r^2 convert to I~1/r^2?
It doesn't. When it is stated that I ~ 1/R^2 that simply means that the sound intensity increases and decreases as the distance decreases and increases and the the proportion is 1/R^2 (10 times closer results in an intensity 100 times larger).
Thanks for clearing that up! You make the best videos for understanding Physics!
The intensity of a point source of sound, measured at a distance of 6m from the source, is 3.4 * 10^-6 W/m^2, So Calculate the intensity of the sound at a distance 20m from the source. please help?
Multiply the initial answer by 6^2 and divide by 20^2
And, still get the wrong answer.
Any analysis of sound that starts with the premise that sound is some 'energy' that's generated that can be conceptualized as passing through an aperture and then be calculated mathematically, will give the wrong answer.
Would the sonic boom be as loud at 100,000 ft here on earth?
The intensity of sound does diminish over distance.
but... the R = 1000, not 100. i thought R was supposed to stand for distance, not... multiplier of the distance with respect to the original reference point.
If the distance is 1000 m instead of 10 m, then the distance is 100 times farther and the surface area of the sphere at that distance is 10,000 times larger.
oh okay, that makes a lot more sense, thanks.
Hi , What will be the dB if at 10M given is 66dB...?
so if at 40M, Can it be 63dB or what the actual value?
My equipment is 40M away from a school.
Across 2 road before the school.
You have to convert to intensity first. At 40 m the intensity will be 1/4^2 = 1/16 th of the original intensity which is a drop of 12 dB.
66 dB - 12 dB = 54 dB
sorry sir , i don't get it because the camera is too far . i can't see what you are writing in front there :(
Nurul,
The main purpose of this video was to show the relationship between sound intensity and distance from the source. (The same applies to light intensity)
If you are twice as far from a source the sound will be only 1/4 as loud.
If you are three times as far from the source the sound will only be 1/9 as loud
If you are four times as far from the source the sound will only be 1/16 as loud
If you are 10 times as far from the source the sound will only be 1/100 as loud
Then converting from sound intensity from watts / m^2 to db (decibels) you will have to use the conversion I in dB = 10 log (I / I0) (there are other videos that show you how to do that.
If you watch the videos on a computer you can expand the view to fill the whole screen.
Hope that helps
Thankyou sir. Really appreaciate it
your math is wrong