Hi! Great video! Is there a mathematical equational relationship between the area of a horn's mouth and its decibel output? Or an equational relationship between the length of a horn's throat and its decibel output? I need it for a project I am doing. Thanks!
This is fantastic. I am a bit perplexed though, although the math shows that ultrasound would be reflected by air to me it seems a bit counterintuitive, could someone explain? Thank you so much!!
@@PhysicsHigh Thank you so much for the response. Yes, I totally understand the impedence... I just can't seem to wrap my head around air reflecting soundwaves, Id expect them to refract... I guess I have poor imagination lol. Any thoughts on how you look at it?
The air doesn’t reflect sound waves. In fact if the medium is consistent you should not be able to see any detail as the detail comes from reflection where the acoustic impedance changes. Hope that helps.
@@PhysicsHigh Do you know why its called reflection when the medium goes from say a soft tissue to air... for example the pluera to inside of the lung?
You can think of it this way. If the difference in acoustic impedance is more, the solid is very much denser than air. So, when the sound wave reaches the edge, at the moment of contact, all the air surrounding the edge goes away from the solid because , as the density of solid is very high, obviously the energy of the sound waves is more in solid for it to even exist in the first place. As the air goes away most of the energy has no choice but to reflect itself again in the solid.
I have a question: say Z2= 4 and Z1=6 , R would come up as 1/25 so would mean that the reflection coefficient would be 1/25 for this type of equation..now if I invert the values to as Z2= 6 and Z1=4 , the final value would be the same, 1/25 right? In this case , if I assume Z2 would be the bone impedance and Z1 the soft tissue impedance , does it mean that the reflection coefficient is the same either if I hit bone first and then soft tissue to if I hit soft tissue first and then bone? I know is a little confusing , but let me know if you understand my question. Thanks
@@PhysicsHigh this is correct according to the maths but why should it be? The equation describes the phenomena but why is the phenomena happening? What is physically happening at the boundary that causes the same fraction of reflection when sound travels from concrete to water as from water to concrete?
Since commenting above I found an answer in Technology for Diagnostic Sonography. My summary is: · The reason lies in what conditions allow transfer of pressure energy and momentum across a boundary of materials. Energy that is traveling fast in a high density medium (high Z) has high momentum and can’t readily be transferred to a medium that conveys energy with low momentum (low speed and low density [low Z]. The opposite is true as well. Energy with low momentum that is traveling slowly in a low density medium (low Z) can’t readily be transferred to a medium that conveys energy with high momentum (high density and high speed [high Z]. The law of conservation of momentum means that, in both cases, because there is Z mismatch, the energy transmitted across the boundary is reduced and the surplus reflect.
Can some explain how bone-tissue has the same reflection % as tissue-bone. I cant wrap my head around it, since, tissue is less dense, so more sound energy should pass through it than be reflected compared to dense bone...
this is the exact question I asked as well... it seems the formula they use for the impedance is not able to answer this ...It would be hard to believe that IRC would be the same whether you hit bone first and then soft tissue to viceversa
With due respect,the higher speed of sound in liquids and metals is not because of their density,it is for an variable called bulk modulus.... Water density= 1 and the speed of sound is 1480 m per sec in it mercury density is 13/6 and the speed of sound is 1450m/s
Thanks! Could someone explain if refraction would always occur where reflection occurs due to acoustic impedance mismatching? Since there is a change in densities?
Reflection occurs when you have a difference in impedance(Impedance = density x propagation speed). Refraction does not necessarily occur just because the impedance is different. Refraction may occur when you have an oblique angle and you have two mediums with different propagation speeds. Density and propagation speed are two separate items.
Sir, why the transmitted wave in the second boundary became large and the reflected wave became smaller? I hope you can see this because im so close to getting this course. Thank you sir.
I think it's because of the differences in the acoustic impedances and so when the sound wave went through the first barrier there was a high reflection coefficient (R) but in the second R was smaller and hence, the difference. (its all comes down to the materials)
I found a mistake in this video.Density and speed are inversely related so when speed increases density does not increase but decreases. Stiffness would increase with sound.
According to Edelman’s physics book: as the density of a medium increases, propagation speed would decrease. Stiffness and propagation speed are directly related. Density and propagation speed and inversely related. I think this person wasn’t inferring that the density of the medium decreases and you increase speed since neither of those are adjustable by a sonographer. A more correct thought is that the less dense an object is, the faster a sound wave can propagate through it.
Great video! I never understood this phenomenon until now, but it makes sense, to oversimplify it, its like a small ball hitting a huge ball.
Wow, best teacher ever. 15 episodes down idk how many remaining
What a great explanation!!!
GREAT explanation!! Thank you so much!
beautiful wallahi
Very well explained..
Hi! Great video!
Is there a mathematical equational relationship between the area of a horn's mouth and its decibel output? Or an equational relationship between the length of a horn's throat and its decibel output? I need it for a project I am doing.
Thanks!
Thank you so much sir.
Thanks a lot sir, very helpful.
This is fantastic. I am a bit perplexed though, although the math shows that ultrasound would be reflected by air to me it seems a bit counterintuitive, could someone explain? Thank you so much!!
It’s not the air that reflects but the boundary between the air and skin. The bigger the difference in acoustic impedance the greater the reflection.
@@PhysicsHigh Thank you so much for the response. Yes, I totally understand the impedence... I just can't seem to wrap my head around air reflecting soundwaves, Id expect them to refract... I guess I have poor imagination lol. Any thoughts on how you look at it?
The air doesn’t reflect sound waves. In fact if the medium is consistent you should not be able to see any detail as the detail comes from reflection where the acoustic impedance changes. Hope that helps.
@@PhysicsHigh Do you know why its called reflection when the medium goes from say a soft tissue to air... for example the pluera to inside of the lung?
You can think of it this way. If the difference in acoustic impedance is more, the solid is very much denser than air. So, when the sound wave reaches the edge, at the moment of contact, all the air surrounding the edge goes away from the solid because , as the density of solid is very high, obviously the energy of the sound waves is more in solid for it to even exist in the first place. As the air goes away most of the energy has no choice but to reflect itself again in the solid.
Thank you Sir, its very helpful
Could you please make video of acoustic mechanical impedance (AMI)??
I have a question: say Z2= 4 and Z1=6 , R would come up as 1/25 so would mean that the reflection coefficient would be 1/25 for this type of equation..now if I invert the values to as Z2= 6 and Z1=4 , the final value would be the same, 1/25 right? In this case , if I assume Z2 would be the bone impedance and Z1 the soft tissue impedance , does it mean that the reflection coefficient is the same either if I hit bone first and then soft tissue to if I hit soft tissue first and then bone? I know is a little confusing , but let me know if you understand my question. Thanks
Correct. They value is the same regardless of the direction: more to less dense or vice versa.
@@PhysicsHigh this is correct according to the maths but why should it be? The equation describes the phenomena but why is the phenomena happening? What is physically happening at the boundary that causes the same fraction of reflection when sound travels from concrete to water as from water to concrete?
Since commenting above I found an answer in Technology for Diagnostic Sonography. My summary is:
· The reason lies in what conditions allow transfer of pressure energy and momentum across a boundary of materials. Energy that is traveling fast in a high density medium (high Z) has high momentum and can’t readily be transferred to a medium that conveys energy with low momentum (low speed and low density [low Z]. The opposite is true as well. Energy with low momentum that is traveling slowly in a low density medium (low Z) can’t readily be transferred to a medium that conveys energy with high momentum (high density and high speed [high Z]. The law of conservation of momentum means that, in both cases, because there is Z mismatch, the energy transmitted across the boundary is reduced and the surplus reflect.
Can some explain how bone-tissue has the same reflection % as tissue-bone. I cant wrap my head around it, since, tissue is less dense, so more sound energy should pass through it than be reflected compared to dense bone...
this is the exact question I asked as well... it seems the formula they use for the impedance is not able to answer this ...It would be hard to believe that IRC would be the same whether you hit bone first and then soft tissue to viceversa
With due respect,the higher speed of sound in liquids and metals is not because of their density,it is for an variable called bulk modulus....
Water density= 1 and the speed of sound is 1480 m per sec in it
mercury density is 13/6 and the speed of sound is 1450m/s
Thanks! Could someone explain if refraction would always occur where reflection occurs due to acoustic impedance mismatching? Since there is a change in densities?
Reflection occurs when you have a difference in impedance(Impedance = density x propagation speed).
Refraction does not necessarily occur just because the impedance is different. Refraction may occur when you have an oblique angle and you have two mediums with different propagation speeds. Density and propagation speed are two separate items.
How do you derive the impedance-intensity equation?
Sir, why the transmitted wave in the second boundary became large and the reflected wave became smaller? I hope you can see this because im so close to getting this course. Thank you sir.
I think it's because of the differences in the acoustic impedances and so when the sound wave went through the first barrier there was a high reflection coefficient (R)
but in the second R was smaller and hence, the difference. (its all comes down to the materials)
Thank you
I found a mistake in this video.Density and speed are inversely related so when speed increases density does not increase but decreases. Stiffness would increase with sound.
Alyson Crane well they’re not in this case. The speed of sound in a metal is much faster than in water.
Is this valid, reliable and accurate though? Where are your sources?
According to Edelman’s physics book: as the density of a medium increases, propagation speed would decrease.
Stiffness and propagation speed are directly related.
Density and propagation speed and inversely related.
I think this person wasn’t inferring that the density of the medium decreases and you increase speed since neither of those are adjustable by a sonographer. A more correct thought is that the less dense an object is, the faster a sound wave can propagate through it.
Yes, because c = (κ/ρ)1/2, when the guy said it it confused the fuck out of me!
Thanks a lot
thanks.
What happen if Z1>Z2