Doppler, echo, attenuation

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Hello everyone

Let me address, after a long time, again the experts in the truly bizarre domain which goes by the name "physics":

In my course, my professor says:

$\alpha = 10\cdot \log{\frac{I_0}{I_x}}$

With $\alpha$ the attenuation of ultrasonic sound waves in a tissue, expressed in $\text{dB}\cdot\text{cm}^{-1}$, $I_0$ the intensity of the ultrasonic sound wave upon entrance of the tissue (or rather, right before it) and $I_x$ the remainder intensity after passage through the tissue of width $x$.

How is this even legal in physics? He basically states that $\text{dB}\cdot\text{cm}^{-1}$ is dimensionless. Which clearly isn't the case.

Then, he states that alpha is about 20 dB/cm in bone tissue. I can understand that per cm progression of the sound waves in bone, their volume decrease with 20 dB.

But:

$20 \text{ dB}\cdot\text{cm}^{-1}=10\cdot\log{\frac{I_0}{I_x}}$

$\Leftrightarrow \frac{I_0}{I_x}=100$

Which insinuates the invariability of the intension as the sound wave penetrates the tissue.

Ergo, I don't find it possible for me to solve the question by what factor the original intensity is divided when the sound wave travels 2 cm in bone.

Intuition says: 10,000. But if the formula is correct, and the attenuation is indeed completely independent of the depth, it should be, and remain forever, 100. Please don't tell me that's true.

Thanks;

F

Edited by Function

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How is this even legal in physics? He basically states that $\text{dB}\cdot\text{cm}^{-1}$ is dimensionless. Which clearly isn't the case.

Nope. The constant out in front has units. It's not just 10, it's 10 dB/cm

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Nope. The constant out in front has units. It's not just 10, it's 10 dB/cm

Explains half of my problem. Thank you very much for that.

So what about the apparently indifferent factor of 100 = I0/Ix? Is intensity invariable through a medium? That'd be very counter-intuitive ...

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The equation for intensity through the medium is going to look like I = I0 10^-ax

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The equation for intensity through the medium is going to look like I = I0 10^-ax

How did you get that?

Edited by Function

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How did you get that?

Physics class.