Function Posted January 26, 2017 Share Posted January 26, 2017 (edited) 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: [math]\alpha = 10\cdot \log{\frac{I_0}{I_x}}[/math] With [math]\alpha[/math] the attenuation of ultrasonic sound waves in a tissue, expressed in [math]\text{dB}\cdot\text{cm}^{-1}[/math], [math]I_0[/math] the intensity of the ultrasonic sound wave upon entrance of the tissue (or rather, right before it) and [math]I_x[/math] the remainder intensity after passage through the tissue of width [math]x[/math]. How is this even legal in physics? He basically states that [math]\text{dB}\cdot\text{cm}^{-1}[/math] 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: [math]20 \text{ dB}\cdot\text{cm}^{-1}=10\cdot\log{\frac{I_0}{I_x}}[/math] [math]\Leftrightarrow \frac{I_0}{I_x}=100[/math] 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 January 26, 2017 by Function Link to comment Share on other sites More sharing options...
swansont Posted January 26, 2017 Share Posted January 26, 2017 How is this even legal in physics? He basically states that [math]\text{dB}\cdot\text{cm}^{-1}[/math] 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 Link to comment Share on other sites More sharing options...
Function Posted January 26, 2017 Author Share Posted January 26, 2017 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 ... Link to comment Share on other sites More sharing options...
swansont Posted January 26, 2017 Share Posted January 26, 2017 The equation for intensity through the medium is going to look like I = I0 10^-ax Link to comment Share on other sites More sharing options...
Function Posted January 26, 2017 Author Share Posted January 26, 2017 (edited) The equation for intensity through the medium is going to look like I = I0 10^-ax How did you get that? Edited January 26, 2017 by Function Link to comment Share on other sites More sharing options...
swansont Posted January 26, 2017 Share Posted January 26, 2017 How did you get that? Physics class. Link to comment Share on other sites More sharing options...
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