Problem Statement
Sound intensity decreases as $I = I_0 e^{-2\alpha x}$ due to absorption (coefficient $\alpha$, in m$^{-1}$). Find the absorption coefficient in terms of the medium properties.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Viscous absorption: the main mechanism is viscous dissipation. The classical absorption coefficient:
Step 2 — Apply the relevant physical law or equation: $$\alpha = \frac{\omega^2}{2\rho v^3}\left(\frac{4\eta}{3} + \eta_v\right)$$
Step 3 — Solve algebraically for the unknown: where $\eta$ is shear viscosity and $\eta_v$ is bulk viscosity.
Step 4 — Substitute numerical values with units: The attenuation in dB per meter:
Step 5 — Compute and check the result: $$\alpha_{\rm dB} = 20\log_{10}e\cdot\alpha \approx 8.686\alpha$$
Step 6: Key features: $\alpha \propto f^2$ — absorption grows rapidly with frequency. This is why:
Worked Calculation
$$\alpha = \frac{\omega^2}{2\rho v^3}\left(\frac{4\eta}{3} + \eta_v\right)$$
$$\alpha_{\rm dB} = 20\log_{10}e\cdot\alpha \approx 8.686\alpha$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
Viscous absorption: the main mechanism is viscous dissipation. The classical absorption coefficient:
$$\alpha = \frac{\omega^2}{2\rho v^3}\left(\frac{4\eta}{3} + \eta_v\right)$$
where $\eta$ is shear viscosity and $\eta_v$ is bulk viscosity.
The attenuation in dB per meter:
$$\alpha_{\rm dB} = 20\log_{10}e\cdot\alpha \approx 8.686\alpha$$
Key features: $\alpha \propto f^2$ — absorption grows rapidly with frequency. This is why:
- High-frequency ultrasound (MHz) attenuates strongly in tissue (medical imaging limited to 5–15 cm)
- Low-frequency bass sounds travel farther in air (thunder rumbles are heard at great distances)
Answer
$$\boxed{\alpha_{\rm dB} = 20\log_{10}e\cdot\alpha \approx 8.686\alpha}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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