Let's break the solution into **three steps**, with explanations at each step. --- ### **Step 1: Find the Sound Intensity at Each Distance** **Explanation:** In shallow water where depth << distance, sound spreads cylindrically (not spherically). For a **cylindrical wavefront**, the area over which the energy spreads is \( A = 2\pi r h \), where \( r \) is the horizontal distance and \( h \) is the water depth. The **intensity** \( I \) at a distance \( r \) is: \[ I = \frac{P}{A} = \frac{P}{2\pi r h} \] Where: - \( P \) = Sound power of the source (10 W) - \( h \) = Water depth (10 m) - \( r \) = Distance from the source (100, 200, 400 m) Calculate \( I \) for each \( r \): - **At \( r = 100\,\text{m} \):** \[ I_{100} = \frac{10}{2 \pi \times 100 \times 10} = \frac{10}{6283} \approx 0.00159 \text{ W/m}^2 \] - **At \( r = 200\,\text{m} \):** \[ I_{200} = \frac{10}{2 \pi \times 200 \times 10} = \frac{10}{12566} \approx 0.000796 \text{ W/m}^2 \] - **At \( r = 400\,\text{m} \):** \[ I_{400} = \frac{10}{2 \pi \times 400 \times 10} = \frac{10}{25133} \approx 0.000398 \text{ W/m}^2 \] --- ### **Step 2: Calculate the Sound Intensity Level (SIL) in dB re \(10^{-12}\) W/m²** **Explanation:** The **Sound Intensity Level** is given by: \[ L_I = 10 \log_{10}\left(\frac{I}{I_0}\right) \] Where: - \( I \) = Calculated intensity - \( I_0 = 10^{-12} \) W/m² (reference intensity in underwater acoustics) Calculate \( L_I \) for each distance: - **At 100 m:** \[ L_{I,100} = 10 \log_{10}\left(\frac{0.00159}{10^{-12}}\right) = 10 \log_{10}(1.59 \times 10^9) \approx 10 \times 9.20 = 92 \text{ dB} \] - **At 200 m:** \[ L_{I,200} = 10 \log_{10}\left(\frac{0.000796}{10^{-12}}\right) = 10 \log_{10}(7.96 \times 10^8) \approx 10 \times 8.90 = 89 \text{ dB} \] - **At 400 m:** \[ L_{I,400} = 10 \log_{10}\left(\frac{0.000398}{10^{-12}}\right) = 10 \log_{10}(3.98 \times 10^8) \approx 10 \times 8.60 = 86 \text{ dB} \] --- ### **Step 3: Convert to Sound Pressure Level (SPL) in dB re 1 μPa** **Explanation:** The relationship between intensity and pressure in water is: \[ I = \frac{p^2}{\rho c} \] Where: - \( p \) = RMS sound pressure (in Pa) - \( \rho \) = Density of seawater (\(\approx 1000 \) kg/m³) - \( c \) = Speed of sound in water (\(\approx 1500\) m/s) The reference pressure is \( p_0 = 1 \) μPa = \( 10^{-6} \) Pa. The **Sound Pressure Level** is: \[ L_p = 20 \log_{10}\left(\frac{p}{p_0}\right) \] But since \( I \propto p^2 \), we can relate intensity level to pressure level: \[ L_p = L_I + 10 \log_{10}(\rho c) - 10 \log_{10}(I_0) + 20 \log_{10}(p_0) \] But in practice, for underwater acoustics: - \( I_0 = 10^{-12} \) W/m² - \( p_0 = 1\, \mu\text{Pa} \) - \( \rho c \approx 1.5 \times 10^6 \) kg/(m²·s) So: \[ L_p = L_I + 10 \log_{10}(\rho c) \] \[ 10 \log_{10}(1.5 \times 10^6) \approx 10 \times 6.176 = 61.76 \text{ dB} \] - **At 100 m:** \( 92 + 61.8 = 153.8 \) dB re 1 μPa - **At 200 m:** \( 89 + 61.8 = 150.8 \) dB re 1 μPa - **At 400 m:** \( 86 + 61.8 = 147.8 \) dB re 1 μPa --- ## **Final Answers** | Distance (m) | Intensity Level (dB re \(10^{-12}\) W/m²) | SPL (dB re 1 μPa) | |--------------|-------------------------------------------|-------------------| | 100 | 92 | 154 | | 200 | 89 | 151 | | 400 | 86 | 148 | --- ### **Summary of Steps** 1. **Calculate intensity** at each distance using cylindrical spreading. 2. **Convert intensity to dB** re \(10^{-12}\) W/m². 3. **Convert intensity level to SPL** in dB re 1 μPa using water parameters. **This method uses cylindrical spreading because the water depth is much less than the horizontal range, so the sound energy spreads out in a cylinder, not a sphere.**