Let's solve the problem step by step: --- ## **Given Data** - **Equation:** \[ L_{10}(\text{dB(A)}) = 85. - 10\log_{10}\left(\frac{d}{13.5}\right) + 10\log_{10}\left(\frac{\theta}{180}\right) + C_b \] - **Road segment length** = 720 m - **Barrier length** = 240 m - **r1 distance from barrier** = 60 m - **r2 distance from barrier** = 80 m - **Both points are on the middle line** - **Barrier attenuation, \(C_b\):** - **Inside shadow zone:** \(C_b = -10\) dB(A) - **Outside shadow zone:** \(C_b = \) dB(A) --- ## **Step 1: Calculate \(d\) for r1 and r2** Since both r1 and r2 are on the middle line, the perpendicular distance from the effective source to the receiver = (distance from barrier) + (perpendicular distance from barrier to source). From the diagram: - Distance from effective source to barrier = 3.5 m - So, - **r1:** \(d_1 = 3.5 + 60 = 63.5\, \text{m}\) - **r2:** \(d_2 = 3.5 + 80 = 83.5\, \text{m}\) --- ## **Step 2: Calculate \(\theta\) for r1 and r2** \(\theta\) is the angle subtended by the road segment at the receiver (arc tangent of half the road segment length over distance): \[ \theta = 2 \arctan\left(\frac{L}{2d}\right) \] where \(L = 720\, \text{m}\) (road segment length). - For r1: \(d_1 = 63.5\) m - For r2: \(d_2 = 83.5\) m Calculate \(\theta_1\) and \(\theta_2\): \[ \theta_1 = 2 \arctan\left(\frac{720/2}{63.5}\right) = 2 \arctan\left(\frac{360}{63.5}\right) \] \[ \theta_1 = 2 \arctan(5.669) \approx 2 \times 80.^\circ = 160.^\circ \] \[ \theta_2 = 2 \arctan\left(\frac{360}{83.5}\right) \] \[ \theta_2 = 2 \arctan(4.31) \approx 2 \times 77.^\circ = 154.^\circ \] --- ## **Step 3: Determine \(C_b\) for r1 and r2** - Both r1 (60 m) and r2 (80 m) are further than the barrier (which is only 240 m long, centered on the road). Since both receivers are in the shadow zone (behind the barrier), **\(C_b = -10\) dB(A)** for both. --- ## **Step 4: Calculate \(L_{10}\) for r1 and r2** ### For r1: \[ d_1 = 63.5\, \text{m}, \quad \theta_1 = 160.^\circ, \quad C_{b1} = -10 \] \[ L_{10,1} = 85. - 10 \log_{10} \left(\frac{63.5}{13.5}\right) + 10 \log_{10} \left(\frac{160.}{180}\right) - 10 \] \[ \frac{63.5}{13.5} = 4.704 \] \[ \log_{10} (4.704) = .673 \] \[ -10 \times .673 = -6.73 \] \[ \frac{160.}{180} = .889 \] \[ \log_{10} (.889) = -.051 \] \[ 10 \times -.051 = -.51 \] \[ L_{10,1} = 85. - 6.73 - .51 - 10 = 67.76\, \text{dB(A)} \] --- ### For r2: \[ d_2 = 83.5\, \text{m}, \quad \theta_2 = 154.^\circ, \quad C_{b2} = -10 \] \[ L_{10,2} = 85. - 10 \log_{10} \left(\frac{83.5}{13.5}\right) + 10 \log_{10} \left(\frac{154.}{180}\right) - 10 \] \[ \frac{83.5}{13.5} = 6.185 \] \[ \log_{10} (6.185) = .792 \] \[ -10 \times .792 = -7.92 \] \[ \frac{154.}{180} = .856 \] \[ \log_{10}(.856) = -.068 \] \[ 10 \times -.068 = -.68 \] \[ L_{10,2} = 85. - 7.92 - .68 - 10 = 66.40\, \text{dB(A)} \] --- ## **Step 5: Find the Change in Noise Level** \[ \Delta L_{10} = L_{10,2} - L_{10,1} = 66.40 - 67.76 = -1.36\, \text{dB(A)} \] --- ## **Final Answer** ### **The total noise level decreases by 1.36 dB(A) when the assessment point is moved from r1 to r2.** --- ### **Summary Table** | Point | \(d\) (m) | \(\theta\) (deg) | \(C_b\) (dB) | \(L_{10}\) (dB(A)) | |-------|----------|------------------|--------------|--------------------| | r1 | 63.5 | 160. | -10 | 67.76 | | r2 | 83.5 | 154. | -10 | 66.40 | --- **The change in noise level due to moving from r1 to r2 is a decrease of 1.36 dB(A).**