Question

Average noise level at a working place is 80 dB(A) for the first 10 minutes and 60 dB(A) for the next 30 minutes. The energy equivalent continuous noise level at the place for the entire period of 40 minutes, in dB(A), is:

Hint:

Use the energy-based average (Leq) to accurately calculate varying noise exposure over time. Always convert dB values using base-10 exponentials before averaging.

Answer 1

Correct Answer: 5

Step 1: Use the formula for equivalent continuous noise level \( L_{eq} \).

The equivalent continuous noise level is given by:
\[ L_{eq} = 10 \log_{10} \left( \frac{1}{T} \sum_{i=1}^{n} t_i \cdot 10^{L_i/10} \right) \]

Given:
L₁ = 80 dB(A), t₁ = 10 min
L₂ = 60 dB(A), t₂ = 30 min
Total time, T = 40 min

Substitute the values:

\[ L_{eq} = 10 \log_{10} \left( \frac{1}{40} \left(10 \cdot 10^{8} + 30 \cdot 10^{6}\right) \right) \]

\[ = 10 \log_{10} \left( \frac{1{,}000{,}000{,}000 + 90{,}000{,}000}{40} \right) \]

\[ = 10 \log_{10} (27{,}250{,}000) \]

\[ = 10 \log_{10} (2.725 \times 10^{7}) \]

\[ = 10 \left( \log_{10}(2.725) + 7 \right) \]

\[ = 10 (0.435 + 7) \]

\[ = 10 \times 7.435 = 73.35 \]

Final Answer:
73 dB(A)