Question

The RMS sound pressure is increased by 50%. This increases the sound pressure level, in dB, nearly by

• A. 3.52
• B. 8.11
• C. 100.21
• D. 43.52

Hint:

Use the 20 log10 relation between pressure ratio and decibel change.

Answer 1

The Correct Option is A

Step 1: Recall that sound power is proportional to the square of pressure.
Since intensity $I \propto p^2$, the level in dB can also be written as $L = 10 \log_{10}(I/I_{ref}) = 10 \log_{10}(p^2/p_{ref}^2)$.

Step 2: Build the intensity ratio for a 50% pressure rise.
$p_2 = 1.5 p_1$, so $I_2/I_1 = (p_2/p_1)^2 = 1.5^2 = 2.25$.

Step 3: Convert the intensity ratio to a dB change.
$\Delta L = 10 \log_{10}(2.25) = 10 \times 0.3522 = 3.52\ \text{dB}$.

Final Answer:
Both the pressure-ratio and intensity-ratio routes give the same rise, about 3.52 dB. \[ \boxed{\Delta L \approx 3.52\ \text{dB}} \]