Question

If the sound level in a room is increased from \(50\,\text{dB}\) to \(60\,\text{dB}\), by what factor is the pressure amplitude increased?

• A. \(\sqrt{5}\)
• B. \(\sqrt{10}\)
• C. \(\sqrt{2}\)
• D. \(\sqrt{3}\)

Hint:

A \(10\,\text{dB}\) rise means \(10\times\) intensity; pressure amplitude goes as \(\sqrt{I}\).

Answer 1

The Correct Option is B

Step 1: The sound level in decibels is \[\beta=10\log_{10}\!\left(\frac{I}{I_0}\right)\] The change in level is \[\Delta\beta=10\log_{10}\!\left(\frac{I_2}{I_1}\right)\]
Step 2: Here \(\Delta\beta=60-50=10\,\text{dB}\), so \[10=10\log_{10}\!\left(\frac{I_2}{I_1}\right)\Rightarrow\log_{10}\!\left(\frac{I_2}{I_1}\right)=1\Rightarrow\frac{I_2}{I_1}=10\]
Step 3: Sound intensity is proportional to the square of the pressure amplitude: \[I\propto p_0^{2}\quad\Rightarrow\quad\frac{p_{0,2}}{p_{0,1}}=\sqrt{\frac{I_2}{I_1}}\]
Step 4: Therefore \[\frac{p_{0,2}}{p_{0,1}}=\sqrt{10}\] \[\boxed{\sqrt{10}}\]