Question:easy

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?

Show Hint

A \(10\,\text{dB}\) rise means \(10\times\) intensity; pressure amplitude goes as \(\sqrt{I}\).
Updated On: Jul 2, 2026
  • \(\sqrt{5}\)
  • \(\sqrt{10}\)
  • \(\sqrt{2}\)
  • \(\sqrt{3}\)
Show Solution

The Correct Option is B

Solution and Explanation

Work backwards from the pressure amplitude to the level.

Intensity scales as the square of pressure amplitude, $I\propto p_0^2$. So in decibels \[\beta=10\log_{10}\frac{I}{I_0}=20\log_{10}\frac{p_0}{p_{0,\text{ref}}}\] The rise in level is \[\Delta\beta=20\log_{10}\frac{p_{0,2}}{p_{0,1}}=10\ \text{dB}\] Solving, \[\log_{10}\frac{p_{0,2}}{p_{0,1}}=\tfrac{10}{20}=\tfrac12\quad\Rightarrow\quad\frac{p_{0,2}}{p_{0,1}}=10^{1/2}=\sqrt{10}\] So the pressure amplitude grows by a factor $\sqrt{10}$ (about $3.16$). \[\boxed{\sqrt{10}}\]
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