Question
If the sound level in a room is increased from 50dB to 60dB, by what factor is the pressure amplitude increased?
✓
Answer
$\beta_1=50\text{dB},\ \beta_2=60\text{dB}$$\therefore\text{I}_1=10^{-7}\text{W/m},\ \text{I}_2=10^{-6}\text{W/m}^2$
$\Big($because $\beta=10\log_{10}\big(\frac{\text{I}}{\text{I}_0}\big),$ where $\text{I}_0=10^{-12}\text{W/m}^2\Big)$
Again, $\frac{\text{I}_2}{\text{l}_1}=\Big(\frac{\text{P}_2}{\text{P}_1}\Big)^2$
$=\Big(\frac{10^{-6}}{10^{-7}}\Big)$
$=10$ (where p = pressure amplitude)
$\therefore\Big(\frac{\text{P}_2}{\text{P}_1}\Big)=\sqrt{10}.$
$\Big($because $\beta=10\log_{10}\big(\frac{\text{I}}{\text{I}_0}\big),$ where $\text{I}_0=10^{-12}\text{W/m}^2\Big)$
Again, $\frac{\text{I}_2}{\text{l}_1}=\Big(\frac{\text{P}_2}{\text{P}_1}\Big)^2$
$=\Big(\frac{10^{-6}}{10^{-7}}\Big)$
$=10$ (where p = pressure amplitude)
$\therefore\Big(\frac{\text{P}_2}{\text{P}_1}\Big)=\sqrt{10}.$
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