Question

If the sound level in a room is increased from $50dB$ to $60dB$, by what factor is the pressure amplitude increased?

Answer 1

Step 1: Given data

Sound level in a room is increased from $50dB$ to $60dB$

Step 2: To find

Increase in the amplitude of pressure

Step 3: Calculation

As we know

$$\begin{gathered} \beta =10\log \left(\frac{I}{I_0}\right) \\ {I}_0={10}^{-12}W{m}^{-2} \end{gathered}$$

where $I$ is intensity

When $\beta =50dB$,

$$\begin{gathered} \Rightarrow 50=10\log \left(\frac{I_1}{{10}^{-12}}\right) \\ \Rightarrow 5=\log \left(\frac{I_1}{{10}^{-12}}\right) \\ \Rightarrow {10}^5=\frac{I_1}{{10}^{-12}} \\ \Rightarrow I_1={10}^{-7} \end{gathered}$$

When $\beta =60dB$,

$$\begin{gathered} \Rightarrow 60=10\log \left(\frac{I_2}{{10}^{-12}}\right) \\ \Rightarrow 6=\log \left(\frac{I_2}{{10}^{-12}}\right) \\ \Rightarrow {10}^6=\frac{I_2}{{10}^{-12}} \\ \Rightarrow I_2={10}^{-6} \end{gathered}$$

We know,

$I\propto P^2$

where $P$ is pressure

$$\begin{gathered} \frac{I_1}{I_2}=\frac{P_1^2}{P_2^2} \\ \frac{{10}^{-7}}{{10}^{-6}}=\frac{P_1^2}{P_2^2} \\ 10={\left(\frac{P_2}{P_1}\right)}^2 \\ {P}_2=\sqrt{10}{P}_1 \end{gathered}$$

Hence, if the sound level increases from $50dB$ to $60dB$ their pressure amplitudes get $\sqrt{10}$ times the original value.