Question

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

Answer 1

Given:
Initial sound level \[\beta_1\] = 50 dB
Final sound level 

\[\beta_2\]= 60 dB
Constant reference intensity \[I_0\]=\[10^{-12}\]W/m²
We can find initial intensity I₁ using:

\[\beta_1  = 10 \log_{10} \left( \frac{I_1}{I_0} \right)\] 

\[\Rightarrow 50   = 10 \log_{10} \left( \frac{I_1}{{10}^{- 12}} \right)\] 

On Solving , We get:

\[I_1 =10^{-7} \]W/m^2_.
Similarly,

\[\beta_2  = 10 \log_{10} \left( \frac{I_2}{I_0} \right)\] 

On substituting the values and solving, we get:

\[I_2  =  {10}^{- 6}   \text { W/ m }^2\]

As the intensity is proportional to the square of pressure amplitude (p),
we have:

\[\frac{I_2}{I_1} =  \left( \frac{p_2}{p_1} \right)^2  = \left( \frac{{10}^{- 6}}{{10}^{- 7}} \right) = 10\]

\[\therefore    \left( \frac{p_2}{p_1} \right)^2  = 10\] 

\[ \Rightarrow   \frac{p_2}{p_1} = \sqrt{10}\]

Hence, the pressure amplitude is increased by 

\[\sqrt{10}\] factor .