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प्रश्न
If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?
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उत्तर
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/m2.
Similarly,
\[\beta_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right)\]
On substituting the values and solving, we get:
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
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