Question

The pressure amplitude in a sound waveform radio receiver is $2.0\times {10}^{-3}M/m^2$ and the intensity at a point is ${10}^{-8}W/m^2$. If by turning the "Volume" knowb the presure amplitude is increased to $3\times {10}^{-3}N/m^2$. The intensity is equal to $x\times {10}^{-10}W/m^2$ where $x$ is :

Answer 1

The intensity of sound wave is directly proportional to the square of the pressure amplitude ,

$I\propto P_0^2$ ,

by this we have , $I_1/I_2=P_{01}^2/P_{02}^2$ ,

given $P_{01}=2.0\times {10}^{-3}N/m^2,P_{02}=3\times {10}^{-3}N/m^2,I_1={10}^{-8}W/m^2$ ,

therefore , $I_2=I_1\left(P_{02}^2/P_{01}^2\right)$ ,

$I_2={10}^{-8}(3\times {10}^{-3}/2.0\times {10}^{-3}{)}^2=2.25\times {10}^{-8}W/m^2$