Question

How many times the pressure amplitude is increased if sound level is increased by 40 dB?

Answer 1

The sound level $L$ of a wave of intensity $I$ is given by ,

$L=10lo{g}_{10}\left(I/I_0\right)$ , ........................eq1

where $I_0=$ zero level of intensity ,

now ehen sound level is increased by 40dB , then increased sound level $L^{\prime }$ is given by ,

$L^{\prime }=10lo{g}_{10}\left(I^{\prime }/I_0\right)$ , ........................eq2

given , sound level $L^{\prime }$ is greater than $L$ by 40dB , then $L-L^{\prime }=40dB$ ,

therefore ,subtracting eq1 from eq2 ,

$L^{\prime }-L=10\left[lo{g}_{10}\right(I^{\prime }/I_0)-lo{g}_{10}(I/I_0\left)\right]=10lo{g}_{10}\left(I^{\prime }/I\right)$ ,

or $40=10lo{g}_{10}\left(I^{\prime }/I\right)$ ,

or $I^{\prime }/I={10}^4$ ,

we have , $I\propto P_0^2$ , where $P_0=$ pressure amplitude ,

or $P_0^{\prime }/P_0=\sqrt{I^{\prime }/I}$ ,

therefore , $P_0^{\prime }=P_0\sqrt{I^{\prime }/I}=P_0\sqrt{{10}^4}=100P_0$ ,

pressure amplitude is increased by 100 times .