Question

the sound intensity level at a point 4m from the point source is 10dB, then the sound level at a distance 2m from the same source will be ?

Answer 1

ds may be the answer ,n i just hope its going to help u

Answer 2

The sound from the 2m distance would be 16.02 dB

Solution:

The sound level which is at a distance of 2 m from the same source as given, will be calculated as follows,

Given: Sound intensity $\beta_{1}$= 10 dB

$\beta_{2}$= ?

Distance d1 = 4 m

Distance d2 = 2 m

From the formula of loudness, we have -  

$\bold{\beta=\frac{10 \log I}{I_{0}}}$

$\beta_{1}$= loudness at 4 m and $\beta_{2}$= loudness at 2m

$\begin{array}{l}{\beta_{1}=\frac{10 \log I}{I_{0}}} \\ {\beta_{2}=\frac{10 \log I_{2}}{l_{0}}} \\ {\beta_{1}-\beta_{2}=10\left(\frac{\log I}{I_{0}}-\frac{\log I_{2}}{l_{0}}\right)}\end{array}$

$\begin{array}{l}{\beta_{1}-\beta_{2}=10\left(\log \frac{l}{I_{2}}\right)} \\ {\beta_{1}-\beta_{2}=10\left(\log \frac{d 2^{2}}{d 1^{2}}\right)} \\ {\beta_{1}-\beta_{2}=10\left(\log \frac{4}{16}\right)}\end{array}$

$\begin{array}{l}{\beta_{1}-\beta_{2}=10 \log \frac{1}{4}} \\ {\beta_{1}-\beta_{2}=10(0-0.602)} \\ {\beta_{1}-\beta_{2}=10(-0.602)}\end{array}$

$\begin{array}{l}{\beta_{2}-\beta_{1}=6.02} \\ {\beta_{2}=10+6.02} \\ \bold{{\beta_{2}=16.02 d B}} \end{array}$