Question

# The sound level at a point $$5.0\ m$$ away from a point source is $$40\ dB$$. What will be the level at a point $$50\ m$$ away from the source ?

Solution

## We know that $$\beta =10\log_{10}\left( \dfrac{I}{I_0}\right)$$$$\beta_A= 10 \log \dfrac{I_A}{I_O},$$             $$\beta_B =10 \log \dfrac{I_B}{I_O}$$$$\Rightarrow I_A/ I_0 =10^{(\beta_A /10)}$$       $$I_B /I_O =10^{(\beta_B /10)}$$$$\Rightarrow \dfrac{I_A}{I_B}=\dfrac{r_B^2}{r_A^2}=\left( \dfrac{50}{5}\right)^2$$$$\Rightarrow 10^{\dfrac{(\beta_A -\beta_B)}{10}}=10^2$$$$\Rightarrow \dfrac{\beta_A -\beta_B}{10}=2$$$$\Rightarrow \beta_A- \beta_B=20$$$$\Rightarrow \beta_B=40+20=60\ dB$$.Physics

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