Wave Mechanics Question 1103
Question: Intensity level 200 cm from a source of sound is 80 dB. If there is no loss of acoustic power in air and intensity of threshold hearing is $ {{10}^{-12}}W{{m}^{-2}} $ then, what is the intensity level at a distance of 400 cm from source
Options:
A) Zero
B) 54 dB
C) 64 dB
D) 44 dB
Show Answer
Answer:
Correct Answer: B
Solution:
$ I\propto \frac{1}{r^{2}}\Rightarrow \frac{I _{2}}{I _{1}}=\frac{r _{1}^{2}}{r _{2}^{2}}=\frac{2^{2}}{{{(40)}^{2}}}=\frac{1}{400} $
Therefore $ I _{1}=400I _{2} $ Intensity level at point 1, $ L _{1}=10{\log _{10}}( \frac{I _{1}}{I _{0}} ) $ and intensity at point 2, $ L _{2}=10{\log _{10}}( \frac{I _{2}}{I _{0}} ) $ \ $ L _{1}-L _{2}=10\log \frac{I _{1}}{I _{2}}=10{\log _{10}}(400) $
Therefore $ L _{1}-L _{2}=10\times 2.602=26 $
$ L _{2}=L _{1}-26=80-26=54\ dB $
BETA
