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The sound level at a point is increased by 30 dB. What is factor is the pressure amplitude increased? |
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Answer» The sound level is dB is `L=10log_(10)((I)/(I_0))` If `L_1` and `L_2` are the sound levels and `I_1` and `I_2` are the coresponding intensities in the two cases, then `L_2-L_1=10[log_(10)((I_2)/(I_0))-log_(10)((I_1)/(I_0))]` `implies30=10log_(10)((I_2)/(I_1))` `implies(I_2)/(I_1)=10^3` As the intensity is proporional to the square of the pressure amplitude thus we have `(trianglep_2)/(trianglep_1)=sqrt((I_2)/(I_1))=sqrt(1000)=32` |
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