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When comparing the sound measured in decibles, a normal convo between two people measured 60 dB while the threshold of pain from the loud noise is ofter concidered to be 130 dB.

The formula is β=10log(I/Io)

Io =10-12 w/m2

a.) determine how many times louder the intensity of the thrreshold of pain is compaired to the conversation

  I need help solving... 60dB=10log(I/10-12)

b.) if the convo intensity was trippled louder, determine what the new decibel level would be.

 Jun 5, 2014

Best Answer 

 #1
avatar+33661 
+10

With β = 10*log(l/lo)  then l = lo*10(β/10) 

a) lconv = lo*10(60/10) or lconv = lo106       lpain = lo*10(130/10) or lpain = lo*1013

Therefore lpain/lconv = 1013/106 = 1013-6 = 107 .  That is, the intensity at the pain threshold is ten million times greater than at conversation level.

 

See if you can now do part b)

 Jun 6, 2014
 #1
avatar+33661 
+10
Best Answer

With β = 10*log(l/lo)  then l = lo*10(β/10) 

a) lconv = lo*10(60/10) or lconv = lo106       lpain = lo*10(130/10) or lpain = lo*1013

Therefore lpain/lconv = 1013/106 = 1013-6 = 107 .  That is, the intensity at the pain threshold is ten million times greater than at conversation level.

 

See if you can now do part b)

Alan Jun 6, 2014
 #2
avatar+154 
0

This helped so much actually thanks ! I got 65dB for the seccond answer

 Jun 6, 2014

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