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# What is the answer for question 18) ? http://postimg.org/image/6g52noj6f/

lkballer24 | Student, Grade 11 | Valedictorian

Posted May 28, 2013 at 11:37 PM via web

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What is the answer for question 18) ?

http://postimg.org/image/6g52noj6f/

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llltkl | College Teacher | Valedictorian

Posted May 29, 2013 at 1:21 AM (Answer #1)

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The sensation of loudness is approximately logarithmic to human ear. The unit used to express the acoustic intensity or power, therefore, is logarithmic. Decibel (dB) is a logarithmic unit that indicates the ratio of a physical quantity (usually power or intensity) relative to a specified or implied reference level. A ratio in decibels is ten times the logarithm to base 10 of the ratio of two power quantities.

Thus, the ratio of a power value `I_1` to another power value `I_0` is represented by `L_(dB)` , that ratio expressed in decibels,which is calculated using the formula:

`L_(dB) = 10 log (I_1/I_0)`

The faintest audible sound having acoustic intensity of `1.0 *10^(-12) ` W/m^2 is the reference, `I_0` .

So, for the sound which is assigned 60 dB has actual acoustic intensity I, then

`60 = 10 log(I_1/(1.0 *10^(-12)))`

`=>`` log (I_1/(1.0 *10^(-12))) = 6`

`rArr` ` ``I_1/(1.0*10^-12)=10^6`

`rArr`  `I_1= 10^6*(1.0 *10^(-12))`

= `1.0 *10^(-6)` W/m^2

Now, for the sound assigned 120 dB has actual acoustic intensity `I_2` , then

`120 = 10 log(I_2/(1.0 *10^(-12)))`

`rArr`  `log (I_2/(1.0 *10^(-12))) = 12`

`rArr`  `I_2/(1.0*10^-12)=10^12`

`rArr`

`I_2 = 10^12*(1.0 *10^(-12)) `

`= 1.0` W/m^2

So, `I_2/I_1 = 10^6`

Therefore a 120 dB sound, usually associated with rock concerts, is actually one million times more powerful than a 60 dB sound of a conversation.

This could damage human audible system in a matter of minutes! This is how Jacob would probably be able to convince Anderson to take the ear protection plugs in the concert.

Sources:

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