You are standing in the middle of a large room listening to a cacaphony of sound

Question

You are standing in the middle of a large room listening to a cacaphony of sounds. Rank the intensity level of the sounds from each source (1=loudest, 2= next loudest, ...) Consider all objects to be right next to you. If two values are within 0.1 dB from each other, consider them to be the same.

Ten buzzers each producing a sound with intensity level of 80dB.

An object emitting a sound with an intensity level of 90dB.

Two buzzers each producing a sound with intensity level of 87dB.

One hundred buzzers each producing a sound with intensity level of 70dB.

Four buzzers each producing a sound with intensity level of 84dB.

Explanation / Answer

we know the relation between the sound intensity level and the intensity is

S.L = 10 log(I/I_0)       I sound intensity , I_0 is threshold sound intensity is 1*10^-12 W/m^2

( I / I_0 )= e^(B/10)

given sound intensity level are

ten buzzersof each with sound 80 dB = 80*10 = 800 dB ==> ( I / I_0 )= e^(800/10)=e^80

object emitting 90 dB = ( I / I_0 )= e^(B/10) = => ( I / I_0 )= e^(90/10)= e^9

2 buzzers with sound 70 dB =70*2= 140 dB ==> ( I / I_0 )= e^(B/10) ==> ( I / I_0 )= e^(140/10)= e^14

100 buzzers each with sound 70 dB = 70*100 = 7000 ==>( I / I_0 )= e^(B/10) = e^(7000/10)=e^700

4 buzzers each with sound 84 dB = 84*4 =336 dB ==>( I / I_0 )= e^(B/10) = e^(336/10)= e^33.6

Ranling of sond levels is 100 buzzers> 10 buzzers > 4 buzzers >2 buzzers> object

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