Really need help...by Jayhawk1 Tags: None 

#1
Apr1805, 07:25 PM

P: 44

I tried getting some help before... but I couldn't figure out the logs if anyone could help me (especially with the manipulation of the logs) I'd appreciate it. Here's the problem:
11) [3.0/4.0] A tornado warning siren on top of a tall pole radiates sound waves uniformly in all directions. At a distance of 15 m the sound intensity of the siren is 0.39 . Neglect any effects from reflection of the sound waves from the ground. a) At what distance from the siren is the intensity 0.19 ? b) What is the total acoustical power output of the siren? c) At what distance is the sound intensity reduced by 15 dB from its level at 15 m? Now I did get parts A and B: a) 21.5 m b) 1100 Watts Now.. part C I can not get at all, and the help I have had has not made sense to me. Please help. Thanks! 



#2
Apr1805, 07:34 PM

HW Helper
P: 2,566

The decibels of a magnitude is defined in relation to some base 0 dB level:
[tex] V(dB) = 20 log(V/V_0) [/tex] But this will cancel out when finding the difference between the the decibel level of two magnitudes: [tex]V_2(dB)  V_1(dB) = 20 log(V_2/V_0)  20 log(V_1/V_0) = 20 log((V_2/V_0)\cdot(V_0/V_1))=20 log(V_2/V_1)[/tex] 



#3
Apr1805, 07:38 PM

Sci Advisor
HW Helper
P: 3,033

You have the equation for this, I know. When you have something in a log that you have to figure out, you can eliminate the log by isolating it and using both side of the equation as a power of 10. The log is the inverse of 10 to a power
a = b log(c) (a/b) = log(c) 10^(a/b) = c Use this on the db level equation you already have Specifically, [tex] L = 10log({\frac{I}{I_0}}) [/tex] [tex] 15/10 = log({\frac{I}{I_0}}) [/tex] [tex] 10^{1.5} = {\frac{I}{I_0}} [/tex] [tex] I_0 10^{1.5} = I [/tex] You have the intensity at one point [tex] I_0 [/tex]. Now you can find it at the second point. Then you can calculate the distance you need to find. The factor 20 or 10 in the dB equation depends on whether you are talking about amplitude or energy. In your case, you are working with intensity (energy/area) so the factor is 10. 



#4
Apr1805, 07:48 PM

P: 44

Really need help...
This doesn't make any sense to me. How does intensity work into the equation?




#5
Apr1805, 07:50 PM

Sci Advisor
HW Helper
P: 3,033

See the addition to the previous reply
From above [tex] I_0 10^{1.5} = I [/tex] [tex] .39 * 10^{1.5} = I [/tex] [tex] .0123 = I [/tex] Now it is just like part a) 


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