
#1
Feb2408, 04:32 AM

PF Gold
P: 820

1. The problem statement, all variables and given/known data
Suppose that the sound level of a conversation between two men is initially at an angry 70 dB before dropping to a soothing 50 dB. (a) Assuming that the frequency of the sound is 500 Hz, determine the initial and final sound intensities. (b) What are the initial and final sound wave amplitudes? (c)The men dropped the sound level of their conversation because they noticed an “observer” with a small microphone 200 m away. Assume that the sound waves reaching the “observer” were emitted by a point source. What is the sound intensity of these waves? (d) The microphone held by the “observer” has a crosssectional area of 0.750 cm2. Calculate the power intercepted by the microphone. 2. Relevant equations [tex] \beta = 10log (I/I_o) [/tex] [tex] \wp= 1/2 \rho v (\omega s_{max})^2 [/tex] [tex] I= \wp_{avg}/ A = \wp_{avg} / 4\pi r^2 [/tex] 3. The attempt at a solution a) Assuming that the frequency of the sound is 500 Hz, determine the initial and final sound intensities. [tex]\beta_i = 70dB [/tex] [tex]\beta_f= 50dB [/tex] [tex] I_o= 1.00x10^{12} W/m^2 [/tex] [tex] \beta_i = 10log (I_i/I_o) [/tex] [tex] 70dB= 10log (I_i/(1.00x10^{12}W/m^2)) [/tex] [tex] I_i= 1x10^{5} W/m^2 [/tex] [tex] \beta_f = 10log (I/I_o) [/tex] [tex] 50dB= 10log (I_f /(1.00x10^{12}W/m^2)) [/tex] [tex] I_f= 1x10^{7} W/m^2 [/tex] b) What are the initial and final sound wave amplitudes? I'm not sure what amplitude they are refering to is it: A, [tex]s_{max}[/tex], or even [tex]\Delta P_max [/tex] which on is it?? c)The men dropped the sound level of their conversation because they noticed an “observer” with a small microphone 200 m away. Assume that the sound waves reaching the “observer” were emitted by a point source. What is the sound intensity of these waves? I don't know how to find this and use which equation...confused as to the amplitude and which one they want. (d) The microphone held by the “observer” has a crosssectional area of 0.750 cm2. Calculate the power intercepted by the microphone. I really don't know how to incorperate the cross sectional area and find the power intercepted. Please help me. Thank you 



#2
Feb2408, 05:07 PM

Mentor
P: 40,905





#3
Feb2408, 05:46 PM

PF Gold
P: 820

[itex]s_{max}[/itex] [tex]I= 1/2 \rho v (\omega s_{max})^2 [/tex] Ii= 1x10^5W/m^2 [tex] 1x10^5 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex] [tex] s_{max}= 7.005x10^1m [/tex] If= 1x10^7W/m^2 [tex] 1x10^7 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex] [tex] s_{max}= 7.007x10^1m [/tex] Is the average power found from the distance? (since I have Ii and If ?) As for the distance isn't it the same since the men are not mentioned to move? and the microphone is 200m away? Then I'd find the I ? so I guess it'd be ... I= P/A A= 0.750cm^2 I = ? => I guess I'd find that from above part. P = ? Thanks Doc Al 



#4
Feb2408, 06:21 PM

Mentor
P: 40,905

Waves in a conversation 



#5
Feb2408, 07:11 PM

PF Gold
P: 820

[tex]I= 1/2 \rho v (\omega s_{max})^2 [/tex] Ii= 1x10^5W/m^2 [tex] 1x10^5 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex] [tex] s_{max}= 7.007x10^5m [/tex] If= 1x10^7W/m^2 [tex] 1x10^7 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex] [tex] s_{max}= 7.007x10^6m [/tex] [tex]I= \wp_{average} / A= \wp_{average} / 4 \pi r^2 [/tex] thus in my view... [tex] (Ii + If)/ 2= I_{average} [/tex] thus use that here since I can find A [tex]I= \wp_av / A [/tex] I'm guessing that I'm incorrect but I'm just not getting what you're saying because the speaking men do not move their own position and the microphone is a stationary 200mm away. The men just lower their voice...=( Thank you 



#6
Feb2408, 07:22 PM

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P: 40,905

For example, if the intensity was I_1 at a distance of 10 m, the intensity at 200 m would be I_2 = (10/200)^2*I_1. Make sense? 



#7
Feb2408, 07:31 PM

PF Gold
P: 820

but how would I find the sound intensity at 1 m ? would I use the fact that it was 70dB and then find A using 1m and then use that to find [tex]I_1[/tex]? I assume so and then use that to find the [tex] I_2[/tex] by the ratio.. so I assume that it was measured at 1m? Thanks 



#8
Feb2408, 07:47 PM

Mentor
P: 40,905

Just out of curiosity: What textbook are you using? 



#9
Feb2408, 08:19 PM

PF Gold
P: 820

the intensity I found for 50dB [tex]I_f= 1x10^7W/m^2 [/tex] and I'm lost.....on where I'd use this since I need intensity at r= 200m and I know that [tex]I= \wp_{average} / A = \wp_{average}/ 4 \pi r^2 [/tex] need I but don't have Pav ...basically I'm not sure where I use the I_f = 1x10^7 N/m^2 to find the I for what they ask for.... I really need to learn how to find Pav since that' what getting me confused here basically where do I use the intensity for 50dB? since I need intensity..for 200m Thanks 



#10
Feb2508, 08:39 AM

Mentor
P: 40,905

[tex]P = IA = I_1 4\pi R_1^2 = I_2 4\pi R_2^2[/tex] R_1 = 1 m (we're just guessing); R_2 = 200 m (we're given this). 



#11
Feb2508, 09:30 AM

PF Gold
P: 820

but looking at that ratio I'd deduce that P is the same for both (why do they label it Pav )?!?! thus the average power is just: [tex]P = IA = I_1 4\pi R_1^2 = I_2 4\pi R_2^2[/tex] [tex]P_{average}= I_2 4\pi R_2^2=(1.0x10^7N/m^2)(4 \pi)(200m)^2 = 0.05026W [/tex] But if this is true then the intensity at 200m is the same as I originally calculated (I_f= 1x10^7N/m^2) hopefully I get it now... Thanks alot 



#12
Feb2508, 07:28 PM

PF Gold
P: 820

Okay..well since no one said anything..I'll hope it's correct and use it for the last part.
I_f= 1x10^7N/m^2 since the Intensity is I=P/A A of microphone is: 0.750 cm2 then the power intercepted by it is... P= (1x10^7N/m^2)(7.5 × 105 m^2) P= 7.5e12W That's pretty small. I'm not sure if my thought process is wrong but...is it? Thanks Doc 



#13
Feb2508, 08:13 PM

Mentor
P: 40,905

What I wanted you to do is to take that power equation and find the relationship between the intensities as a function of distance from the source: [tex]I_1 R_1^2 = I_2 R_2^2[/tex] We know I_1 (that's the I_f you had calculated), R_1, and R_2. Find I_2. 



#14
Feb2608, 06:52 AM

PF Gold
P: 820

[tex]I_1R_1^2= I_2R_2^2 [/tex] (1x10^7N/m^2)*(1m)= I_2 (200m) I_2= 5x10^10 N/m^2 d) I= P/A A= 0.750cm^2 = 7.5x10^5 m^2 I= 5x10^10N/m^2 P= 3.75x10^14 W Power intercepted by microphone. 



#15
Feb2608, 07:14 AM

Mentor
P: 40,905





#16
Feb2608, 08:22 AM

PF Gold
P: 820

[tex]I_1R_1^2= I_2R_2^2 [/tex] (1x10^7N/m^2)^2*(1m)= I_2 (200m)^2 I_2= 2.5x10^12N/m^2 d) I= P/A A= 0.750cm^2 = 7.5x10^5 m^2 I= 2.5x10^12N/m^2 P= 1.875x10^16W Thank you Doc Al 



#17
Feb2608, 09:46 AM

Mentor
P: 40,905

Looks good!
In my opinion, this problem left out an essential piece of information which we had to supply (R_1 = 1 m). So if your instructor ever reveals the textbook solution, I'd be curious as to what they use. 


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