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Waves in a conversation

by ~christina~
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~christina~
#1
Feb24-08, 04:32 AM
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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 cross-sectional 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 cross-sectional 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
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Doc Al
#2
Feb24-08, 05:07 PM
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Quote Quote by ~christina~ View Post
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]
Looks good.

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??
I'd say they are looking for the displacement amplitude, what you call [itex]s_{max}[/itex].


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.
Sound intensity drops off as the inverse square of the distance from the source. (Look at the power vs intensity equation.) What's unclear to me is the distance from the men when the power level is 50 db.

(d) The microphone held by the “observer” has a cross-sectional 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.
Intensity is power/area.
~christina~
#3
Feb24-08, 05:46 PM
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Quote Quote by Doc Al View Post
I'd say they are looking for the displacement amplitude, what you call [itex]s_{max}[/itex].
Oh okay I got the answer for both but wasnt' sure so I didn't post it.
[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]



Sound intensity drops off as the inverse square of the distance from the source. (Look at the power vs intensity equation.) What's unclear to me is the distance from the men when the power level is 50 db.
true however I forgot to mention what has me confused was the average power as the equation is [tex]I= \wp_{average} / A= \wp_{average} / 4 \pi r^2 [/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 ?

Intensity is power/area.
Oh darn..I thought that the A they had in the book was Amplitude ...they didn't specify what it was...now I remember my prof saying something about this.. and it messed me up on the above part as well. (I'm writing that in my book now)

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

Doc Al
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Feb24-08, 06:21 PM
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Waves in a conversation

Quote Quote by ~christina~ View Post
Oh okay I got the answer for both but wasnt' sure so I didn't post it.
[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]
Redo these calculations, paying better attention to the exponents.

true however I forgot to mention what has me confused was the average power as the equation is [tex]I= \wp_{average} / A= \wp_{average} / 4 \pi r^2 [/tex]

Is the average power found from the distance? (since I have Ii and If ?)
If you knew the distance and the intensity, then you could calculate the average intensity. (Assuming a point source.)
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 ?
The problem is that you're not given the initial distance from the men. The sound intensity is given as 50 db. But where? 1 m away? 100 m away? Lacking any info, just for grins I'd assume that they are using a distance of 1 m.

Oh darn..I thought that the A they had in the book was Amplitude ...they didn't specify what it was...now I remember my prof saying something about this.. and it messed me up on the above part as well. (I'm writing that in my book now)

so I guess it'd be ... I= P/A
A= 0.750cm^2
I = ? => I guess I'd find that from above part.
P = ?
Right.
~christina~
#5
Feb24-08, 07:11 PM
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Quote Quote by Doc Al View Post
Redo these calculations, paying better attention to the exponents.
fixed.

[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]

If you knew the distance and the intensity, then you could calculate the average intensity. (Assuming a point source.)The problem is that you're not given the initial distance from the men.
The sound intensity is given as 50 db. But where? 1 m away? 100 m away? Lacking any info, just for grins I'd assume that they are using a distance of 1 m.
But I'll say it again..isn't the distance from the microphone the same? (200m) from the begining and thus since I have Ii and If from when they went from 70dB to 50dB while staying in the same spot thus I'm not understanding what your saying....

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
Doc Al
#6
Feb24-08, 07:22 PM
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Quote Quote by ~christina~ View Post
fixed.

[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]
Much better.
But I'll say it again..isn't the distance from the microphone the same? (200m) from the begining and thus since I have Ii and If from when they went from 70dB to 50dB while staying in the same spot thus I'm not understanding what your saying....

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...=(
It's certainly true that the distance to the microphone doesn't change. What we need to know is how far away is the microphone compared to the where the 50 db level was measured.

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?
~christina~
#7
Feb24-08, 07:31 PM
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Quote Quote by Doc Al View Post

It's certainly true that the distance to the microphone doesn't change. What we need to know is how far away is the microphone compared to the where the 50 db level was measured.

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?
Oh...well it sort of makes sense but I don't get how you got that equation...but it looks familiar..(for some reason or another)

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
Doc Al
#8
Feb24-08, 07:47 PM
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Quote Quote by ~christina~ View Post
Oh...well it sort of makes sense but I don't get how you got that equation...but it looks familiar..(for some reason or another)
It comes from the power equation. The average power doesn't change as you get further away from the source, but it's spread out over a greater area ([itex]4\pi r^2[/itex]). And intensity is power/area.

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?
Assuming that is better than doing nothing. But you are given the intensity (or at least the dB level); it's the same numbers that you used in part 1. We are just assuming that it was measured at 1 m. I'd use the intensity for the 50 db sound level.

Just out of curiosity: What textbook are you using?
~christina~
#9
Feb24-08, 08:19 PM
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Quote Quote by Doc Al View Post
It comes from the power equation. The average power doesn't change as you get further away from the source, but it's spread out over a greater area ([itex]4\pi r^2[/itex]). And intensity is power/area.

Assuming that is better than doing nothing. But you are given the intensity (or at least the dB level); it's the same numbers that you used in part 1. We are just assuming that it was measured at 1 m. I'd use the intensity for the 50 db sound level.
oh my goodness...I'm mixing up all the intensities and powers.....basically getting confused here...lets see..

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

Just out of curiosity: What textbook are you using?
I'm using or attempting to use...Physics for Scientists and Engineers 7th edition by Serway & Jewett


Thanks
Doc Al
#10
Feb25-08, 08:39 AM
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Quote Quote by ~christina~ View Post
oh my goodness...I'm mixing up all the intensities and powers.....basically getting confused here...lets see..

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 or so I think..basically I'm not sure where I use the I_f to find the I for the 200m.....
Read this: Inverse Square Law for Sound

[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).
I'm using or attempting to use...Physics for Scientists and Engineers 7th edition by Serway & Jewett
I think I have the 6th edition floating around somewhere. Not a bad book.
~christina~
#11
Feb25-08, 09:30 AM
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Quote Quote by Doc Al View Post
Read this: Inverse Square Law for Sound

[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).
Originally I thought: average usually mean that you divide by 2? thus I thought that you would find P for I_i and P for I_f and then add the 2 P's then divide by 2 and that would be your average...

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...

I think I have the 6th edition floating around somewhere. Not a bad book.
I'd have to disagree for the 7th edition. The 7th edition (I don't know about the 6th) is particularly bad for projectile motion. (They give 2 example which isn't like any problem in the end of chapter questions) I've looked at another physics text and they give numerous useful examples for projectile motion which is better than this book in my opinion.

Thanks alot
~christina~
#12
Feb25-08, 07:28 PM
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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 × 10-5 m^2)

P= 7.5e-12W

That's pretty small.

I'm not sure if my thought process is wrong but...is it?

Thanks Doc
Doc Al
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Feb25-08, 08:13 PM
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Quote Quote by ~christina~ View Post
Originally I thought: average usually mean that you divide by 2? thus I thought that you would find P for I_i and P for I_f and then add the 2 P's then divide by 2 and that would be your average...

but looking at that ratio I'd deduce that P is the same for both (why do they label it Pav )?!?!
Yes, it's confusing. They mean for you to assume the power is uniformly emitted in all directions.

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]
No, you used R_2 (200 m) but assumed that I_2 was 1.0x10^-7N/m^2. But I_2 is what you are asked to find! (You didn't need to calculate the power, just relate the two intensities.)

But if this is true then the intensity at 200m is the same as I originally calculated (I_f= 1x10^-7N/m^2)
No. That I_f is the intensity at some unknown distance close to the men--certainly not 200 m away (that's where the microphone is). We are assuming that unknown distance is 1 m.

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.

I'd have to disagree for the 7th edition. The 7th edition (I don't know about the 6th) is particularly bad for projectile motion. (They give 2 example which isn't like any problem in the end of chapter questions) I've looked at another physics text and they give numerous useful examples for projectile motion which is better than this book in my opinion.
Sorry to hear that! (I've read some of Jewitt's pedagogical articles and thought they were pretty good. But those are meant for teachers, not students!)
~christina~
#14
Feb26-08, 06:52 AM
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Quote Quote by Doc Al View Post
Yes, it's confusing. They mean for you to assume the power is uniformly emitted in all directions.
I'll remember that.

No, you used R_2 (200 m) but assumed that I_2 was 1.0x10^-7N/m^2. But I_2 is what you are asked to find! (You didn't need to calculate the power, just relate the two intensities.)


No. That I_f is the intensity at some unknown distance close to the men--certainly not 200 m away (that's where the microphone is). We are assuming that unknown distance is 1 m.

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.
Oh...yep I didn't get it before.

[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.

Sorry to hear that! (I've read some of Jewitt's pedagogical articles and thought they were pretty good. But those are meant for teachers, not students!)
Doc Al
#15
Feb26-08, 07:14 AM
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Quote Quote by ~christina~ View Post
Oh...yep I didn't get it before.

[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
Careful. Those distances from the source should be squared. (The famous inverse square law.)
~christina~
#16
Feb26-08, 08:22 AM
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Quote Quote by Doc Al View Post
Careful. Those distances from the source should be squared. (The famous inverse square law.)
corrected.

[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
Doc Al
#17
Feb26-08, 09:46 AM
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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.
~christina~
#18
Feb26-08, 04:21 PM
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Quote Quote by Doc Al View Post
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.
My instructor posts the answers online after we do the problems so I'll let you know when I find out.

Thanks for your help Doc.


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