Sound Intensity of a howler monkey

AI Thread Summary
The discussion centers on calculating the distance at which the sound intensity of a howler monkey's call is 49 dB, given its loudness can be heard up to 1.4 km. Participants emphasize the use of the inverse square law and the conversion of decibels to intensity ratios. The correct formula for sound intensity in decibels is highlighted, along with the need to determine the reference intensity level. Confusion arises regarding the values used for intensity and the application of the formulas, but a solution approach is confirmed to yield a distance of approximately 5.0 meters for the given intensity level. The conversation concludes with a similar problem involving a different intensity level, demonstrating the application of the same principles.
Soaring Crane
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Homework Statement


The howler monkey is the loudest land animal and can be heard up to a distance of 1.4 km. Assume the acoustic output of a howler to be uniform in all directions. The distance at which the intensity level of a howler's call is 49 dB, in SI units, is closest to:

a. 6.0---b.9.9----c. 8.4----d.7.5---e. 5.0


Homework Equations


The only formula that I think is adequate in the inverse squares law in which:

I/I_o = r_2^2/r_1^2, where r is the distance and I is intensity in W/m^2



The Attempt at a Solution



I only know of two values being given:
I = 49 dB = 0.4083 W/m^2
r_1 = 1.4 km

Is the third value I_o suggested anywhere in the problem? How do I use the above formula?

Thanks.
 
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Soaring Crane said:

Homework Equations


The only formula that I think is adequate in the inverse squares law in which:

I/I_o = r_2^2/r_1^2, where r is the distance and I is intensity in W/m^2
Yes, you'll need the inverse square law. But you'll also need to know how to go from decibels to intensity ratios. Look up the definition of decibel.

Also: Does the problem state 29 or 49 db? (I suspect that the 29 db was a typo.)
 
Sorry, it's 49 dB.

I converted the dB into W/m^2, but I don't know what is meant by corresponding it to the intensity ratio.
 
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Let me retype the law again:

I_1/I_2 = r_2^2/r_1^2

The reason why I was confused before is that must I use I_o for I_1?

I_o is standarad, so, since another I value is not given, must I use I_o?
 
I_0 would be the intensity at 1.4 km--where the sound can barely be heard.
 
This is what I did so far:

I_1= 10^-12 W/m^2
I_2 = .4083 W/m^2
r_1 = 1.4 km
r_2 = ?

sqrt{(10^-12)/(.4083)W/m^2]*(1400 m)^2} = r_2

r_2 = .00219 km??/

What did I do wrong?
 
Soaring Crane said:
This is what I did so far:

I_1= 10^-12 W/m^2
I_2 = .4083 W/m^2
Where did you get that value for I_2?

All you really need is the ratio of the intensities. That's where the dB value comes in.
 
I divided 49 dB by 120 dB according to the units for W/m^3.
 
  • #10
Soaring Crane said:
I divided 49 dB by 120 dB according to the units for W/m^3.
Sorry, but I don't understand this.

Did you look at the link I gave you in post #4? The equation that you need (in addition to the inverse square law that you already know) is the definition of the decibel scale of intensity:
I(in dB) = 10 log(I/I_0)
 
  • #11
i need to no whether its 49 dbspl or dbsil or swl...
 
  • #12
this is just an attempt takin 0dbspl i.e., the threshold of hearing at the 1.4th km or 1400m... wil get 4.96m .. i.e., approx option (e)5...this can be done using the inverse square law...
L2=L1-20logD2/D1 where L2=49dbspl, L1=0dbspl, D2=1400 and D1 is got as 4.96.. i hope this is rite
 
  • #13
Looks good to me.
 
  • #14
Hey, I am taking Physics II and this is a problem on an old exam. Follow this setup with your own numbers and you should be golden

Question:
The howler monkey is the loudest animal and can just be heard at a distance of 3.6km (threshold of hearing s 10^-12 /m^2). Assume the acoustic output of a howler monkey to be uniform in all directions. How far are you from the monkey if the intensity level is 28 dB at your location?

a: 210m
b: 140m
c: 290m
d: 170m
e: 240m

Solution:

1st Step: Definition of Intensity Levels
/beta=10dBlog(I/I_o) ---> rearrange: I=I_o(10^(/beta/10))---> I=(1x10^-12)(10^2.8)= 6.31x10^-10 W/m^2

2nd Step: Inverse Law

(I/I_0)=(r_1^2/r_2^2)--->rearrange: r_2=Square((r_1^2)(I_o)/(I))
I=6.31x10^-10 W/m^2
I_o= 1x10^-12 W/m^2
r_1= 3.6km~~3600m

r_2=Square((3600m^2)(1x10^-12 W/m^2)/(6.31x10^-10 W/m^2))
r_2=146.5m

Answer is closest to (B)
 
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