Sound Intensity of a howler monkey

1. Dec 4, 2006

Soaring Crane

1. The problem statement, all variables and given/known data
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

2. Relevant 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

3. 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.

Last edited: Dec 4, 2006
2. Dec 4, 2006

Staff: Mentor

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.)

3. Dec 4, 2006

Soaring Crane

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.

Last edited: Dec 4, 2006
4. Dec 4, 2006

Staff: Mentor

Last edited by a moderator: Apr 22, 2017
5. Dec 4, 2006

Soaring Crane

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?

6. Dec 4, 2006

Staff: Mentor

I_0 would be the intensity at 1.4 km--where the sound can barely be heard.

7. Dec 4, 2006

Soaring Crane

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?

8. Dec 4, 2006

Staff: Mentor

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.

9. Dec 4, 2006

Soaring Crane

I divided 49 dB by 120 dB according to the units for W/m^3.

10. Dec 4, 2006

Staff: Mentor

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. Aug 10, 2007

vijaymaruthi

i need to no whether its 49 dbspl or dbsil or swl...

12. Aug 10, 2007

vijaymaruthi

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. Aug 10, 2007

Staff: Mentor

Looks good to me.

14. Feb 4, 2010

Brojito

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