"Sound Intensity Decreases with Distance

In summary, The decibel is a unit of measurement used to handle large ratios. The equation for power in dB based on the ratio of intensities is: 10 \log_{10} \left( \frac{P}{A_1}}{\Phi_{ref}} \right)
  • #1
Aoiumi
23
0

Homework Statement


At 1 m away from the source of a sound, the intensity of the sound is 90 dB. At 10 m away the intensity is…?

The answer is 70 dB but I don't know how...

Homework Equations



I = Power/Area


The Attempt at a Solution


My thought was to set up the equations:

90dB = P/4pi
XdB = 4pir^2

I'm not sure where to go from here.
 
Physics news on Phys.org
  • #2
Aoiumi said:

Homework Statement


At 1 m away from the source of a sound, the intensity of the sound is 90 dB. At 10 m away the intensity is…?

The answer is 70 dB but I don't know how...

Homework Equations



I = Power/Area


The Attempt at a Solution


My thought was to set up the equations:

90dB = P/4pi
XdB = 4pir^2

I'm not sure where to go from here.

You are on the right track.

What is the ratio of the surface area of the two spheres?

And what is the equation for power in dB based on the ratio of intensities?
 
  • Like
Likes 1 person
  • #3
The ratio of the surface area of the two spheres would be

4pi/4pi(10)^2 or
1/100

I'm not sure how to set up the quation for power in dB based on the ratio of intensities

I0/I = (p1/a1)/(p2/a2)?
 
  • #4
Aoiumi said:
The ratio of the surface area of the two spheres would be

4pi/4pi(10)^2 or
1/100

I'm not sure how to set up the quation for power in dB based on the ratio of intensities

I0/I = (p1/a1)/(p2/a2)?

You use dB to handle large ratios.

For voltage, V(dB) = 20 * log(V/Vo) where Vo is some reference voltage

For power P(dB) = 10 * log(P/Po) where Po is some reference power

So for this question, you will be using the 2nd equation...
 
  • #5
First, allow me to offer some general guidance regarding decibels.

Decibels always imply power ratios. Always. No matter what.

The decibel is defined, relating the ratio of two powers P2 and P1 as:

[tex] 10 \log_{10} \left( \frac{P_2}{P_1} \right) [/tex]
So one might ask, "well then how can we use decibels with voltage? Voltage isn't power," or "where does the '20' come from when we use ratios of voltages?"

The answer is that we make the pseudo-assumption that the input and output impedances of a given circuit are equal; i.e., Rin = Rout = R. (By the way, this is a realistic pseudo-assumption, since it is often the case for circuits that are optimized for maximum power transfer, signal to noise ratio, etc.)

Also we know that the power through a resistor is V2/R.

Plugging that into our decibel equation, and noting that log(x2) = 2logx,

[tex] 10 \log_{10} \left( \frac{ \frac{V_2^2}{ R}}{ \frac{V_1^2}{ R}} \right) = 10 \log_{10} \left( \frac{V_2^2}{V_1^2} \right) = 10 \log_{10} \left( \left[ \frac{V_2}{V_1} \right]^2 \right) = 20 \log_{10} \left( \frac{V_2}{V_1} \right) [/tex]

----------------------------

Now allow me to move on to something more specific: Sound/acoustics.

Sound decibels are defined in terms of power flux* ratios.

*(Power flux is power through a unit area -- you may think of the unit area as the cross sectional area of your ear canal if it helps visualize it -- or if you want to stick with SI units, the sound power propagating through a square meter.)

The reference power flux is typically defined as the power flux that corresponds to the human threshold of hearing.

If we define the reference power flux, that of the threshold of human hearing, as [itex] \Phi_{ref} [/itex], and the power of the source as P, then we know from the problems statement that

[tex] 90 \ \mathrm{dB} = 10 \log_{10} \left( \frac{\frac{P}{A_1}}{\Phi_{ref}} \right) [/tex]
where [itex] A_1 [/itex] is the surface area of a circle with a 1 meter radius. So now the question is, what is

[tex] 10 \log_{10} \left( \frac{\frac{P}{A_{10}}}{\Phi_{ref}} \right) \ ,[/tex]
where [itex] A_{10} [/itex] is the surface area of a circle with radius 10 meters?

Hint: There is an easy way to do this such that you won't even need a calculator to solve this problem (you might even be able to solve it in your head). Do this by noting that log(x/y) = logx - logy, and asking yourself, Does the area of a circle increase proportionally with the radius, or does it increase proportionally with the square of the radius?" :wink:
 
Last edited:
  • #6
Thanks for the clarification, collinsmark. Nice post :smile:
 
  • Like
Likes 1 person

1. How does sound intensity decrease with distance?

Sound intensity decreases with distance because as sound travels, it spreads out in all directions, causing the sound waves to become less concentrated. This results in a decrease in the amount of energy that reaches our ears, making the sound seem quieter.

2. What is the relationship between distance and sound intensity?

The relationship between distance and sound intensity is inverse. This means that as distance increases, sound intensity decreases. This is due to the spreading out of sound waves as they travel, resulting in a decrease in amplitude and energy.

3. How is sound intensity measured?

Sound intensity is measured in decibels (dB). This unit is used to quantify the loudness of a sound, with higher decibel levels corresponding to louder sounds. Sound intensity can also be measured using sound level meters, which measure the pressure of sound waves in a given area.

4. What factors can affect the decrease in sound intensity with distance?

Several factors can affect the decrease in sound intensity with distance. These include the type of sound source, the medium through which the sound travels, and any obstacles or obstructions in the path of the sound waves. The temperature and humidity of the environment can also affect the speed and intensity of sound waves.

5. How can sound intensity be increased with distance?

In order to increase sound intensity with distance, the sound source must have a higher initial intensity or be placed closer to the listener. Additionally, using amplifiers, reflectors, or other acoustic technologies can help to increase sound intensity over longer distances. However, these methods may also affect the quality and clarity of the sound being transmitted.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
789
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top