Solving Sound Wave Confusion: Finding Intensity at Distance d_0 from Source

In summary, if the sound has an intensity of beta dB at 1 m from the source, it will decrease to 0 dB at a distance of d_0.
  • #1
HeLLz aNgeL
30
0

Homework Statement


In general, if a sound has intensity of beta dB at 1 m from the source, at what distance d_0 from the source would the decibel level decrease to 0 dB? Since the limit of hearing is 1 dB this would mean you could no longer hear it.
Express the distance in terms of beta. Be careful about your signs!

I know that the relationship between the intensity and distance is an inverse square relationship, but I'm not sure what exactly the question is looking for. Do i have to find the distance when it goes to 0 or to 1 ? Because if that is zero, then even after 100m, the intensity wouldn't be zero ?

im confused !
 
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  • #2
anyone ?
 
  • #3
You are right, the intensity won't fall to zero. But there is a log in the relation between dB and intensity.
 
  • #4
so how do i get it to zero ? i mean keep on increasing distance till log of it = 0, and how do i write it in terms of beta ? :S
 
  • #5
Ok, let 'I' be the intensity at 1m and 'I0' be the intensity at 0dB. dB=10*log(I/I0) (all logs base 10). So beta=10*log(I/I0). You know at distance Rm the intensity becomes I/R^2. So 0=10*log((I/I0)/R^2). Now use a property of logs.
 
  • #6
HeLLz aNgeL said:

Homework Statement


In general, if a sound has intensity of beta dB at 1 m from the source, at what distance d_0 from the source would the decibel level decrease to 0 dB? Since the limit of hearing is 1 dB this would mean you could no longer hear it.

im confused !

Why do you say the limit of hearing is 1 dB?
 
  • #7
Dick said:
Ok, let 'I' be the intensity at 1m and 'I0' be the intensity at 0dB. dB=10*log(I/I0) (all logs base 10). So beta=10*log(I/I0). You know at distance Rm the intensity becomes I/R^2. So 0=10*log((I/I0)/R^2). Now use a property of logs.


thanks a ton ! finally got it ... this one was a bugger ! thanks again...
 

1. What is a sound wave?

A sound wave is a type of mechanical wave that is created when an object vibrates and transfers energy through a medium, such as air, water, or solids. This energy travels in the form of longitudinal waves, causing particles in the medium to vibrate and create areas of high and low pressure, which our ears interpret as sound.

2. How does sound intensity change with distance from the source?

According to the inverse square law, sound intensity decreases as the distance from the source increases. This means that the further you are from the source, the weaker the sound will be. This is because the energy from the sound waves spreads out over a larger area as it travels, making it less intense at a greater distance.

3. What is the formula for calculating sound intensity?

The formula for sound intensity at a given distance from the source is I = P/(4πd2), where I is the sound intensity in watts per square meter (W/m2), P is the power of the sound source in watts (W), and d is the distance from the source in meters (m).

4. How can I use the formula to find the sound intensity at a specific distance from the source?

To find the sound intensity at a specific distance, you will need to know the power of the sound source and the distance from the source. Plug these values into the formula I = P/(4πd2) and solve for I. The resulting value will be the sound intensity at that distance.

5. Why is it important to understand sound intensity at different distances from the source?

Understanding sound intensity at different distances is important for a variety of reasons. For example, it can help us determine the safe exposure levels for workers in loud environments, assess the impact of noise pollution on wildlife, and ensure that audio equipment is set up properly for optimal sound quality. It also allows us to better understand the behavior of sound waves and how they interact with their surroundings.

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