Sound intensity and the hearing distance

In summary: I used the following formula to calculate the power:P= I* Area → P= 4πr²*IThe intensity used in this formula is the one that I calculated earlier, I= 1.02 x 10^-6 W/m²So the power is :P= 1.02 x 10^-6 W/m² * 4π(22.3)^² m²P= 0.025 WIs this correct?And for the distance calculation, I used the intensity that I calculated earlier, I= 1.02 x 10^-6 W/m², and the limiting intensity, I= 10^-12 W/m².Is this correct?Yes,
  • #1
mmoadi
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Homework Statement



A tiny speaker emits sound with a frequency of 1000Hz evenly on all sides. On
a distance of 1m from the loudspeaker is an amplitude variation of the particles of air 0.005 μm. On what distance we can barely hear the sound, if the limit of the audible strength or intensity is 10^-12 W/m2? The speed of sound is 340m/s, air density is 1.2 kg/m3. What is the sound power of the loudspeaker at a distance of 1m?


Homework Equations



I= P/Area

The Attempt at a Solution



For the second part: What is the sound power of the loudspeaker at a distance of 1m?

I= P/ 4πr² → P= 4πr²*I
P= 1.256 x 10^-11 W

Can someone please give me a hint for the first part?
Thank you for helping!
 
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  • #2
mmoadi said:

The Attempt at a Solution



For the second part: What is the sound power of the loudspeaker at a distance of 1m?

I= P/ 4πr² → P= 4πr²*I
P= 1.256 x 10^-11 W
I don't think this is correct, because I = 10-12 W/m2 is not the sound intensity at 1 m. It is the limiting sound intensity below which humans can't hear anything, which in this case will occur at some unknown distance from the source that you are supposed to solve for. What you need is a relation that will tell you the intensity in a sound wave at a given density, sound speed, and amplitude of oscillation (all of which you have been given). I'm sure that this relation must have been derived somewhere in your notes or your book, right?
 
  • #3
You need the formula that relates intensity of sound to the wave's parameters:
I =C f^2*rho*A^2
Where rho is density, f is frequency, A is the amplitude (maximum displacement).
C is a constant, I don;t remember the exact form. It shoul be in your book.
 
  • #4
So, the intensity of the tiny speaker is:
I= C*f²*ρ*A²
I= 340 m/s * (1000 Hz)² * 1.2 kg/m³ * (0.0000005 m) ²
I= 1.02 x 10^-6 W/m²

Is this correct?

But how do I continue now? I still have to calculate the distance at which we can barely hear the sound. Help?!
 
  • #5
mmoadi said:
So, the intensity of the tiny speaker is:
I= C*f²*ρ*A²
I= 340 m/s * (1000 Hz)² * 1.2 kg/m³ * (0.0000005 m) ²
I= 1.02 x 10^-6 W/m²

Is this correct?


Probably not! So you just used the formula that nasu gave you? :rolleyes: The point of what both of us were saying to you was that you should look it up in your book or notes. He didn't give you the exact form because wasn't 100% sure of it, and he specifically said that you should find out exactly what the relation is. Also, the "C" he wrote was just meant to be some sort of constant, not the speed of sound. He said as much in his post. You need to read things more carefully.

As for the second part, calculating the distance at which the sound becomes inaudible, you already *know* how intensity varies with distance. You wrote the equations for this in your first post.
 
  • #6
Here we go again:

I looked it up and Found the equation for the sound intensity is: I= ½ ρ*v*ω²*A², where ρ is the density of the air, v is the velocity of the sound, ω is the angular frequency, and A is the amplitude of the air oscillations

I= ½ ρ*v*ω²*A²
I= ½ (1.2 kg/m³)*(340 m/s)*(2π*1000 Hz)²*(0.0000005 m)²
I= 2.01 x 10^-3 W/m²

Is this calculation for the intensity of the tiny speaker now correct?

And I know that the relationship between the intensity and the distance is: I → 1/d² (inverse square law relationship), but I still can’t figure out how to calculate this distance.
 
  • #7
So, if I understood the relationship correctly, this is how to calculate the distance:

I= x/r² → r= sqrt(x/I)
r= 22.3 m

And for the second part: What is the sound power of the loudspeaker at a distance of 1m?

I= P/Area
P=I* Area → P= 4πr²*I
P= 0.025 W

Are my calculations correct?
Thank you for helping!:smile:
 
  • #8
mmoadi said:
So, if I understood the relationship correctly, this is how to calculate the distance:

I= x/r² → r= sqrt(x/I)
r= 22.3 m

Depends what you used for x.
Here x should be the power of the source. It is NOT the intensity that you calculated.

You don't really need this power.
You know the intensity at 1m and the intensity at some unknown distance r.
The ratio between intensities is same as the inverse ratio of the distances squared.
You coul also calculate the power of the source and go for there. Maybe this is what you have done?
 

FAQ: Sound intensity and the hearing distance

How does sound intensity affect hearing distance?

Sound intensity refers to the amount of energy that a sound wave carries per unit area. As sound travels, it spreads out and becomes less intense. This decrease in intensity affects the hearing distance, as the further away you are from the source of the sound, the lower the intensity will be, making it harder to hear.

What is the relationship between sound intensity and decibels?

Decibels (dB) are a unit of measurement used to quantify sound intensity. The higher the decibel level, the louder the sound. Sound intensity is measured on a logarithmic scale, meaning that a small increase in decibels represents a large increase in sound intensity. For every 10 dB increase, the sound intensity doubles.

How does distance affect sound intensity?

As sound waves travel through the air, they spread out and decrease in intensity. This means that the further away you are from the source of the sound, the lower the sound intensity will be. In fact, sound intensity decreases by a factor of four for every doubling of distance.

What is the threshold of hearing and how does it relate to sound intensity?

The threshold of hearing is the lowest sound intensity that can be detected by the human ear. It is measured at 0 decibels and represents the minimum amount of energy required to stimulate the auditory system. As sound intensity increases above the threshold of hearing, the sound becomes louder and easier to hear.

How does sound intensity affect the perception of loudness?

Loudness is a subjective perception of sound intensity. It is influenced by various factors, such as the frequency of the sound and the sensitivity of an individual's hearing. However, in general, the higher the sound intensity, the louder the sound will be perceived. This is why a sound at 100 dB will sound much louder than a sound at 50 dB, even though it is only a 10 dB difference in intensity.

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