Isotropic loudspeaker - calc based

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In summary, the loudspeaker system emits sound isotropically with a frequency of 2.00 x 103 Hz and an intensity of 1.00 x 10-3 W/m2 at a distance of 7.00 m. Assuming there are no reflections, the displacement amplitude at 7.00 m is 1.91 m, the pressure amplitude is 1.74 MPa, and the intensity at 25.0 m is .92 W/m2.
  • #1
sapiental
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1) Isotropic loudspeaker A certain loudspeaker system emits sound isotropically
with a frequency of 2.00 x 103 Hz and an intensity of 1.00 x 10-3 W/m2 at a distance of 7.00 m.
Assume there are no reflections. Use 344 m/s for the velocity of sound in air and 1.21 kg/m3 for
the density of air.
a) What is the displacement amplitude at 7.00 m?
b) What is the pressure amplitude at 7.00 m?
c) What is the intensity at 25.0 m?

Hi, does the term isotropic change the way I calculate the displacement amplitude or does the wave still propagate in the +x direction?

I want to use these formulas,

v = sqrt(B/p) B is the bulk modulus and p is the density.
v = (f)(wavelength)

I = 1/2(sqrt(pB)(w^2)(A^2) = (p_max^2)/(2pv)

= (p_max^2)/(2(sqrt(pB))

for the intensity formula denoted I. p = density and p_max is the pressure amplitude which is directly proporional to A, the displacement amplitude.


please let me know if I can use these formulas for this case.. thank you!
 
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  • #2
Isotropic just means that the sounds travels equally in all directions, so the intensity will drop off as 1/r^2.

Those equations are OK, but you should be able to write the intensity as a function of pressure amplitude or displacement amplitude without needing the bulk modulus. (Since you have density and speed as given.)
 
  • #3
do you mean something like this

p_max(x) = sqrt((I)(2pv))sin(kx)
 
Last edited:
  • #4
No. Start with your equation:
sapiental said:
I = 1/2(sqrt(pB)(w^2)(A^2) = (p_max^2)/(2pv)

The right hand term gives you intensity as a function of pressure amplitude.

To find intensity as a function of displacement amplitude, eliminate "sqrt(pB)" from the middle term.

Neither equation will have B.
 
  • #5
ok,

a ) 1.00 x 10-3 W/m2 = (p_max^2)/(2(1.21 kg/m3)(344m/s)
(p_max) = sqrt((1.00 x 10-3 W/m2)(2(1.21 kg/m3)(344m/s))
= .91m

b) then the equation just becomes I = (w^2)(A^2) = y_max?

w = vk
k = 2pi/wavelength
v = (f)wavelenth
 
  • #6
sapiental said:
a ) 1.00 x 10-3 W/m2 = (p_max^2)/(2(1.21 kg/m3)(344m/s)
(p_max) = sqrt((1.00 x 10-3 W/m2)(2(1.21 kg/m3)(344m/s))
= .91m
That's pressure amplitude, not displacement amplitude. (Units!)

b) then the equation just becomes I = (w^2)(A^2) = y_max?
Huh? y_max is A! You already had the equation:
I = 1/2(sqrt(pB)(w^2)(A^2)
Just replace sqrt(pB) to eliminate B from the expression. (Hint: v = sqrt(B/p)... so what's sqrt(pB)?)
 
  • #7
hmm

I = 1/2pv(w^2)(A^2)

?

thank you!
 
  • #8
sapiental said:
hmm

I = 1/2pv(w^2)(A^2)

?
That's the one. (Relate w to frequency and you're done. Then you can solve for A.)
 
  • #9
w = vk
k = 2pi/wavelength
v = (f)wavelenth

ok i just want to make sure that this is right,

v/f = wavelength
344/2.00 x 10^3 = .172m

so k = 2pi/.172 = 36.53

w = vk = 344 * 36.53 = 12566.3

so (A) = sqrt((1.00 x 10-3 W/m2)/(1/2(344m/s)(1.21kg/m^3)(12566.3^2)))
A = 1.74 * 10 ^-7

this seems wrong..
 
  • #10
I didn't check over your last calculations, but frequency and omega are related by: [itex]\omega = 2 \pi f[/itex]. Combine that with your equation from post #7.
 

Related to Isotropic loudspeaker - calc based

1. What is an isotropic loudspeaker?

An isotropic loudspeaker is a type of speaker that emits sound equally in all directions. This means that the sound waves produced by the speaker have the same intensity and quality in all directions, resulting in a more immersive and balanced sound experience.

2. How does an isotropic loudspeaker work?

An isotropic loudspeaker uses a combination of multiple drivers and a spherical enclosure to produce sound waves that spread out evenly in all directions. This is achieved through careful placement and design of the drivers and enclosures to ensure that the sound waves cancel out any directional biases.

3. What are the advantages of using an isotropic loudspeaker?

An isotropic loudspeaker offers a more natural and immersive sound experience, as the sound is evenly distributed in all directions. This can be especially beneficial in larger spaces or rooms with irregular shapes, where traditional speakers may have difficulty filling the entire space with sound.

4. How is an isotropic loudspeaker calculated?

The calculation for an isotropic loudspeaker is based on the principles of sound wave propagation and the design of the speaker's drivers and enclosures. This involves mathematical equations and simulations to determine the optimal placement and design for the speaker components to achieve isotropic sound dispersion.

5. Can an isotropic loudspeaker be used for all types of audio?

Yes, an isotropic loudspeaker can be used for a variety of audio applications, including music, movies, and live performances. However, it is important to note that the sound quality and effectiveness of the speaker may vary depending on the specific audio source and environment it is being used in.

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