Intensity of a sound wave problem

[Chase11]

Homework Statement

Two loudspeakers are placed beside each other and produce sound of the same intensity at the position of a listener. One speaker produces a low note of 40 Hz and the other produces a high note of 2560 Hz. What is the ratio of the maximum displacements of the speakers vibrating cones?

Homework Equations

1) $I$=P/4piR^2
2) I=1/2BωkA^2
3) I=1/2$\sqrt{ρB}$ω^2A^2

The Attempt at a Solution

I understand that I am supposed to use equation 3 for both frequencies and set them equal to each other to come up with a ratio. I just don't understand how equation 3 is derived from equation 1, or how equation 2 is derived from equation 1 for that matter. If I could see how to manipulate these equations I would understand this type of problem much better. (The only equation I am given on my equation sheet is the first one).

 

[Simon Bridge]

Well what do each of the terms in the equations mean?

 

[Chase11]

Well I is intensity of the wave, p is the pressure, r is radius, ω is angular frequency, k is 2pi/$\lambda$, A is amplitude, and ρ is density. Right?

 

[Simon Bridge]

Well I is intensity of the wave, p is the pressure, r is radius, ω is angular frequency, k is 2pi/λ, A is amplitude, and ρ is density. Right?

This is incomplete, and you have not been consistent in your notation.

Taken in order:
- Intensity of the sound wave - good;
- there is no "p" in your equations. Do you mean "P" here?
$\qquad$... sound is a pressure wave, so there are lots of pressures all over the place so which pressure does P refer to? Or is that P for "power"?
- there is no "r" in your equations, do you mean "R"? What is R the radius of?
- $\small{\omega}$ = angular frequency of the wave
$\qquad \small{\omega = vt = 2\pi f}$ where v is the wave-speed and f is the frequency of the wave. $\small{k(x-vt)=kx-\omega t}$
- $\small{k=2\pi/\lambda}$ good, it's called the wave number.
- What is A the amplitude of
$\qquad$- if "the sound wave" then is it a pressure or a displacement or something else?
- there are lots of different kinds of density - what is $\small{\rho}$ the density of?
- what is B? You missed it out.

If you don't know what the terms refer to then you won't be able to understand the equations.
I think you need to check your ideas about what sound intensity means:
http://en.wikipedia.org/wiki/Sound_intensity
... the intensity of the sound is the rate that energy is delivered to the listeners location per unit area.
Energy rate = energy per unit time = Power, so $I=P/A$ i.e. is power per unit area.
Revisit the equation list in post #1 with that in mind.

 

[HalfLostAndSearching]

Firstly, it is important to note that the equations provided are for different scenarios and cannot be directly compared. Equation 1 calculates the intensity of a sound wave at a certain distance from a single source, while equations 2 and 3 are for calculating the intensity of a sound wave produced by a vibrating object (such as a loudspeaker).

Now, for this problem, we can use a different approach to find the ratio of the maximum displacements of the speakers' vibrating cones. We know that the intensity of a sound wave is directly proportional to the square of the amplitude (maximum displacement) of the source. Therefore, we can set up the following equation:

I1/I2 = (A1/A2)^2

Where I1 and I2 are the intensities of the low and high notes respectively, and A1 and A2 are the corresponding amplitudes of the speakers' vibrating cones.

Since both speakers produce sound at the same intensity at the listener's position, we can equate the two intensities:

I1 = I2

Substituting this into our equation, we get:

1 = (A1/A2)^2

Taking the square root of both sides, we get:

1 = A1/A2

Therefore, the ratio of the maximum displacements of the speakers' vibrating cones is 1:1. This means that the amplitudes of the speakers' vibrating cones are equal, regardless of the frequency of the sound they produce.