Solving Marine Acoustics Integral: Integrating 2nd Order ODE

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In summary, the book provides an equation for finding the amplitude of a sound ray as a function of the arc length, s. By integrating the equation, we get a solution for A_0(s). However, the book does not provide any intermediate steps and it may be unclear how to perform the integration. To help with this, the book also provides equations for the sound speed (c) and Jacobian determinant (J) at a given arc length s. By using the chain rule and some algebraic manipulation, we can transform the original equation into a form that is easier to integrate, resulting in the solution for A_0(s).
  • #1
Luminous Blob
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I'm trying to follow a derivation in a book on marine acoustics for finding the amplitude along a sound ray, as a function of the arc length, s, of the ray.

The book gives the following equation:

[tex]
2\frac{dA_0}{ds}+ \left[ \frac{c}{J}\frac{d}{ds}\left(\frac{J}{c}\right) \right]A_0 = 0
[/tex]

The book then says that by integrating the above equation, we get:

[tex]
A_0(s)= A_0(0)\left| \frac{c(s)J(0)}{c(0)J(s)} \right|^{1/2}
[/tex]

The book doesn't give any intermediate steps, and I'm not really sure how the integration is actually done. I gather that the limits are from 0 to s, but I don't know how you deal with something like this where J, c and A_0 all seem to depend on s.

In these equations, c is the sound speed at a given arc length s along the ray, and J is the Jacobian determinant, given as:

[tex]
J = r\left[\frac{dr}{ds}\frac{dz}{d\theta} - \frac{dz}{ds}\frac{dr}{d\theta}\right]
[/tex]

or alternatively:

[tex]
J = r\left[\left(\frac{dz}{d\theta}\right)^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{1/2}
[/tex]

So...can anyone explain to me the steps involved in going from the first equation to the second? Any help would be appreciated.
 
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  • #2
You can get it in the form...

[tex]\frac{2}{A_0}\frac{dA_0}{ds}+\frac{1}{J}\frac{dJ}{ds}+\frac{1}{C}\frac{dC}{ds}=0[/tex]

where [tex]C(s)=1/c(s)[/tex]
 
  • #3
J77 said:
You can get it in the form...

[tex]\frac{2}{A_0}\frac{dA_0}{ds}+\frac{1}{J}\frac{dJ}{ds}+\frac{1}{C}\frac{dC}{ds}=0[/tex]

where [tex]C(s)=1/c(s)[/tex]

As a follow up to J77's excellent hint, think of chain rule (this is like reversing chain rule).
 
  • #4
Aah, thanks guys! I see what's happening now.
 

1. How does marine acoustics integral differ from other types of integrals?

Marine acoustics integral is specifically used for solving problems related to underwater sound propagation, while other types of integrals may be used for a variety of mathematical applications.

2. What is the significance of solving 2nd order ODE in marine acoustics integral?

Second order ordinary differential equations (ODEs) are commonly used to model physical systems, including underwater sound propagation. By solving these equations, we can gain a better understanding of the behavior of sound in marine environments.

3. Can marine acoustics integral be used for real-world applications?

Yes, marine acoustics integral is used in a variety of real-world applications, such as predicting the behavior of underwater sound in different environments and designing sonar systems for marine navigation.

4. What are the steps involved in solving a marine acoustics integral?

The first step is to set up the integral and identify the variables involved. Then, the integral is solved using appropriate methods, such as the method of separation of variables or the Laplace transform. The solution is then checked for accuracy and interpreted in the context of the problem.

5. How can marine acoustics integral be applied to environmental conservation efforts?

By understanding how sound behaves in marine environments, we can better monitor and protect underwater ecosystems. Marine acoustics integral can be used to model the effects of noise pollution on marine animals and help mitigate its impact on their habitats.

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