How does the vibration of a sphere relate to the pressure field it generates?

[enc08]

Hi,

Attached is the equation relating the vibration of a sphere radius R, to the pressure field it generates. ρ is the density of the medium in which the sphere sits.

The article I got this from just states the equation - I haven't been able to find anywhere that derives this equation.

Any help with how this equation comes about would be appreciated a lot!

Thanks,

enc08

 

[Chestermiller]

Hi,

Attached is the equation relating the vibration of a sphere radius R, to the pressure field it generates. ρ is the density of the medium in which the sphere sits.

The article I got this from just states the equation - I haven't been able to find anywhere that derives this equation.

Any help with how this equation comes about would be appreciated a lot!

Thanks,

enc08

I can guess how this was derived. The only parameter in the equation is the density, so the analysis must involve only inertial forces. The radius of the sphere is oscillating, so this imposes a kinematic boundary condition. Within each radial shell, Newton's second law is satisfied on a differential basis. Maybe they've linearized the equation for small deviations from R, and it isn't clear whether R is the average radius or an actual function of t.

chet

 

[bahamagreen]

Something like this?

 

[enc08]

Thanks. So I've tried to proceed starting from the momentum and continuity equations, and I get quite a close answer, but missing two things as explained below (I did it in an online Latex editor hence it's an image).

 

[mfb]

Small detail: your last equation should have $\dot R^2$ instead of $\dot R$.

I don't see where your momentum conservation comes from. Where did the derivative with respect to r in the pressure vanish?

 

[enc08]

Thanks for noting that.

Re momentum conservation: I make the assumption that the fluid in which the sphere sits is incompressible, so the partial derivative of velocity wrt radius becomes zero. That's how my momentum equation simplifies.

Thanks

 

[mfb]

so the partial derivative of velocity wrt radius becomes zero.

That can't be true (and it is not, if you check your velocity equation). The product v*r^2 is constant for incompressible fluids (I agree with that assumption).

 

[enc08]

I see, I had misread your question and looked at another part (you're right regarding the above).

For the pressure equation, I got rid of the derivative with respect to r by saying that grad(p) is a body force, and the scattered pressure is therefore r * grad(p), so that the scattered pressure Ps is actually Ps = r * grad(P).

I now see why I have an extra 1/r term; I haven't multiplied the momentum equation throughout by r according to my above reasoning. That would remove the extra r in the denominator.

I think my reasoning is right?

If so, that leaves only one outstanding difference to the textbook equation: where does their 1/r^4 term come from?

Thank you :)

 

[Chestermiller]

The easiest way to do this problem is to look up the Euler equations in spherical coordinates and substitute the relationship you obtained from the continuity equation (velocity varying inversely with r₂) into the radial component equation . You then integrate the radial pressure derivative from the sphere to infinity to get the pressure at the sphere.

Chet

 

[Delta2]

Hmmm, i used Newtons'2 2nd law for the volume of fluid between two imaginary spheres of radius r and r+dr and end up in a differential equation for pressure and velocity. Replacing the velocity obtained from the continuity equation and solving it yields an expression for pressure that doesn't include the 1/r^4 term. Whats wrong with this approach anyone can tell me?

 

[Chestermiller]

Hmmm, i used Newtons'2 2nd law for the volume of fluid between two imaginary spheres of radius r and r+dr and end up in a differential equation for pressure and velocity. Replacing the velocity obtained from the continuity equation and solving it yields an expression for pressure that doesn't include the 1/r^4 term. Whats wrong with this approach anyone can tell me?

Please show us the details. The Euler equation is simply:

ρ\left(\frac{∂u}{∂t}+u\frac{∂u}{∂r}\right)=-\frac{∂p}{∂r}

Is this what your Newton's 2nd law equation boils down to? If you use this equation, do you get the "right" answer?

Chet

 

[Delta2]

Please show us the details. The Euler equation is simply:

ρ\left(\frac{∂u}{∂t}+u\frac{∂u}{∂r}\right)=-\frac{∂p}{∂r}

Is this what your Newton's 2nd law equation boils down to? If you use this equation, do you get the "right" answer?

Chet

Ehm nope this isn't the equation i end up with. Apparently my mistake is that i didnt take into count the flow of momentum across the imaginary spheres. The equation you give is verified by the solution given at the OP.