One form of the equation is this: There's a simpler form, which is not applicable in your case as it requires that M1 and M3 are identical, and together are identical with M2, as well as D1 being identical to D3. If those conditions are met, then:Waka wrote: Hi Stuart, that's very interesting. Can I ask which formulae you use for your 3 leaf systems? I'm struggling to find a good one.
f+ = (1 / (2 x pi) ) x SQRT( (3.6 x Rho x c^2) / ( m x d ) )
f- = f+ / ( SQRT(2) )
Where:
Rho= air density
c= speed of sound in air
m=the mass of leaf 1 (or leaf3) by itself (or half of M2, if you prefer)
d=the gap between M1 and M2 (which is the same as the gap between M2 and M3)
This does NOT apply to your situation for another reason too: it assumes that the spring is air. In your case, the spring is rubber, so you have to replace that term with the spring constant for your rubber.
An interesting fact: in this configuration, the equivalent 2-leaf system (having the same total mass and total thickness), provides BETTER isolation than the 3-leaf system, up to a frequency that is FOUR TIMES the MSM resonant frequency of the 2-leaf system. Above that point, 3 leaf is better, but below that point 3 leaf is worse, and 2 leaf wins. This is why it's never a good idea for studios to have 3-leaf systems, since you need much greater mass and cavity depth, just to get the same isolation for low frequencies.
Another interesting fact: For this configuration f+ will always be exactly twice the MSM resonant frequency of the equivalent 2-leaf system. So if you can't be bothered to do the three-leaf math, then you can just figure out f0 for the 2-leaf, multiply by 2 to get f+, then work backwards from there to get f-. That's the simplest method... but it's only really valid for this specific configuration, where m1=m3= (m2*0.5) =m, and d1=d2=d. For your system, where those conditions are not true, you have to use the full equations.
The math gets a little more complicated here.
- Stuart -