Ape wrote:This is propably a rather silly question, showing my complete lack of understanding of what you're trying to relate here, but, well, here goes ; How can dfr be >1?
What did I miss?

Hi Ape,
No harm done in asking.
For the sake of illustration, let's assume my pump runs at 1800 rpm.
f = (1800 rev/min) / (60 sec/min) gives 30 rev/sec
During 1 rev the pump cycles 1 time so
f = 30 cycles/sec for the pump frequency
let fn = resonant frequency of the valve
I have defined driving frequency ratio as
dfr = w / wn = (2pi/2pi) x w / wn = f / fn
So, the dfr is dependent on the resonant frequency of the valve, fn
If fn = 10 cps, then dfr = 30/10 or 3.0
If fn = 30 cps, then dfr = 30/30 or 1.0
If fn = 60 cps, then dfr = 30/60 or 0.5, etc.
One way to change the fn of the valve is to vary its thickness.
CAUTION: I think the fn is proportional to the thickness raised to the 3rd power so it won't take much change in thickness to radically alter its value. IOW if you double the thickness, you will increase its resonant frequency by 8 times!