The duplication of an assumed frequency f1=20 kHz to f2=40 kHz at constant power intensity, would result to a halving of the amplitude in an optimal system free of losses. The maximum velocity of the oscillating surface would remain constant, due to the shorter way to be taken.

An increase in viscosity of the sonicated liquid dampens the movement of the sonotrode. At the sonication of water or liquids with even higher viscosity, more power intensity is needed to reach a constant amplitude, compared to the free movement in air. The frequency relationship should generally not be affected by the fluid, but that should be subjected to a practical examination to include the structure of the sonication device (Sonotrode, Flat transducer etc.)

Best regards,

Jochen

Jochen --

If we assume a constant velocity (resulting in halving the amplitude in going from 20 kHz to 40 kHz), it remains unclear to me if the power intensity then remains constant. Using an electrical analogy, the power developed in a linear resistor is P = I^2 R where I is the current and R is the resistance. If the current I is held constant (analogous to constant mechanical velocity) then the power P remains constant; however, this is only true if R is independent of frequency. (As a further analogy, the hysteresis losses in an electrically-driven magnetostrictive material increase rapidly with frequency. This is why most magnetostrictive transducers are limited to an upper frequency of about 30 kHz.) Thus, this gets to my basic question -- is the acoustic resistance of the cavitating fluid actually independent of frequency since both the bubble size and the bubble density (the cavitation cloud) depend on the frequency? (I'm looking for a reference study with verifiable results.)

Regards.