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Re: [Phys-l] Oscillator amplitude(Q) question.



On 5/8/2011 5:07 PM, Bernard Cleyet wrote:
[bc] If one wishes to increase the amplitude of an harmonic mechanical oscillator when driven at its resonance frequency, one increases the Q. Not necessarily. Please comment. Is this not necessarily true of an electrical oscillator, also? [RLC oscillator] bc wonders about cavity oscillators, also.
[bw]If one wishes to increase the amplitude of an oscillator, one can increase input power, or decrease the power dissipation. Decreasing dissipation, other things equal = increased Q
Power dissipation can be reduced by reducing the resonant frequency at constant energy loss per cycle, Constant energy loss per cycle, other things equal = constant Q or by reducing the energy loss per cycle at constant frequency. Reduced energy loss per cycle, other things equal = increased Q
Is this what you meant?

I was hoping there would be comments about the Q in both cases. What you write is very true.
bc disappointed.


Is this what you were hoping for?


Nope. What I hoped was a discussion on the general relationship of Q and amplitude of a driven harmonic oscillator at resonance.
Both mechanical and electric (RCL).

Hint: Does increasing the Q result in increased amplitude, always?


bc has given up and given it away.

p.s. for pendula, horologists have known (experimentally) the answer to this for centuries.
The answer is obvious from the particular solution of the inhomogenous diff. eq.

It's amusing to suppose that the maker of the World's most accurate chronometer design of its day, or the maker of the World's most accurate observatory pendulum clock design of its day would have considered differential equations in the design of their respective masterworks. Amusing, but inaccurate.

But getting back to bc's puzzlement: one can provide increased amplitude for a mechanical or electrical oscillator with higher, same or lower Q than before.
Is that plain enough? :-)

Brian W