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3.6 The resonance peak - qualitatively
We see from Fig. 10 above and Eqns (50) and (63), that higher Q's result in taller and more narrow resonance peaks. Why is this? In this section we explore an alternate way to view the process of a sinusoidal force exciting a resonator, a view that will qualitatively explain the effect that Q has on the resonance peaks.
A series of impulses
This way to view a sinusoidal driving force is that it is similar to a long string of impulses. Each cycle of the sinusoidal excitation is equivalent of a shock impulse to the resonator as discussed in an earlier posting and reviewed above. We saw there, that a single impulse will result in the resonator ringing at a frequency of ωdecay given by Equation (24) and being described by a damped sinusoidal equation, of the form of Equation (23). Because we have a linear system, the series of ringings made by a series of impulses will simply add up to give the total response of the resonator. See an earlier posting for more on the addition of sinusoidal functions.
We illustrate this in the following animations where we show each ringing separately as well as their sum (the actual response of the resonator). Read the captions and experiment with the animations to understand this concept further.
Frequency shift due to phase slippage
The above discussion brings up the question: How do the ringing curves, all oscillating at a frequency ωdecay, add up to a total response that oscillates at a different frequency ω = ωdrive ?
This question relates to a resonator being driven at a frequency different than the resonator's resonant frequency. In reference to Fig. 15e, this means that we have dragged the yellow rectangle to the left or right of the peak, perhaps to frequencies of 2.6 or 3.0Hz. It also relates to the fact that in the steady state response, i.e. after the resonance is nicely oscillating in a steady fashion, the resonator must oscillate at whatever frequency it is driven at and not at the frequency it may "wish" to oscillate at. You can see this as you drag the yellow rectangle to the extreme right or left, that the peaks and valleys in the "total resonator response" window bunch up or become drawn out in spite of the fact that ω0 and ωdecay do not change.
The change in frequency from that of the individual ringing curves (all oscillating at ωdecay) to that of the sum response (which is oscillating at ω = ωdrive ) is due to the fact that off-resonance, the ringing curves do not all line up. There is a phase slip between successive ringing curves. Since frequency is defined as the rate of change of phase, this slippage in phase changes the overall frequency. In fact, the amount of phase slippage per time exactly equals the frequency difference between ωdecay and ωdrive.
Another question that might come up is: why is the "total resonator response" in Fig. 15e rather jagged when the drive frequency is 2Hz and not a smooth sinusoid? The reason is that when pushed to this limit our series of impulses is a rather poor approximation to a sinusoidal drive force. We will do better in our approximation in the next posting (which uses impulses in two directions) and still better yet in the postings after that (which use calculus to sum an infinite number of impulses that vary in amplitude and direction to perfectly model a sinusoidal driving force).
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