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3.7 More on the impact model
Impulses in two directions
We further illustrate the concepts of the previous posting, the rough equivalence of a sinusoidal drive source to a series of impacts, now with the impacts in both positive and negative directions, representing each half cycle of the drive. This is a better approximation to a sinusoidal drive force than the purely negative impacts shown above in Fig. 15.
Since we have a linear set of defining equations, the response from each "impulse" will go on unaffected by the other impulses. The total response we actually observe will be the sum of the responses of all the single impulses. We see this illustrated in Figs. 16a and 16b below.
↑ Fig. 16a. In the top of this figure we see the sinusoidal force (shown in red) applied to the resonator. We can view this force as a series of impulses (shown as blue vectors) directed at the resonator and alternating in sign. Each "impulse" results in a damped ringing, shown in various colors in the lower half of the figure. To avoid an overly cluttered illustration, we have only shown the ringings from four of the impulses, but in fact the ringings would continue to be started with each impulse. Each ringing slowly dies out.
In this figure we have chosen the frequency of the driving force (and consequently of the impulses) to equal the ringing frequency of the resonator. This makes the up and down motions of the ringings line up with each other so that their peaks and valleys occur at the same time. This lining up of the oscillations is an example of constructive interference.
In an actual mechanical resonator, we will not see each individual ringing function, but instead only the effect of all of them added together. Since they all line up for this choice of drive frequency, the sum of these ringings will be a maximum and we will observe a maximum combined oscillation of the mass on the spring.
|↑ Fig. 16b. This figure is the same as Fig. 16a except that the sinusoidal force frequency is considerably higher than the ringing frequency of the resonator. This makes the ringings in the lower part of the graph NOT line up with each other. The ringing oscillations destructively interfere with each other. In this case, the various ringings would not add up to the maximum total oscillations as they do in Fig. 16a.|
We see that this is an interference phenomenon where the ringing from each half cycle interferes with the ringing of the other half cycles. Only when the drive frequency is close to the resonate frequency do the ringings of all these half cycles line up and produce strong constructive interference.
In the next posting, we will use calculus to mathematically sum up these ringings. In the remainder of this posting we will qualitatively discuss this view of resonance and make some crude approximations as to the resonance curve height and peak width.
Reason for taller peak with higher Q
We see from Figs. 16a-d that if the ringings of the sucessive impulses line up (constructively interfere) then the total amplitude of oscillation should be proportional to the average amplitude of these ringings multiplied by the number of impulses in a decay time length. The average amplitude is f/2 and the number of half cycles in a time period τ is equal to 2τ/T = 2×(2Q/ω)×(ω/2π) = 2Q/π . Multiplying these two factors together we have:
where A is the total amplitude and f/2 is the value of one impulse.
So we expect the peak height at the frequency of maximum constructive interference to be proportional to the impulse amplitude and the Q.
Even more intuitively, we would guess that if all the ringings were lined up properly, the response amplitude should be proportional to the time each ringing hangs around, i.e. to the decay time constant, and thus to the Q (since τ = 2Q/ω ).
Reason for narrower peak with higher Q
At a very intuitive level, the higher the Q the longer each ringing persists. Thus, the higher the Q, the longer the driving oscillations need to keep in step with the ringings and therefore higher Q requires a closer match between drive frequency ω and the decay resonant frequency ωdecay .
Let's do this a little better. First, we need to think about the "best" phase between driver and oscillations for the driver to transfer power to the oscillations. Because the power delivered to the bob by the driver is proportional to the driver's force times the bob's velocity, the condition for maximum power transfer (and maximum amplitude) is such that the driver force be maximum when the bob's velocity is maximum. Since there is a π/2 phase difference between the bob's position and its velocity, the condition for producing maximum oscillations is that the oscillations of the force and those of the bob's displacement differ in phase by π/2.
If each ringing starts off with the proper π/2 phase difference between driver and displacement, then how long should the phase difference remain approximately this way? You would guess that the time period of good phase should be about a decay time period, after which the ringing gets small. Let's also assume that the phase should remain within 0.5 radian of this best phase. Now to do the simple math. The phase drift due to a frequency error of half the peak height is given by:Setting phase drift to half a radian, we have:
Solving for the allowed bandwidth:
Amazingly this is within a fact of two of what we arrived at in Equation (63) after a page of calculations.
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