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Fig. 7. This animation shows the motion of a **driven** mass-spring resonator. The driver is made to change colors as it pulls up and pushes down to drive the resonator at a particular frequency. Mouse over the figure to see the motion.

3.3 Driven oscillations

The above discussion relates to the free decay of a resonator where the resonator is shocked into oscillation by some impulse or quick action. Another mode of oscillation is where the resonator is connected to an external source of sinusoidal oscillations. This type of oscillation is called "driven", "forced" or "continuous".

A question that comes to mind is: "at what frequency do you drive the resonator?" The instinctive answer is "at the resonant frequency." However, actually we are free to drive the resonator at any frequency: at the resonant frequency or at a different frequency. We next derive the response of the resonator as a function of the driving frequency.

With a driver force connected to the mass (22) becomes:

There are a number of ways to apply an oscillating force to a moving bob. Probably the easiest is to move the upper end of the spring up and down as shown in the animation. In this case we replace the *f _{spring}* in (2) with

*k*(

*x*

_{0}−

*x*) where

*x*

_{0}is the (now moving) equilibrium bob position with the effect of gravity included. With this (21) becomes: which makes (22) become: Comparing this with (30), we see that the driver force is given by

*kx*

_{0}, i.e.:

*f*

_{driver}=

*kx*

_{0}, where

*x*

_{0}is oscillating because of the driver.

To minimize the algebraic tedium, we will use the complex forms for oscillations (see an earlier posting on complex phasors and adjacent postings). We substitute *f _{driver}* =

*F e*and the last of the forms shown in (23)

^{iωt}*x*=

*A e*into (30):

^{iωt}where *F* is the complex amplitude of the oscillating driving force and *A* is the complex amplitude of the bob's position. Taking the derivatives and canceling out the common *e ^{iωt}* factors, we have:

We solve for *A*:

Fig. 8. This animation shows the motion of a mass-spring resonator when the drive frequency is swept through a range of frequencies. The animation starts with an angular frequency of 1Hz and sweeps up through 5Hz. The resonant frequency of this particular animation is 2.5Hz. The displacement is graphed in red, the amplitude of the the motion (i.e. the envelope) in blue and the phase difference between the driver and the bob's motion in purple. In a sweep of the frequency it is important to sweep slowly enough that equilibrium is established at each frequency. In actual practice, the frequency would need to be swept somewhat slower than is illustrated here. Mouse over the figure to see the motion. Click on the figure to restart it.

Factoring out the *iωR* from the denominator and using the relations in (24), we get:

Fig. 9. The graph in the above animation with features better labeled. The frequency *f* in the graph is related to the angular frequency *ω* in Equations (34) and (36) by the definition of angular frequency: *ω* = 2*πf* .

Looking at the graph, we see that the phase difference between the driver and the bob changes as we go through resonance. At a frequency lower than resonance, the phase difference is roughly zero meaning that the driver and bob are in phase, i.e. move together. At the frequencies higher than resonance, the phase difference approaches −*π* or 180 degrees, so that the driver and bob are moving opposite each other. At resonance, there is a −*π*/2 or 90 degree phase difference. If you look carefully at the above animation, you can see this phase dependence.

Finally, we use the definition of the quality factor *Q* from an earlier posting:

*Q* = *τ _{energy}*

*ω*

_{0}=

*τω*

_{0}/2 , (35)

where the decay time for energy is half of the decay time for amplitude, i.e. *τ _{energy}* =

*τ*/2 . An earlier posting indicated that

*τ*= 2

*A*/

*B*where

*A*and

*B*are the coefficients of the second and first time derivatives in the defining differential equation, i.e. in (31) above where

*A*=

*m*and

*B*=

*R*. Thus,

*τ*= 2

*m*/

*R*or

*R*= 2

*m*/

*τ*=

*mω*/

_{0}*Q*. Substituting this into (34), we have:

Figure 9 shows the graph from the above animation. We see the amplitude (blue) and the phase (purple) of the bob's oscillations relative to the driver. The parameters of this resonance are *Q* = 5 and *f*_{0} = 2.5 . The angular resonant frequency is given by *ω* = 2*πf*_{0} .

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