17.12: A Driven System
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It would probably be useful before reading this and the next section to review Chapters 11 and 12.
Figure XVII.12 shows the same system as figure XVII.2, except that, instead of being left to vibrate on its own, the second mass is subject to a periodic force F = ˆFsinωt. For the time being, we’ll suppose that there is no damping. Either way, it is not a conservative force, and Lagrange’s equation will be used in the form of Equation 13.4.12. As in Section 17.2, the kinetic energy is
T=12m1˙x21 + 12m2˙x22
Lagrange’s equations are
ddt∂T∂˙x1 − ∂T∂x1 = P1
and
ddt∂T∂˙x2 − ∂T∂x2 = P2.
We have to identify the generalized forces P1 and P2.
In the nonequilibrium position, the extension of the left hand spring is x1 and so the tension in that spring is f1 = k1x1. The extension of the right hand spring is x2 − x2 and so the tension in that spring is f2 = k2(x2−x1). If x1 were to increase by δx1, the work done on m1 would be (f2−f1)δx1 and therefore the generalized force associated with the coordinate x1 is P1 = k2(x2−x1)−k1x1. If x2 were to increase by δx2, the work done on m2 would be (F−f2)δx2 and therefore the generalized force associated with the coordinate x2 is P2=ˆFsinωt−k2(x2−x1). The lagrangian equations of motion therefore become
m1¨x1 + (k1+k2)x1 −k2x2 = 0
and
m2¨x2 + k2(x2−x1) = ˆFsinωt.
Seek solutions of the form ¨x1=−ω2x1 and ¨x2=−ω2x2. The equations become
(k1+k2−m1ω2)x1 − k2x2 = 0
and
−k2x1 + (k2 − m2ω2)x2 = ˆFsinωt.
We do not, of course, now equate the determinants of the coefficients to zero (why not?!), but we can solve these equations to obtain
x1 = k2ˆFsinωt(k1+k2−m1ω2)(k2−m2ω2)−k22
and
x2 = (k1+k2−m1ω2)ˆFsinωt(k1+k2−m1ω2)(k2−m2ω2)−k22.
The amplitudes of these motions (and how they vary with the forcing frequency ω) are
ˆx1 = k2ˆFm1m2ω4 − (m1k2+m2k1+m2k2)ω2+k1k2
and
ˆx2 = (k1+k2−m1ω2)ˆFm1m2ω4 − (m1k2+m2k1+m2k2)ω2+k1k2
where I have re-written the denominators in the form of a quadratic expression in ω2.
For illustration I draw, in figure XVII.13, the amplitudes of the motion of m1(continuous curve, in black) and of m2(dashed curve, in blue) for the following data:
ˆF=1, k1=k2=1, m1=3, m2=2,
when the equations become
ˆx1=16ω4−7ω2+1=1(6ω2−1)(ω2−1)
and
ˆx1=2−3ω26ω4−7ω2+1=2−3ω2(6ω2−1)(ω2−1)
Where the amplitude is negative, the oscillations are out of phase with the force F. The amplitudes go to infinity (remember we are assuming here zero damping) at the two frequencies where the denominators of Equations ??? and ??? are zero. The amplitude of the motion of m2 is zero when the numerator of Equation ??? is zero. This is at an angular frequency of √(k1+k2)m1, which is just the angular frequency of the motion of m1 held by the two springs between two fixed points. In our numerical example, this is ω = √23 = 0.8165. This is an example of antiresonance.