Question 15.8: Find the position of stable equilibrium of the pendulum of E...
Find the position of stable equilibrium of the pendulum of Exercise 15.7 if the hinge oscillates with the frequency ω ≫\sqrt{g/l}.
We first rewrite the Lagrangian of the pendulum of Exercise 15.7 as follows: The terms
\frac{mω^{2}}{2} h^{2}_{0} sin^{2} ωt and −mgh_{0} cos ωtcan be written as total differentials with respect to time:
\frac{mω^{2}}{2} h^{2}_{0} sin^{2} ωt = \frac{d}{dt} \left(−\frac{1}{4} mωh^{2}_{0} sin ωt cos ωt\right) + C,−mgh_{0} cos ωt = \frac{d}{dt} \left(−\frac{mgh_{0}}{ω} sin ωt\right).
We can omit these terms, since Lagrangians that differ only by a total derivative with respect to time, according to the Hamilton principle δ \int_{t_{1}}^{t_{2}}{Ldt} = 0, are equivalent.
Hence,
= \frac{m}{2} [\dot{ϑ}^{2}l^{2} + 2ωh_{0} \dot{ϑ}l sin ϑ sin ωt −2gl cos ωt]. (15.19)
Another transformation yields
mωh_{0} \dot{ϑ}l sin ϑ sin ωt =− \frac{d}{dt} (mωh_{0}l cos ϑ sin ωt)+mω^{2}h_{0}l cos ϑ cos ωt,so that the Lagrangian finally reads
L = \frac{m}{2} [\dot{ϑ}^{2}l^{2} +2ω^{2}h_{0}l cos ϑ cos ωt −2gl cos ϑ]. (15.20)
From this, one obtains of course the equation of motion as in Exercise 15.7.
We consider ϑ as a generalized coordinate with the appropriate mass coefficient ml^{2}. The equation of motion then reads
ml^{2}\ddot{ϑ} = mgl sin ϑ − mω^{2}h_{0}l sin ϑ cos ωt
=−\frac{du}{dϑ} + f (15.21)
with u = mgl cosϑ and f =−mω^{2}h_{0}l sin ϑ cos ωt. The additional force f is due to the motion of the hinge. For very fast oscillations of the hinge, we assume that the motion of the pendulum in the potential u is superposed by quick oscillations ξ :
ϑ(t) =\widetilde{ϑ}(t)+ξ(t).The average value of the oscillations over a period 2π/ω equals zero, while \widetilde{ϑ} changes only slowly; therefore,
\widetilde{ϑ}(t) = \frac{ω}{2π} \int\limits_{0}^{2π/ω}{ϑ(t) dt} =\widetilde{ϑ}(t). (15.22)
Equations (15.21) with (15.22) can then be written as
ml^{2} \ddot{\widetilde{ϑ}}+ ml^{2} \ddot{ξ}(t)=−\frac{du}{dϑ}+f (ϑ).Because f (ϑ) = f (\widetilde{ϑ}+ ξ) = f (\widetilde{ϑ}) + ξdf/dϑ, an expansion up to first order in ξ yields
ml^{2} \ddot{\widetilde{ϑ}} +ml^{2} \ddot{ξ} =−\frac{dU}{d \widetilde{ϑ}} − ξ \frac{d^{2}U}{d\widetilde{ϑ}^{2}}+f (\widetilde{ϑ})+ξ \frac{df}{d\widetilde{ϑ}}. (15.23)
The dominant terms for the oscillations are ml^{2} \ddot{ξ} and f (\widetilde{ϑ}):
ml^{2} \ddot{ξ} = f (\widetilde{ϑ})⇒ \ddot{ξ} =−\frac{ω^{2}h_{0}}{l} sin \widetilde{ϑ} cos ωt,
and from this, we obtain
ξ = \frac{h_{0}}{l} sin \widetilde{ϑ} cos ωt =− \frac{f}{mω^{2}l^{2}} . (15.24)
We now calculate an effective potential created by the oscillations, and for this purpose we average (15.23) over a period 2π/ω (the mean values over ξ and f vanish):
ml^{2} \ddot{\widetilde{ϑ}} =−\frac{dU}{d\widetilde{ϑ}}+\overline{ξ \frac{df}{d\widetilde{ϑ}} }=−\frac{dU}{d \widetilde{ϑ}} − \frac{1}{mω^{2}l^{2}} \overline{f \frac{df}{d \widetilde{ϑ}}} .This can be written as
ml^{2} \ddot{\widetilde{ϑ}} =−\frac{dU_{eff}}{d \widetilde{ϑ}} with U_{eff} = U + \frac{1}{2mω^{2} −l^{2}} \overline{f^{2}}. (15.25)
Because \overline{cos^{2} ωt} = 1/2, we get
U_{eff} = U + \frac{mω^{2}h^{2}_{0}}{4} sin^{2} ϑ
= mgl cos ϑ + \frac{mω^{2}h^{2}_{0}}{4} sin^{2} ϑ. (15.26)
The minima of U_{eff} give the stable equilibrium positions:
\frac{dU_{eff}}{dϑ} =− mgl sin ϑ + \frac{mω^{2}h^{2}_{0}}{4} sin ϑ cos ϑ \overset{!}{=} 0
⇒ sin ϑ = 0 or cos ϑ = \frac{2gl}{ω^{2}h^{2}_{0}}. (15.27)
From this, it follows that for any ω the position vertically downwards (ϑ = π) is stable. ϑ = 0 is excluded because U_{eff}(ϑ = 0) = mgl. Additional stable equilibrium positions arise for ω^{2} > 2gl/h^{2}_{0} with the angle given above.