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Suppose we have a pendulum tied to the ceiling of an elevator which is at rest. The pendulum is oscillating with a time period $T$, and it has an angular amplitude, say $\beta$. Now at some time when the bob of the pendulum was at the mean position(equilibrium position/center position), the elevator suddenly starts moving up with some acceleration $a$. I need to find out what is the new angular amplitude of the pendulum, in terms of $a$,$\beta$ and if needed $T$, and $l$(length of the string).

I'm completely stuck at this concept. I absolutely do not know how to proceed. My ideas are that we could use some conservation, maybe the energy of the bob. Any hints?

P.S. I'm sorry if anything is too elementary.


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  • $\begingroup$ Here energy is not conserved because of the external force. But I suppose you should try to replace the $g$ acceleration (gravity) by $g+a$ in your expression for the amplitude. $\endgroup$ – Ana S. H. May 5 '13 at 15:05
  • $\begingroup$ shouldn't the amplitude be constant?what is the answer? $\endgroup$ – ABC May 5 '13 at 15:21
  • $\begingroup$ Energy isn't conserved, but momentum in the $x$ direction is, since the external force acts only in the $y$ direction. Maybe that's enough to work out the answer? $\endgroup$ – Nathaniel May 5 '13 at 15:31
  • $\begingroup$ @Nathaniel: That may work, but how? I mean, If I conserve the momentum, what do I write? $mv=mv'$ gives $v=v'$, and leads me to nothing. Any other hints? $\endgroup$ – Ashish Gaurav May 5 '13 at 15:33
  • $\begingroup$ @007: The amplitude is not constant. Actually the question is a numerical, I've modified the parameters to get a general expression. $\endgroup$ – Ashish Gaurav May 5 '13 at 15:34

Energy is conserved , because at the moment of exertion of the force , the motion is perpendicular to direction of the force. So , after that moment , the mass simply , as before continues to exchange it's kinetic energy with potential energy while oscillating (this time with $g'=g+a$) . So , we write an equation for conservation of energy: $$mgl(1-\cos \beta)=m(g+a)l(1- \cos \beta')$$

because $\beta <<1$ we expand the cosine: $$\cos x = 1 - \frac{x^2}{2}$$

so $$g\beta^2=(g+a) \beta'^2 \to \beta'=\beta \sqrt{\frac{g}{g+a}}$$

special cases:

  • $a=0 :\beta'=\beta$
  • $a \to \infty : \beta'=0 $
  • $\begingroup$ Well this is correct , but giving the momentum argument to support energy conservation will help. $\endgroup$ – ABC May 6 '13 at 4:47

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