A simple pendulum has a point like bob of mass $m$ and a light inextensible string. Suppose, by means of some arrangement, we shorten the string extremely slowly, so that at any instant, the ratio of length to rate at which it changes is much greater than the period of the pendulum. Suppose the length of the string is $L$ and the angular amplitude (assumes to be small) is $\theta$. Now, at this moment, the length of the string is changed by $\Delta L(<0)$ , what would be the corresponding change in angular amplitude $\theta$.

I believe the amplitude would increase (based on horrible imagination so I might be wrong). I am unable to convincingly work out this problem. I am having difficulty in incorporating the fact that "at any instant, the ratio of length to rate at which it changes is much greater than the period of the pendulum."

Could anyone please give an elementary approach(highlighting the mathematics) or a hint of some kind?

Edit: I tried analysing the forces on the bob using polar coordinates as I could see terms like $\dot r$ and $\dot \theta$. But I couldn't make significant progress towards the original question.

(I am not familiar with the Lagrangian as yet.)


2 Answers 2


If the string (fixed at the top) passes through a hole in a movable mounted rod, and you move the rod downward (when the string is straight), then you have done no work on the pendulum and its maximum kinetic or potential energy do not change. The maximum height of swing will not change, but with a shorter string, the angle of swing will increase. If you pull the string up through the hole, you are doing work on the pendulum. It will swing to a greater height, but the bottom position is also rising. I'm guessing the maximum angle will be the same as in the previous situation.


Let us assume that when you shorten the pendulum you are not doing any work on the bob.We know that that total energy of a pendulum is equal to the potential energy possessed by it at the highest point.Also the Maximum height reached by the pendulum is $l(1-cos\theta)$ where l is effective length and $\theta$ is the angular amplitude. So total mechanical energy of pendulum can be written as $mgl(1-cos\theta)$ . So when you reduce l, $ \theta$ increases such that $cos\theta$ decreases and $1-cos\theta$ increases keeping total mechanical energy same.So when you reduce l Angular amplitude increases.But you cannot reduce l indefinitely because a point will be reached when angular amplitude will be as high as $\pi$ which is the maximum possible angular amplitude. Let this limiting value of l be $l_{limiting}$.

Now the question arises : What happens when you reduce effective length such that it becomes less than $l_{limiting}$.Well in that case the total mechanical energy won't change (since we haven't done any work on the bob) but potential energy at the maximum height i.e at angular displacement of $\pi$ with respect to vertical will be less than Total mechanical energy so the body will have some kinetic energy at this point i.e It will execute motion in vertical circle.(Refer to diagrams provided);

Here's the math: initial Mechanical energy = $mgl_1(1-cos\theta)$

Final mechanical energy = $mgl_2(1-cos\phi)$

But since work done = 0 , Initial mechanical energy = Final mechanical energy

$mgl_1(1-cos\theta)$ = $mgl_2(1-cos\phi)$

where $l_1$ is initial length , $l_2$ is final length , $\theta$ is initial angular amplitude and $\phi$ is final angular amplitude .

Solving this equation we can calculate the value of $\phi$.

If $l_2$ < $l_1$ then $\phi$ > $\theta$

Let me know if this was clear enough.

Reference diagram

  • $\begingroup$ I'll ponder more upon your hint. Thanks. I just have one question-where/how have you used the fact that "at any instant, the ratio of length to rate at which it changes is much greater than the period of the pendulum"? $\endgroup$ Commented Apr 15, 2021 at 13:50
  • $\begingroup$ According to me if rate of change of length is large wrt pendulum's time period it means that change happened almost instantaneously without changing kinetic energy of body i.e without doing work on the body.let me know if you have meant something else by this statement. $\endgroup$
    – Möbius
    Commented Apr 15, 2021 at 16:34

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