The problem that I am thinking of is phrased as follows:

A person on a swing is holding a sandbag and is moving with some initial velocity $v_0$ at the bottom of the swing of length $l$. The person, who weighs $m$, drops the sand bag, which weighs $\epsilon$ at the bottom of the swing. What happens to the amplitude and frequency of the system?

I know that frequency in this case is proportional to $\omega = \sqrt{\frac{g}{l}}$, since the SHO equation for simple pendulums is: $$\frac{\partial^2 \alpha(t)}{\partial t^2} + \frac{g}{l}\alpha(t) = 0$$

Since frequency is independent of mass, we have that the frequency does not change.

However, I wrongly suspected that the amplitude decreases, since the mass of the system is decreased($m+\epsilon$ to $m$) and thus the kinetic energy of the system is decreased, leading to a lower maximal amplitude. What's wrong with my reasoning here?

Also, I am curious about what happens if one were to drop the sand bag at the max amplitude; would this make a difference in our solution?

  • $\begingroup$ Try looking into Landau&Lifshitz, Volume 1, Chapter 27 (download it from LibGen - google...) $\endgroup$
    – mavzolej
    Mar 16, 2019 at 23:48
  • $\begingroup$ @mavzolej which page exactly? $\endgroup$ Mar 16, 2019 at 23:59
  • $\begingroup$ Never mind, I misread your question and thought you're asking about mass varying in time. $\endgroup$
    – mavzolej
    Mar 17, 2019 at 0:25

1 Answer 1


Both the kinetic energy $\frac 12mv^2$ and gravitational potential energy $mgh$ are proportional to the mass $m$ so changing the mass will change each form of energy in the same ratio.

Another way of looking at the arrangement is to consider two separate pendulums of the same length and amplitude but of differing masses.
They will have the same period.

It so happens that your arrangement starts off with both the masses (person and sandbag) joined together and then one of the masses is ditched.

  • $\begingroup$ I see. Then does that imply that regardless of where the sandbag is dropped, in this simple pendulum example, the amplitude does not change? $\endgroup$ Mar 17, 2019 at 0:14

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