# Oscillations - Mass Change on Simple Pendulum

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?

• Try looking into Landau&Lifshitz, Volume 1, Chapter 27 (download it from LibGen - google...) – mavzolej Mar 16 at 23:48
• @mavzolej which page exactly? – OneRaynyDay Mar 16 at 23:59
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.