# Why the common beam in Barton's pendulum?

N

So the theory is that R vibrates with largest amplitude due to resonance. Because when measured from the common beam its length is the same as that of A, so A's vibration would pass to R the same amount of energy R would have vibrating on its own. Right?

My question is, how come the lengths have to be measured from the common beam? It just oscillates about the point on the string where it was fixed, so shouldn't that be the point from where you measure length if you want to get the right values of energy?

I've been thinking , if you measured the lengths of one of the smaller pendulums, one from the common beam and the other from the string, and also measured its time period, which length would correspond to that time period?

I know that the common beam is the right one, because I saw R vibrating with high amplitude with my own eyes. (It threatened to knock my eyes out ☺) I just want to know why, because theoretically it seems to me that it should be from the string.

• This is the first time I've seen Barton's pendulum so I intend to research further. It seems to me there are potentially multiple modes since there are multiple pivot points for the various pendulums. A pendulum's frequency is independent of the mass. So why then is the 'driver' pendulum provided a heavier mass? If the experimental facts are that R resonates with the driver, then one might assume only the attachment points of the common support swing matter, and the others do not. After all the natural frequency of the simple pendulum is $$\omega= \sqrt{\frac{L}{g}}$$ – docscience Apr 26 '16 at 14:08
• After viewing several demos on YouTube it seems the point behind having a heavier driver pendulum is to demonstrate the fact I mentioned above - that the pendulum frequency is independent of the mass. So two lessons. Sympathetic oscillations (resonance) and the independence of frequency on pendulum mass. – docscience Apr 26 '16 at 14:20
• Driver had a high mass so that its kinetic energy will be higher and thus less lost in transmission. – Jayne Apr 28 '16 at 14:43
• I think you mean total energy, or energy capacity. Remember a pendulum's energy is only fully kinetic at the bottom of its swing. But yes that make's sense. You can store more initial energy with the larger mass and thus allow the system to operate for a longer time against the inevitable energy losses. – docscience Apr 28 '16 at 14:49

## 2 Answers

You have reasoned correctly that one should measure from the point the pendulum pivots about. You missed to see however, that if you are in the the resonance scenario, the 'common support swing' is swinging to and fro too, and the line the string of the pendulum in resonance / the driving one forms lies in the same plane as the common swing. I.e. the pendula which are in resonance actually turn around the support swing's pivots.

Unfortunately I have trouble putting the situation into words; I suggest viewing/imagining the apparatus 90° turned from the view on the sketch and it should become clear.

• Nice to hear. If you are content with the answer you should accept it, so the question disappears from the queue. – caconyrn Apr 25 '16 at 19:34

The driver has a heavier mass in order to not being affected by the much much smaller energy transfered by the other pendulums, so it's oscillations stays constant whatever propagates back from the other pendulums.