Question

Two identical pendulums are connected by a light coupling spring. Each pendulum has a length of 0.4m. With the coupling spring connected,one pendulum is clamped and the period of the other is found to be 1.25 sec.
With neither pendulum clamped,what are the periods of the two normal modes?

Answer 1

For 1 normal mode ,i have the frequency that is √g/l .
But i am unable to get the significance of the clamp thing over here to find out the 2nd normal mode frequency

Answer 2

@irishboy123

Answer 3

.

Answer 4

Hi Sami

[post editted cos i said some dumb things first time round :-(]


what kinda steer do you need?!?!

the physics bit is fiddly but i doubt you'd find it hard. just more assumptions that \(\sin \theta = \theta\) for small oscillations.

the approach for the first bit (forces and FBD's vs Energy) are outlined here in this MIT thing with some nice little drawings...., though there is a mistake in there (!!) and there are neater ways to do it,......, but it is an approach, and the solution is sweet

http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W15D1-7.pdf

for the second it's more of the same, but you now have 2 degrees of freedom and therefore a chaotic system, like the double pendulum. so the angles at which the pendula are hanging are diferent, they are the 2 dof's and you get a connected DE's. i've had a play w/ it and i'd say that **key additional bit** you can assume the spring is always horizontal, ie another small angle/oscillation assumption.

the physics is still well doable but the maths is the fiddly bit. seperating out 2 DE's and using eigenvalues to find solutions. IOW, assuming a solution of form \(\vec{ x} = \vec{ \alpha } e^{i \omega t}\) and forcing a non-null solution.

if you follow an energy approach, i think you'd need a Lagrangian to separate it out


.....so use a FBD/forces and forces would be my go-to

it's all here, and if you need a hand following it.....

http://dluong1.bol.ucla.edu/swarthmore/e12/lab3.htm

PS just for once, can you ask an **easy** question :-))

Answer 5

FWIW, I get \(T_{1,2} = 1.23s, 1.27s\)

which are for eigenvectors \(\left(\begin{matrix}1 \\ -1\end{matrix}\right)\) {I think that's 180 phase out} and \(\left(\begin{matrix}1 \\ 1\end{matrix}\right)\) {think that's in phase }