Two Identical Pendulums Connected by a Light Spring

  • Thread starter Thread starter want2graduate
  • Start date Start date
  • Tags Tags
    Light Spring
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
want2graduate
Messages
10
Reaction score
0

Homework Statement



Two identical pendulums of the same mass m are connected by a light spring. The displacements of the two masses are given, respectively, by xa = Acos( (w2-w1)t/2 )cos( (w2 + w1)t/2 ), xb = Asin( (w2-w1)t/2 )sin( (w2 + w1)t/2 ).

Assume that the sprint is sufficiently weak that its potential energy can be neglected and that the energy of each pendulum can be considered to be constant over a cycle of its oscillation.

Show that the energies of the two masses are are:

Ea = 1/2 m * A^2 ( (w2 + w1)/2 )^2 cos^2( (w2 - w1)t/2 )
Eb = 1/2 m * A^2 ( (w2 + w1)/2 )^2 sin^2( (w2 - w1)t/2 )

Homework Equations



Energy of simple harmonic oscillator = (1/2)mw^2 (amplitude)^2

The Attempt at a Solution



In the book's solution, it says that amplitude = A cos[ (w2 - w1)t/ 2] or A sin[ (w2 - w1)t/2 ]. Where does it get this from? What happened to the other cosine/sine term? Why is it using this particular term?
 
Last edited:
Physics news on Phys.org
want2graduate said:
The displacements of the two masses are given, respectively, by xa = Acos( (w2-w1)/2 )cos( (w2 + w1)/2 ), xb = Asin( (w2-w1)/2 )sin( (w2 + w1)/2 ).
Those are constants. Should there be some occurrences of t in there?
 
Whoops, thanks for catching that. I edited my post.
 
If the two frequencies are similar, you can view the displacement equations as a rapid oscillation (the average of the frequencies) modulated in amplitude by a beat frequency (half the difference). In that view, we can treat the beat frequency factor as a time-dependent amplitude, A(t). Applying the standard formula should then yield the result.
That said, there is nothing in the statement of the question to justify the assumption that the frequencies are so close.