Help   |   Contact Us

 

Lecture 5: Coupled Oscillators

OCW Scholar

« Previous | Next »

Lecture Topics

Two metal poles with a spring suspended between them.
  • Coupled Oscillators
  • Normal Modes

Learning Objectives

By the end of this lecture, you should:

  • understand how to analyze an undamped coupled harmonic oscillator.
  • define normal modes and how they may be combined.
  • find the normal modes of coupled pendulums.
  • determine the motion of coupled pendulums from initial conditions.
  • understand how beats can originate through coupled modes.
  • apply the recipe for determining normal modes in general.
  • describe the normal modes of three coupled bodies.

Lecture Activities

Check Yourself

  • Extending the idea of a double pendulum a bit, consider three identical pendulums set to oscillate in the plane of their support. They are equally spaced on that support and joined by identical springs. Describe in detail but without mathematical analysis two obvious modes in the small angle case.

› View/hide answer

Most simply, all three pendulums can move identically with angular frequency \(\sqrt {\frac{g}{l}}\)and no stretching of the springs. Another simple motion is the two outer pendulums moving out of phase with each other with the inner one stationary. The similarity of each side of this situation to a double pendulum leads us to conclude that the angular frequency would be \[\sqrt {{{\left( {\frac{g}{l}} \right)}^2} + \alpha {{\left( {\frac{k}{m}} \right)}^2}} \] which is higher than the in-phase mode due to the involvement of the springs. The positive constant α would need to be determined by more detailed analysis.

 

  • In the demo near the start of Lecture 5, two pendulums of about 1 m length seem to exchange energy one to the other in about 10 seconds. We will estimate the mass of each bob at 500 g, and neglect the mass of the (fairly hefty) supporting rods. Calculate the spring constant this would imply, but look carefully and suggest why this may not be a very valid result. (Hint for calculation: reverse the sign of the beat term shown in the lecture.)

› View/hide answer

The period to return to the same situation would be 20 seconds if the energy goes from one to the other in 10 seconds and the slow term angular frequency for each bob is \(\frac{{{\omega _ + } - {\omega _ - }}}{2}\), corresponding to \[\frac{{2\pi }}{T} \approx \frac{{6.3}}{{20}} = 0.3rad/s.\] For a one meter spring \[{\omega _0} = {\omega _ - } = \sqrt {\frac{g}{l}} \approx 3.1 rad/s.\] This gives \({\omega _ + } = 2(0.3) + {\omega _ - } = 3.7rad/s.\) But \({\omega _ + } = \sqrt {2\omega _s^2 + \omega _0^2} \) or \[\omega _s^2 = \frac{{\omega _ + ^2 - \omega _0^2}}{2} = \frac{{{{3.7}^2} - {{3.1}^2}}}{2} = 2\] Since \(\omega _s^2 = \frac{k}{m}\) we have \(k = 0.5(2) = 1\) N/m, a rather weak spring as the ones in the video look to be. However, we note that they are also drooping, so their mass moving up and down may also affect the frequency.

 

« Previous | Next »