In Problem show that, for weak coupling, the period at which the energy trades back and fo...

In Problem show that, for weak coupling, the period at which the energy trades back and forth is approximately equal to Ta(2K/K') where Ta = 2π/ωa = 2π/(m/K)1/2 is the period of die symmetric oscillation.

Problem

In the system of two identical coupled oscillators shown in Figure 11.3.1, one oscillator is started with initial amplitude A0, whereas the other is at rest at its equilibrium position, so that the initial conditions are

Show that the amplitude of the symmetric component is equal to the amplitude of the antisymmetric component in this case and that the complete solution can be expressed as follows:

in which  and Δ= (ωb – ωa)/2. Thus, if the coupling is very weak so that K'« K, then ω will be very nearly equal to ωa = (K/m)1/2, and A is very small. Consequently, under die stated initial conditions, die first oscillator eventually comes to rest while the second oscillator oscillates with amplitude A0. Later, die system returns to die initial condition, and so on. Thus, the energy passes back and forth between die two oscillators indefinitely.