(a) Suppose two identical pendulums are coupled by means

Chapter , Problem 47RP

(choose chapter or problem)

Suppose two identical pendulums are coupled by means of a spring with constant k. See Figure 7.R.12. Under the same assumptions made in the discussion preceding Example 3 in Section 7.6, it can be shown that when the displacement angles \(\theta_{1}(t) \text { and } \theta_{2}(t)\) are small, the system of linear differential equations describing the motion is

                                                                       \(\theta_{1}^{\prime \prime}+\frac{g}{l} \theta_{1}=-\frac{k}{m}\left(\theta_{1}-\theta_{2}\right)\)

                                                                       \(\theta_{2}^{\prime \prime}+\frac{g}{l} \theta_{2}=\frac{k}{m}\left(\theta_{1}-\theta_{2}\right)\).

Use the Laplace transform to solve the system when \(\theta_{1}(0)=\theta_{0}, \theta_{1}^{\prime}(0)=0, \theta_{2}(0)=\psi_{0}, \theta_{2}^{\prime}(0)=0\), where \(\theta_{0} \text { and } \psi_{0}\). For convenience let \(\omega^{2}=g / l\), K = k/m.

(b) Use the solution in part (a) to discuss the motion of the coupled pendulums in the special case when the initial conditions are \(\theta_{1}(0)=\theta_{0}, \quad \theta_{1}^{\prime}(0)=0\), \(\theta_{2}(0)=\theta_{0}, \theta_{2}^{\prime}(0)=0\). When the initial conditions are \(\theta_{1}(0)=\theta_{0}, \theta_{1}^{\prime}(0)=0, \theta_{2}(0)=-\theta_{0}, \theta_{2}^{\prime}(0)=0\).

Text Transcription:

theta_1 t

theta_2 t

theta_1^prime prime+g/l theta_1=-k/m(theta_1-theta_2)

theta_2^prime prime+g/l theta_2=k/m(theta_1-theta_2)

theta_1 0=theta_0

theta_1^prime 0=0

theta_2 0=psi_0

theta_2^prime 0=0

theta_0

psi_0

omega^2=g / l

theta_1^prime 0=0

theta_2 0=theta_0

theta_2^prime 0=0

theta_2 0=-theta_0

theta_2^prime(0)=0