Problem 5 Consider two identical simple pendulums moving in a common vertical plane and connected by a spring (see Figure). The distance between the two suspension points is h. The masses of the pendulums are identical and equal to m. We assume that at equilibrium the two pendulums hang vertically, with the spring at its natural length h. The equilibrium position is thus 01 02 = 0. Therefore, the potential due to the spring has the form = Vspring = k > 0, (1) - where P₁P2=√√(x1 − x2)² + (Y1 − 2)² is the distance between the two masses, where (x1,y1) and (x2, y2) are the Cartesian coordinates of the two masses. h 1 P₁ m 12 m P₂ Figure 1: The two pendulums. The spring is depicted as a blue line. (i) How many degrees of freedom does the system have? Choose appropriate generalised coordinates and write the kinetic energy in terms of them. Determine the form of the matrix Tij which appears in the small oscillations approximation. (ii) Write the distance P₁P2 in terms of the angles 01 and 02 (defined in the Figure). Expand the expression of (P1P2)2 as a function of 01 and 02 about the equilibrium position 0₁ = 02 = 0, and check that the term (y1 y2) can be dropped in the small oscillations approximation, as it is fourth order. Use this to prove that the you can drop this term in the spring potential, since we are only interested in quadratic fluctuations. Hence, from now on use the approximated form Vspring → k(x2 - -x1-h)² for the spring potential. (iii) Write the full potential (spring plus gravitational), and expand it up to the quadratic fluctuations about the equilibrium postion. Determine the matrix Vij. (iv) Find the frequencies of small oscillations about the equilibrium position. (v) Find the normal modes associated to the two characteristic frequencies w² (1) and w²(2). 3

Problem 5 Consider two identical simple pendulums moving in a common vertical plane and connected by a spring (see Figure). The distance between the two suspension points is h. The masses of the pendulums are identical and equal to m. We assume that at equilibrium the two pendulums hang vertically, with the spring at its natural length h. The equilibrium position is thus 01 02 = 0. Therefore, the potential due to the spring has the form = Vspring = k > 0, (1) - where P₁P2=√√(x1 − x2)² + (Y1 − 2)² is the distance between the two masses, where (x1,y1) and (x2, y2) are the Cartesian coordinates of the two masses. h 1 P₁ m 12 m P₂ Figure 1: The two pendulums. The spring is depicted as a blue line. (i) How many degrees of freedom does the system have? Choose appropriate generalised coordinates and write the kinetic energy in terms of them. Determine the form of the matrix Tij which appears in the small oscillations approximation. (ii) Write the distance P₁P2 in terms of the angles 01 and 02 (defined in the Figure). Expand the expression of (P1P2)2 as a function of 01 and 02 about the equilibrium position 0₁ = 02 = 0, and check that the term (y1 y2) can be dropped in the small oscillations approximation, as it is fourth order. Use this to prove that the you can drop this term in the spring potential, since we are only interested in quadratic fluctuations. Hence, from now on use the approximated form Vspring → k(x2 - -x1-h)² for the spring potential. (iii) Write the full potential (spring plus gravitational), and expand it up to the quadratic fluctuations about the equilibrium postion. Determine the matrix Vij. (iv) Find the frequencies of small oscillations about the equilibrium position. (v) Find the normal modes associated to the two characteristic frequencies w² (1) and w²(2). 3