First label the forces in Figure 1 . Resolve the forces in linear coordinates to start with. The pendula p1 and p2 are set up as follows: o Both pendula consist of a mass m = 1 kg suspended by a massless. frictionless, in- extensible rod of length L. o A spring connects the two pendula and exerts a force P; (which direction, and where?) o The coupling length (from the fulcrum to the point where the spring connects to the pendulum} is l for both pendula. o For any given experiment the maximal angular displacement is lbl = |1|=|2|. Figure 1'. Schematic of a coupled pendulum. For simplicity concentrate on the motion of one pendulum. The system is symmetric (p1 and p2 have identical set-ups) and thus the derivation for p1 is valid for pg, 1. For the forces in Figure 1 write a general equation for the total force (Newton's second law of motion) of the system. Your equation should have the general form total torque = summation over individual torques\". 2. Write the torques (1') in terms of linear forces (F) and radii (r). This step is important because you can only measure the components of the linear forces5. 3. Now perform the cross product on the R.H.S to introduce an angular term. The small angle approximation may come in handy for simplication\"? At this point you will need to think what. if any. limitations this will have on your experiment. This will need to be explicitly addressed in your report. 4. How can you introduce an angular term into the L.H.S? Hint: it should be an angular acceleration {,5 . 5. Solve for the angular acceleration. Replace the moment of inertia l with known terms. Hint: What is the moment of inertia for a point mass? 6. Solve the differential equations and consider the three initial conditions: a. When the motion of the pendula are in phase. b. When the motion of the pendula are out of phase. c. When the motion of the pendula are beating (the starting condition for beat motion is to maximally displace one pendulum? by (plijr = 1}) = (p while the other pendulum starts at 462 (t = } = 0.) max 2.1 Theoretical background The equations of motion for the coupled pendulum will be derived in your prework. Here we include some additional information which may help you derive your equations of motion. 2.1.1 Solving second order inhomogeneous differential equations: a cheat sheet You may need to solve differential equations of the form: $1 (t) + A2 p1 (t) = -B2 [pz(t) - $1 (t)], (1) $2 (t) + A2 $2 (t) = B2 [$2 (t) - $1 (t)]. (2) Where A and B are constants; 4, (t) and 2 (t) are functions of time. For p1 (t) = $2(t); [$1(t) - $2(t)] = 0 the solution is: $1 (t) = $2 (t) = $max Cos (At). (3) For -$1 (t) = $2(t); [p1 (t) - $2(t)] = 2p, (t) the solution is: $1 (t) = Pmax Cos (VAZ + 2B2 . t), (4) $2 (t) = -Pmax Cos(VAZ + 2B2 . t). (5) If both pendula are initially stationary with $1 (0) = max and $2 (0) = 0 the solution is: $1 (t) = Pmax Cos( VAZ + 2BZ - A VAZ + 2B2 + A (6) 2 . t) . cos ( - . t ) , 2 $2 (t) = -$max sin( VAZ + 2B2 - A t ) . sin( VAZ + 2B2 + A - . t ) (7) 2 2