Question

Question 2 The simple pendulum, discussed in week 4 and lab session 4, has the equation of motion f0/dt2_0? sin θ 9-(g/L)I/2. with The total energy of the pendulum is constant during the motion, and is given by _mgL cos θ, wher dt is the angular speed of the motion in radians per second. Consider the simple pendulum with initial conditions θ(0) and u(0)-wi, 0 i.e. starting from the vertically down position with an initial angular speed w1 (a) Use the energy equation to show analytically that the pendulum will overshoot θ around) if wi>2(g/L)172 π (i.e. whirl (b) Suppose that friction (damping) is present and is proportional to w, so that the equation of motion is where k is the coefficient of friction. By introducing the non-dimensional time t ts -2T(L/g)1/2, show analytically that the equation of motion can be re-written t/ts, where (10) where K tsk, and where a dot denotes differentiation with respect to t. (c) Write a code to solve Eq. (10) using fourth order Runge-Kutta. For k 1.0, determine how many complete revolutions the damped pendulum makes when: i) wi 4(g/L)1/2; (ii) wi - 10(9/L)1/2.

I need help with c).

Answer 1

consider the above given that energy E = mc^2 wi^2/2 mgc as theta = 0 cos theta = 1 To over shoot at theta = pi initial energy E initial > potential energy at top > mgL E (pi) = E (o) - 2 mg L = m L^2 wi^2/2 - 2 mg L > 0 MC^2 wi^2/2 > 2 mg L the energy execution at E (pi) = m L^2 wi^2/2 > 2 mg L d theta/dt = d theta/dz dt/at at F = t/ts d^2 theta/dt^2 = (d^2 theta/d 2^2) (dF/dt)^2 = 1/t_s^2 - d^2 theta/dt^2 r/n/ d theta/dt^2 + k d theta/dt + Ohm^2 sin theta = 0 --- (2) d^2 theta/dz^2 + k t_s d theta/dz + Ohm^2 ts^2 sin theta = 0 d^2 theta/dz^2 + k (1) d theta/dz + (pi)^2 (1)^2 sin theta = 0 d^2 theta + k d theta/lambda > + 4 pi^2 sin theta = 0