Question

5. Use the same mass, damping coefficient, and spring constant as in #4. a. The mass is pulled down 5 cm from rest and just released. Determine the IVP describing the motion of the mass. (6 pts) b. Solve the resulting DE from part a. (6 pts) c. Sketch the graph of the motion. (4 pts) 6 6.Again use the same mass, damping coefficient, and spring constant as in 4. a. Now the mass is pulled down 5 cm from rest and given an upward velocity of 10 cm/sec. Determine the IVP describing the motion of the mass. (6 pts) b. Solve the resulting DE from part a. (6 pts) c. Sketch the graph of the motion. 4 pts) d. Find the maximum displacement of the mass once it passes through the equilibrium. (5 pts) 7. An RLC circuit has a 1.5-volt battery as a voltage source (E(t)-1.5 The values for the components are R 1.5 ohms,L 1 henries, and C2 farads. Initially, the charge on the capacitor is 2 coulombs and the current in the resistor is 4 amps. Find the charge on the capacitor and the current as functions of time. (10 pts)

Answer 1

k = 10g/sec^2 and u = 20 g/sec x = dx/dt mx = -(k x + u x) Rightarrow m x + u x + k x = 0 Rightarrow x + u/m x + k/m x = 0 with initial condition x = 0 at t = 0 x = 5cm at t = 0 x + u/m x + k/m x = 0 Let x = e^rt put in above eq we get r^2 u/m r + k/m = 0 r = -u/m plusminus squareroot (u/m - 4 k/m/2) let m = 1 gm (assume) so r = -20/1 plusminus squareroot (20/1)^2 - 4 (10)/(2)/2 \r\n r = -20 plusminus squareroot 400 - 40/2 solution is x (t) = Ae_0^- 19.4 s t + B e^- 0.51 st x (0) = is so is = A + B - X3 alpha x (t) = -19.48 A e_0^- 19.4 s t - 0.515 B e^- 0.51 st