Question

A brick is attached to a horizontal spring and is able to slide back and forth. The mass of the brick is m= 2 kg., the spring constant is k = 40 N./m., and friction provides a resistive force equal (in N.) to c= 16 times the velocity of the brick (in m./sec.). a) Formulate a differential equation that models this problem, using 1 = time (in secs.) and x(t) = displacement (to the right of the equilibrium position, in m.). b) Use the auxiliary polynomial to find two linearly independent solutions x and x2 of the associated homogeneous equation. c) Use the Wronskian determinant to prove that your answers to part (b) are linearly independent. d) Use your answers to part (b) to write the complementary solution. e) Is the system critically damped, overdamped, or underdamped? Justify your answer. f) Suppose we supply an external force of 10e * N. on the brick. Use either the Undetermined Coefficients or Variation of Parameters method to find a particular solution of the resulting nonhomogeneous equation g) Use your answers to parts (d) and (t) to write the general solution.

Answer 1

K240 w m >*() damping Restroing fone by spring (ka) By Newtons and ma - mdr = dt2 law, spring force & damping force, – kx - cda dt .. moze & Eda & kx = 0 d+2 dt .

2 d²x dt? + 16 dx + 40x co dt + = 0 is the DE (a) I dx dt2 8 dx + 20x dt. (b) Auxillony solution is D² + 8D + 20 = 0 Da- 8 + 64-80 = – 4 7 21 2x1 Di = - 4+2i and D2 = – 4-21

complementary solution is Y = put ( acos (2 t) + (2 sin (2 t)] Y = c.9, (t) + C2 Y₂ (t), are y, A) and Thus two independent solutions (t) where, y,(t) = étacos (21) yu(t) = ente sin (2+) (e) wronskion = 1 4 (t) ly'lt) Ya(t) '(t)

his, we y, (t). Y '(t) - 92(t).y, (t) y (t) = & 4t (-sin (2+). 2) + cos(at).e at 1-4) Lent 2 sin (at) + HCOS (2 t) Y (t) = x 4t I cos (2) 2] + sin(2t). eat (-4) & 4t [2 Los (2 t) - 4 sin (2+)] - 4t W=èat cos(at) (ett (21050t - 4 sinot)) - & sin (2t) e Casin2t+41052t)) Bt 2cos izt) - Lesin(2 t).cos(2+)7-851-2 sin? (27) – 4 COS(2).sin(2+) sét 2005'(2 t) - sin(2t).cos(2+) + 2sin? (20) + Les inl2t).cos(2+) ] = est [ 2[sin (2+)+ cos? (2+)] ] & Zest W to since wly, Y₂) to both y, It) and I (t) are independent solutions. (Here, x, (t) and rart)] (d) Complementary solution is y = cylt) + (2 Y2 (t) 9.2 & 4t [ci cos (2 t) + C2 sin (2 t) ] (e). Here, c² - Link = 16² - 4x2x40 = 256 - 320 --64 since, " c² - Link to system is under-damped.

(f) Method of undefermined md²x coefficients, dt2f cand & Kala F(t). Here, RHS is 10 e 3t. sus, initial guess for particular solution is, Ip (t) = Ae-3t Ypct) = A (-3) e 3 – 30 e 3t p(t) = - 3A (-3) e 3t = a Ae3t substituting in differential eauction, 9 Aest + 8 (-3A é 3 t) + 20 (Ae-3t) = 10 e 3t SA e 3t = 10e3t - SA=10 . A=2. Thus particular solution is. 4p(t) = 2e 3t / 2 + I (9) General solution is X(t) = Y t up xe(t) = 4t (c, cos(2 t) + Ca sin (2 t) ] + 2 e 3t