SIMULTANEOUS EQUATIONS 3.41. Solve the system of equations
3.43. A 20 lb weight suspended from the end of a vertical spring stretches it 6 inches
Assuming no external forces, find the position of the weight at any time if initially the weight is (a) pulled down 2 inches and released, (6) pulled down 3 inches and given an initial velocity of 2 ft/sec downward. Find the period and amplitude in each case.
Let A and B [Fig. 3-4] represent the position of the end of the spring before and after the weight W is put on. B is called the equilibrium position. Call y the displacement of W at any position C from the equilibrium position. Assume that y is positive in the down- ward direction.
By Hooke's law, 20 lb stretches the spring .5 ft, 40 lb stretches it 1 ft and so 40(.5 + y) lb stretches it (.5 + y) ft. Thus when W is at C, the tension in the spring is 40(.5 + y) lb.
By Newton's law, Mass • Acceleration =
or
= Net force on W
= Weight downward
= 20
— Tension upward 40(.5 + y) or
(6) The damping force is given by —8 dx/dt, regardless of where the particle is. Thus for example if x < 0 and dx/dt > 0, then the particle is to the left of O and moving to the right so the damping force must be to the left, i.e. negative. Thus by Newton's law,
Fig. 3-4
which becomes
+ 64j/ = 0 or V — GI cos 8t + C2 sin 8i
(a) Since at t - 0, we have , c2 = 0 and so y - £ cos 8i. The amplitude is ft and the period is 2W8 = W4 sec.
(6) Since and
Then the amplitude is ft approx. and the period is W4 sec.
3.44. Solve Problem 3.43 taking into account an external damping force given in pounds by PV where v is the instantaneous velocity in ft/sec and (a) /3 = 8, (b) ft = 10, (c) /3 = 12.5.
The equation of motion with damping force fiv = /? dy/dt is
(a) If /3 = 8, Solving subject to the conditions y = 1/6, dy/dt = 0 at í = 0, we find
V =
The motion is damped oscillatory with period 2^/4.8 = 5?r/12 sec.
(6) If ¿S = 10, Solving subject to the conditions as in (a), y = i. The motion is called critically damped since any smaller value of /3 would produce oscillatory motion.
(c) If j8 = 1 2 . 5 , a n d we findThe motion is called overdamped.
3.45. (a) Work Problem 3.43 (a) if an external force given by F(t) = 40 cos 8i is applied for t > 0 and (b) give a physical interpretation of what happens as í increases.
(a) The equation of motion in this case becomes
or
The solution of this, subject to y = 1/6, dy/dt = 0 at t = 0, is y — % cos 8i + 4t sin 8*
(b) As t increases, the term 4i sin 8t increases numerically without bound and physically the spring will ultimately break. This illustrates the phenomenon of resonance and shows what can happen when the frequency of the applied force is equal to the natural frequency of the system.
3.46. A rod AOB [Fig. 3-5 below] rotates in a vertical plane about a point O on it with constant angular velocity «. A particle P of mass m is constrained to move along the rod. Assuming no frictional forces, find (a) a differential equation of motion of P, (b) the position of P at any time and (c) the condition under which P describes simple harmonic motion.
(a) Let r be the distance of P from O at time t and suppose that the rod is horizontal at t - 0.
We have
or
Vertical
Fig. 3-5
Net force on P = Centrifugal force + Component force due to gravity
where r — r0, dr/dt = v0 at t = 0 i.e. r0 and v0 are the initial displacement and velocity of P.
(b) Solving (1) subject to (2), we find
(c) Simple harmonic motion along the rod results if and only if r0 = 0 and v0 — g/2o.
(J)
3.47. An inductor of 2 henries, resistor of 16 ohms and capacitor of .02 farads are con- nected in series with a battery of e.m.f. E = 100 sin3í. At t = 0 the charge on the capacitor and current in the circuit are zero. Find the (a) charge and (6) current at £ > 0.
Letting Q and / be the instantaneous charge and current at time t, we find by Kirchhoff's laws
or since / = dQ/dt,
(6)
The first term is the steady-state current and the second, which becomes negligible as time increases, is called the transient current.
Solving this subject to Q - 0, dQ/dt - 0 at t = 0, we find (a)
(2)
3.48. Given the electric network of Fig. 3-6. Find the currents in the various branches if the initial currents are zero.
Kirchhoff's second law states that the algebraic sum of the voltage drops around &ny closed loop is zero. Let us traverse loops KLMNK and JKNPJ in a counterclockwise direction as shown. In traversing these loops we consider voltage drops as positive when we go against the current. A voltage rise is
considered as the negative of a voltage drop. Fig. 3-6
Let 7 be the current in NPJKN. This current divides at junction point K into /i and /2 so that / = Ii + 72. This is equivalent to Kirchhoff's first law.
Applying Kirchhoff's second law to loops JKNPJ and KLMNK respectively, we have
201 - 120 + 2^ + 10/n = O (Í)dl,
d/j d/2
-lO/i - 2^ + 4^ + 20/2 = 0 («)
Putting / = /! + /2 and using the operator D = d/dt, these become
(D + 15)/! + 10/2 = 60 (S)
-(D + 5)/! + (2D + 10)/2 = 0 (4)
Solving these subject to I1 — I2 = 0 at t = 0, we find
/! = 3(1-«-*>«), /2 = f(l-e-2«i), / = |(i_e-20t)
Supplementary Problems
OPERATORS
3.49. Write each of the following in operator form: (a) y" + 4j/' + 5j/ = e~x, (b) 2y'" - x — 2y' + y, (c) xy' + 2y = 1.
3.50. (a) Evaluate (Z>3 + 1) sin 2x and (D + l)(D2 - D + 1) sin 2x. (b) Are the operators I>3 + 1 and (D + 1)(I>2 - D + 1) equivalent? Explain.
3.51. (a) Evaluate (xD - S)(D + 2){x2 + x - 4} and (D + 2)(xD - B){x2 + x - 4}. (6) Are the operators (xD - S)(D + 2) and (D + 2)(xD - 3) equivalent? Explain.
3.52. (a) Evaluate (xD)(xD)(xD){x2 + 2ex} which can be written (xD)3{x* + 2ex}. (b) Is this the same as xsDs{x2 + 2ex}"! (c) Are the operators (xD)t and x3D* equivalent? Explain.
3.53. Under what conditions will the operators (D + a)(D + b) and (D + b)(D + a) be equivalent? Explain.
LINEAR DEPENDENCE AND WRONSKIANS
3.54. (a) Show that the functions x2, Bx + 2, x — \, 2x + 5 are linearly dependent. (6) Are the functions x2, 3x + 2, x — i linearly dependent?
3.55. Investigate the linear dependence of ex, xex, x2ex.
3.56. Show that y = x is a solution of xy" + xy' — y = 0 and find the general solution.
3.57. If î/t is a solution of y" + p(x)y' + q(x)y = r(x), explain how to find the general solution. Illustrate by an example.
THE REDUCED OR HOMOGENEOUS SOLUTION
3.58. (a) Find three linearly independent solutions of (D + 2)(D - 1)(D - S)y = 0 and (6) write the general solution.
3.59. Solve (a) y" + 8y' + I2y = 0, (b) (D* - 4Z> - l)y = 0.
3.60. Solve (a) (D* + 25)y = 0; y(Q) = 2, y'(0) = -5, (6) y" - 8y' + 20y = 0, (c) (D» + 8)y - 0.
3.61. Solve (a) y" + 4y' + % = 0 (d) (D» + D«)y = 0 (6) ley" - 8y' + y = O (e) (£>» + 64)2j/ = 0 (c) (D + 6)4(1) - 3)2j, = 0
3.62. Solve D*(D + 1)2(D2 + W + 5)2(Z>2 + 4)j/ = 0.
THE COMPLETE OR NONHOMOGENEOUS EQUATION
3.63. Solve (a) y" - 5y' + 6y = 50 sin 4w, (6) (Z>3 - 8)y = 16» + 18e~x + 64 cos 2x - 32.
3.64. Solve (a) y" + By' + 2y = 4e~2*, (6) (D» + 3£>2)j/ = 180ws + 24a.
3.65. Solve (I>6 - 2I>5 + D*)y = 12Qx + 8e*.