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8. We use Eq.(33) without R and E(t) (there is no resistor or
impressed voltage) and with L = 1 henry and 1/C = 4x106 since
C = .25x10-6 farads. Thus the I.V.P. is Q" + 4x106 Q = 0,
Q(0) = 10-6 coulombs and Q'(0) = 0.
9. The spring constant is k = (20)(980)/5 = 3920 dyne/cm.
The I.V.P. for the motion is 20u" + 400u' + 3920u = 0 or
u" + 20u' + 196u = 0 and u(0) = 2, u'(0) = 0. Here u is
measured in cm and t in sec. The general solution of the D.E. is u = Ae-10tcos^/6t + Be-10tsin4/6t. The I.C. u(0) = 2 ^ A = 2 and u'(0) = 0 ^ -10A + 4óÒâ = 0. The
solution is u = e-10t[2cos4 1 + 5(sin4 V"6t)^V"6 ]cm.
The quasi frequency is m = 4/6, the quasi period is Td = 2pm = p/^V"6 and Td/T = 7/2ä/6 since T = 2p/14 = p/7. To find an upper bound for t, write u in the form of Eq.(30): u(t) = \J 4+25/6 e-10tcos(^V"6 t-S).
Now, since I cos(4^/~6t-8) I < 1, we have | u(t) | < .05 ^
\J4+25/6 e-10t < .05, which yields t = .4046. A more precise answer can be obtained with a computer algebra system, which in this case yields t = .4045. The original estimate was unusually close for this problem since cos(4\/"6t-8) = -0.9996 for t = .4046.
12. Substituting the given values for L, C and R in Eq.(33),
we obtain the D.E. .2Q" + 3x102 Q' + 105 Q = 0. The I.C.
are Q(0) = 10-6 and Q'(0) = I(0) = 0. Assuming Q = ert, we obtain the roots of the characteristic equation as r — = -500 and r2 = -1000. Thus Q = c—e-500t + c2e-1000t and hence Q(0) = 10-6 ^ c— + c2 = 10-6 and Q'(0) = 0 ^ -500c— - 1000c2 = 0. Solving for c— and c2 yields the solution.
17. The mass is 8/32 lb-sec2/ft, and the spring constant is 8/(1/8) = 64 lb/ft. Hence (1/4)u" + gu' + 64u = 0 or u" + 4gu' + 256u = 0, where u is measured in ft, t in sec and the units of g are lb-sec/ft. We look for solutions of the D.E. of the form u = ert and find r2 + 4gr + 256 = 0, so r—,r2 = [-4g ± V16g2 - 1024 ]/2.
The system will be overdamped, critically damped or
underdamped as (16g2 - 1024) is > 0, =0, or < 0, respectively. Thus the system is critically damped when g = 8 lb-sec/ft.
r t r t
19. The general solution of the D.E. is u = Ae 1 + Be 2
where r1,r2 = [-g ± (g2 - 4km) 1/2]/2m provided
g - 4km Ï 0, and where A and B are determined by the
I.C. When the motion is overdamped, g - 4km > 0 and
r t r t
r1 > r2. Setting u = 0, we obtain Ae 1 = - Be 2 or
(r -r )t
e 1 2 = - B/A. Since the exponential function is a
monotone function, there is at most one value of t
(when B/A < 0) for which this equation can be satisfied.
Hence u can vanish at most once. If the system is critically damped, the general solution is u(t) = (A + Bt)e-gt/2m. The exponential function is never zero; hence u can vanish only if A + Bt = 0. If B = 0
then u never vanishes; if B Ï 0 then u vanishes once at
t = - A/B provided A/B < 0.
20. The general solution of Eq.(21) for the case of critical damping is u = (A + Bt)e-gt/2m. The I.C. u(0) = u0 ^
A = u0 and u'(0) = v0 ^ A(-g/2m) + B = v0. Hence
u = [u0 + (vo + gu0/2m)t]e-gt/2m. If v0 = 0, then
u = u0(1 + gt/2m)e gt/2m, which is never zero since g and
m are postive. By L'Hopital's Rule u^0 as t^rc>.
Finally for u0 > 0, we want the condition which will
insure that v = 0 at least once. Since the exponential function is never zero we require
u0 + (v0 + gu0/2m)t = 0 at a positive value of t. This
requires that v0 + gu0/2m Ï 0 and that
t = -u0(v0 + u0g/2m) -1 > 0. We know that u0 > 0 so we must have v0 + gu0/2m < 0 or v0 < -gu0/2m.
23. From Problem 21: A = ---------- = Tdg/2m. Substituting the
known values we find g = --------------- = 5 lb sec/ft.
24. From Eq.(13) w2 = ------ so P = 2ð/ä/ 2k/3 = p ^ k = 6.
Thus u(t) = c1cos2t + c2sin2t and u(0) = 2 ^ c1 = 2 and
u'(0) = v ^ c2 = v/2. Hence u(t) = 2cos2t + —sin2t =
Section 3 . 8
4 + — cos(2t-g). Thus ] 4 + —
4 V 4
3 and v
27. First, consider the static case. Let À² denote the length of the block below the surface of the water. The weight of the block, which is a downward force, «.