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homework 10 solutions - Section 3.6 4 Noting that there is...

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Unformatted text preview: Section 3.6 4. Noting that there is no ﬁrst—derivative term, we try x = Acos 4t and ﬁnd the particular solution xp = lOcos 4t. Then imposition of the initial conditions on the general solution x(t) = c1 cos 5t + c2 sin 5t + lOcos 4t yields x(t) = (—10 cos 5t+ 18 sin 5t) + 10 cos 4t 2(—5cos5t + 9sin5t) +10cos4t 5 9 ‘ 24106 — cosSt+——————sin5t +10cos4t { 4106 J106 ) 2J106 cos(5t — a) where a = 7t— tan'1(9/5) z 2.0779 is a second—quadrant angle. The following figure shows the graph of x(t). 271 30 —30 16. (S—af)A+4coB =10, —4a)A+(5—a)2)B = 0' _ 10(5—(02) B _ 40a) 25+6w2+w" 25+6w2+a24 C(w) = 10 / J 25 + 6502 + (04 begins with C(O) = 2 and steadily decreases as 0) increases. Hence there is no practical resonance frequency. C 20. Let the machine have mass m. Then the force F = mg = 9.8m (the machine‘s weight) causes a displacement of x = 0.5 cm = 1/200 meters, so Hooke's law F =k:x, that is, mg = k(l/ 200) gives the spring constant is k = 200mg (N/m). Hence the resonance frequency is a) = Jk/m = JZOOg ~4200x9.8 z 44.27 rad/sec 7.05 Hz , R which is about 423 rpm (revolutions per minute). 21. If 6 is the angular displacement from the vertical, then the (essentially horizontal) displacement of the mass is x = L6, so twice its total energy (KE + PE) is m(x')2 + lot2 + 2mgh = mL2(¢9')2 + kLZBZ + 2mgL(1 - cos a) = C. Differentiation, substitution of 6 z sin 0, and simpliﬁcation yields 6"+ (k/m + g/L)9 = 0 so we 2 ,lk/m+g/L. Section 3.7 3. Now the differential equation is 51’+ 251 = 100 cos 60t. Substitution of the trial solution Ip = A cos 60t + B sin 601‘ yields 1p = 4(cos 60! + 12 sin 600/145. The complementary function is IC = ce's'; the solution with 1(0) = 0 is I(t) = 4(cos 60t + 12 sin 60t - e'5‘)/ 145. 4. The solution of the initial value problem 21’ + 401 = 100 e‘w’, [(0) = O is [(1‘) = 5(e‘10' — e'zo‘). To ﬁnd the maximum current we solve the equation I'(t) = — 50e“°’ +100e‘20' = —50e‘2°‘ (em — 2) = O for I = (1n 2)/10. Then Imax = I((ln2)/10) = 5/4. 7. (a) The linear differential equation RQ’+ (l/C)Q = E0 has integrating factor p = MC. The resulting solution with 9(0) = 0 is Q(t) = E0C(1 — NRC). Then [(0 = Q’(t) = (Ea/ma“? (b) These solutions make it obvious that lim Q(t) = EOC and 13210) = 0. t—bw 21. The differential equation 101" + 201' + 1001 = —-1000 sin 51‘ has transient solution Iu(t) = e" (cl cos 3t+c2 sin 3t), and in Problem 13 we found the steady periodic solution Isp(t) = %g-(2cos5t+ 3sin5t). When we impose the initial conditions [(0) = 0, I '(0) = —10 on the general solution I (t) = Iu(t)+Isp(t), we get the equations c1+4O/13 = O, -c1+3c2+300/13 = —10 with solution c1 = — 40/ 13, c2 = —470/ 39. The ﬁgure at the top of the next page shows the graphs of I (t) and Isp(t). ...
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