Module 5: Practice Problems
Instructions: Work through each problem on paper first, then reveal the solution. These 10 problems cover free vibrations, forced vibrations, resonance, RLC circuits, and the mechanical-electrical analogy.
Problem 1: Undamped Free Vibration
A 4-kg mass on a spring with k = 36 N/m is released from x(0) = 0.5 m with x'(0) = 0. Find x(t).
Solution
4x'' + 36x = 0 ⇒ x'' + 9x = 0. ω0 = 3. x = c1cos 3t + c2sin 3t. x(0) = c1 = 0.5, x'(0) = 3c2 = 0, c2 = 0. x(t) = 0.5 cos(3t). Period = 2π/3 s.
Problem 2: Underdamped Free Vibration
Solve x'' + 4x' + 20x = 0, x(0) = 2, x'(0) = -4.
Solution
r = (-4 ± √(16-80))/2 = -2 ± 4i. x = e-2t(c1cos 4t + c2sin 4t). x(0) = c1 = 2. x'(0) = -2c1 + 4c2 = -4 + 4c2 = -4, c2 = 0. x(t) = 2e-2tcos(4t).
Problem 3: Forced Vibration (No Resonance)
Solve x'' + 4x = 6cos(t), x(0) = 0, x'(0) = 0.
Solution
xh = c1cos 2t + c2sin 2t. Guess xp = A cos t: -A + 4A = 6, A = 2. x = c1cos 2t + c2sin 2t + 2cos t. x(0) = c1 + 2 = 0, c1 = -2. x'(0) = 2c2 = 0, c2 = 0. x(t) = -2cos 2t + 2cos t.
Problem 4: Pure Resonance
Solve x'' + 9x = 12cos(3t), x(0) = 0, x'(0) = 0.
Solution
ω = ω0 = 3: resonance. Guess xp = t(A cos 3t + B sin 3t). xp'' + 9xp = -6A sin 3t + 6B cos 3t = 12cos 3t. A = 0, B = 2. x = c1cos 3t + c2sin 3t + 2t sin 3t. ICs: c1 = 0, c2 = 0. x(t) = 2t sin(3t).
Problem 5: Beating
Solve x'' + 100x = 8cos(9t), x(0) = 0, x'(0) = 0. Express using product of sines.
Solution
ω0=10, ω=9. xp=8/(100-81) cos 9t = (8/19)cos 9t. x = -(8/19)cos 10t + (8/19)cos 9t = (8/19)(cos 9t - cos 10t). Using cos A - cos B = 2sin[(B-A)/2]sin[(B+A)/2]: x(t) = (16/19) sin(0.5t) sin(9.5t). Beat freq = 1 rad/s.
Problem 6: Damped Forced Vibration
Find the steady-state solution for x'' + 4x' + 13x = 10cos(3t).
Solution
Guess xp = A cos 3t + B sin 3t. xp'' + 4xp' + 13xp = (4A+12B)cos 3t + (-12A+4B)sin 3t = 10cos 3t. 4A+12B=10, -12A+4B=0. B=3A, 4A+36A=10, A=1/4, B=3/4. xp = (1/4)cos 3t + (3/4)sin 3t.
Problem 7: RLC Circuit (Free Response)
Solve q'' + 10q' + 25q = 0, q(0) = 5, q'(0) = 0 (L=1, R=10, C=1/25).
Solution
(r+5)² = 0: critically damped. q = (c1 + c2t)e-5t. q(0) = c1 = 5. q'(0) = c2 - 5c1 = c2 - 25 = 0, c2 = 25. q(t) = (5+25t)e-5t. i(t) = q' = (25 - 125t - 25)e-5t = -125te-5t. Wait, let me recompute: q' = 25e-5t -5(5+25t)e-5t = (25-25-125t)e-5t = -125te-5t.
Problem 8: RLC with AC Source
Find the steady-state charge for q'' + 2q' + 10q = 20cos(3t).
Solution
Guess qp = A cos 3t + B sin 3t. Substituting: (A+6B)cos 3t + (-6A+B)sin 3t = 20cos 3t. A+6B=20, -6A+B=0, B=6A. A+36A=20, A=20/37, B=120/37. qp = (20/37)cos 3t + (120/37)sin 3t.
Problem 9: Mechanical-Electrical Translation
A spring-mass has m=2, c=8, k=10. Find the analogous RLC circuit and classify the damping.
Solution
L=2 H, R=8 Ω, 1/C=10 so C=0.1 F. Check: R²=64. 4L/C=4(2)(10)=80. 64 < 80: underdamped. Both systems oscillate with decaying amplitude.
Problem 10: Finding Natural and Resonant Frequency
An RLC circuit has L = 0.5 H, R = 0, C = 0.02 F. What driving frequency produces resonance?
Solution
ω0 = 1/√(LC) = 1/√(0.5 · 0.02) = 1/√(0.01) = 1/0.1 = 10 rad/s. Driving at ω = 10 rad/s produces resonance (unbounded charge growth since R = 0).