Lesson 2: Forced Vibrations and Resonance
Estimated time: 40-50 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Set up and solve the forced vibration equation mx'' + cx' + kx = F0cos(ωt)
- Find the particular solution using undetermined coefficients
- Explain resonance and what happens when the driving frequency matches the natural frequency
- Describe the beating phenomenon when ω is close to ω0
- Calculate amplitude and phase of the steady-state response
The Forced Vibration Equation
When an external periodic force F(t) = F0cos(ωt) acts on a spring-mass system, the equation becomes:
Forced Vibration Equation: mx'' + cx' + kx = F0cos(ωt), where ω is the driving (forcing) frequency, ω0 = √(k/m) is the natural frequency, and F0 is the amplitude of the external force.
The general solution is y = yh + yp. The homogeneous part yh is the transient solution (decays if c > 0), and the particular part yp is the steady-state solution.
Undamped Forced Vibrations (c = 0, ω ≠ ω0)
With no damping and driving frequency different from natural frequency: mx'' + kx = F0cos(ωt).
The homogeneous solution is xh = c1cos(ω0t) + c2sin(ω0t). For the particular solution, guess xp = A cos(ωt):
Particular Solution (Undamped, ω ≠ ω0): xp = F0 / [m(ω0² - ω²)] cos(ωt). The amplitude depends on how close ω is to ω0.
Example 1: Undamped Forced Vibration
Solve x'' + 9x = 6cos(2t), x(0) = 0, x'(0) = 0.
Step 1: ω0 = 3, ω = 2. Since ω ≠ ω0, no resonance.
Step 2: xh = c1cos 3t + c2sin 3t.
Step 3: Guess xp = A cos 2t. Then xp'' = -4A cos 2t. Substituting: -4A + 9A = 6, so 5A = 6, A = 6/5.
Step 4: General solution: x = c1cos 3t + c2sin 3t + (6/5)cos 2t.
Step 5: x(0) = c1 + 6/5 = 0, so c1 = -6/5. x'(0) = 3c2 = 0, c2 = 0.
Answer: x(t) = -(6/5)cos 3t + (6/5)cos 2t = (6/5)(cos 2t - cos 3t).
Resonance (ω = ω0, Undamped)
When ω = ω0, the guess xp = A cos(ω0t) duplicates the homogeneous solution. We must multiply by t:
Resonance Solution: When ω = ω0 in the undamped system x'' + ω0²x = (F0/m)cos(ω0t), the particular solution is xp = [F0/(2mω0)] t sin(ω0t). The amplitude grows linearly with time — this is pure resonance.
Example 2: Pure Resonance
Solve x'' + 4x = 10cos(2t), x(0) = 0, x'(0) = 0.
Step 1: ω0 = 2. The forcing frequency ω = 2 = ω0. Resonance!
Step 2: xh = c1cos 2t + c2sin 2t.
Step 3: cos 2t is in xh, so guess xp = t(A cos 2t + B sin 2t).
Step 4: xp' = (A cos 2t + B sin 2t) + t(-2A sin 2t + 2B cos 2t).
xp'' = (-2A sin 2t + 2B cos 2t) + (-2A sin 2t + 2B cos 2t) + t(-4A cos 2t - 4B sin 2t).
= -4A sin 2t + 4B cos 2t - 4t(A cos 2t + B sin 2t).
Step 5: xp'' + 4xp = -4A sin 2t + 4B cos 2t = 10 cos 2t. So A = 0, B = 5/2.
Step 6: x = c1cos 2t + c2sin 2t + (5/2)t sin 2t. ICs: c1 = 0, c2 = 0.
Answer: x(t) = (5/2) t sin(2t). Amplitude grows without bound.
Beating Phenomenon
When ω is close to (but not equal to) ω0, and x(0) = x'(0) = 0, the solution can be written using a trigonometric identity:
Beating: x(t) = [2F0/(m(ω0² - ω²))] sin[(ω0 - ω)t/2] sin[(ω0 + ω)t/2]. The slowly oscillating envelope sin[(ω0 - ω)t/2] modulates the fast oscillation. The beat frequency is |ω0 - ω|.
Example 3: Beating
Solve x'' + 25x = 4cos(4.5t), x(0) = 0, x'(0) = 0.
Step 1: ω0 = 5, ω = 4.5. Close but not equal, so expect beating.
Step 2: xp = 4/(25 - 20.25) cos(4.5t) = 4/4.75 cos(4.5t) = (16/19)cos(4.5t).
Step 3: x = c1cos 5t + c2sin 5t + (16/19)cos(4.5t). x(0)=0: c1 = -16/19. x'(0)=0: 5c2=0, c2=0.
Step 4: x = (16/19)(cos 4.5t - cos 5t).
Step 5: Using cos A - cos B = 2 sin[(B-A)/2] sin[(B+A)/2]:
Answer: x(t) = (32/19) sin(0.25t) sin(4.75t). Beat frequency = |5 - 4.5| = 0.5 rad/s.
Damped Forced Vibrations
With damping (c > 0), the transient yh decays, leaving only the steady-state response:
Steady-State Amplitude: For mx'' + cx' + kx = F0cos(ωt), the steady-state particular solution has amplitude C = F0 / √[(k - mω²)² + (cω)²] and phase lag φ = arctan[cω/(k - mω²)].
The amplitude C is maximized near (but not exactly at) ω = ω0 when damping is present. Unlike the undamped case, the amplitude stays bounded.
Example 4: Damped Forced System
Find the steady-state solution for x'' + 2x' + 5x = 4cos(t).
Step 1: m = 1, c = 2, k = 5, F0 = 4, ω = 1.
Step 2: Guess xp = A cos t + B sin t.
xp' = -A sin t + B cos t. xp'' = -A cos t - B sin t.
Step 3: Substituting: (-A + 2B + 5A)cos t + (-B - 2A + 5B)sin t = 4cos t.
(4A + 2B) = 4 and (-2A + 4B) = 0. From the second: A = 2B. Then 8B + 2B = 4, B = 2/5, A = 4/5.
Answer: xp(t) = (4/5)cos t + (2/5)sin t. Amplitude = √(16/25 + 4/25) = √(4/5) = 2/√5.
Summary: Frequency Response
| Scenario | Condition | Behavior |
|---|---|---|
| No resonance | ω far from ω0 | Bounded oscillation at two frequencies |
| Beating | ω ≈ ω0, c = 0 | Slow amplitude modulation |
| Pure resonance | ω = ω0, c = 0 | Amplitude grows without bound (t sin) |
| Damped resonance | ω ≈ ω0, c > 0 | Large but bounded steady-state amplitude |
Check Your Understanding
1. For x'' + 16x = 5cos(4t), is this resonance? Why or why not?
2. In a damped system, why does the transient solution eventually vanish?
3. If ω0 = 10 and ω = 9.8, what is the beat frequency?
Key Takeaways
- The forced vibration equation mx'' + cx' + kx = F0cos(ωt) combines a transient and steady-state response.
- Pure resonance occurs when ω = ω0 and c = 0, producing unbounded growth xp = (F0/(2mω0)) t sin(ω0t).
- Beating occurs when ω ≈ ω0 and c = 0: a slow amplitude envelope modulates fast oscillations.
- Damping prevents unbounded growth but allows large amplitudes near resonance.
- The steady-state amplitude C = F0/√[(k - mω²)² + (cω)²] is maximized near ω = ω0.