Damping and Resonance
Damping and Resonance
Real vibrations eventually die out (damping), and when driven at the right frequency, they can grow enormously (resonance).
$$x(t) = A e^{-\gamma t} \sin(\omega_d t + \phi)$$
The exponential envelope $e^{-\gamma t}$ makes the amplitude decrease over time. $\gamma$ is the damping coefficient.
- Underdamped: oscillates with decreasing amplitude
- Critically damped: returns to rest fastest, no oscillation
- Overdamped: returns slowly, no oscillation
When a periodic force drives a system at its natural frequency $f_0$, the amplitude reaches a maximum. This is resonance.
$x(t) = 10 e^{-0.5t}\sin(6t)$. After how long has the amplitude dropped to half its initial value?
$10 e^{-0.5t} = 5$ → $e^{-0.5t} = 0.5$ → $-0.5t = \ln(0.5) = -0.693$
$t = 1.386$ s.
A building has natural frequency 2 Hz. An earthquake produces vibrations at 2 Hz. What happens?
Resonance occurs — the building oscillates with maximum amplitude. This is why earthquakes can destroy some buildings but not others: it depends on whether the building's natural frequency matches the earthquake frequency.
A car's suspension has $\gamma = 4$ s⁻¹. After a bump, the car bounces with initial amplitude 5 cm. Amplitude after 1 second?
$A(1) = 5e^{-4(1)} = 5e^{-4} \approx 5(0.0183) = 0.092$ cm. Almost zero — heavy damping.
Practice Problems
Show Answer Key
1. $8e^{-1} \approx 2.94$ cm
2. $t_{1/2} = \ln 2/0.1 = 6.93$ s
3. 0.5 Hz (match the natural frequency)
4. Overdamped (it slowly returns without bouncing)
5. Marching in step could match the bridge's natural frequency and cause resonance.
6. $5/20 = e^{-3\gamma}$ → $e^{-3\gamma} = 0.25$ → $\gamma = \ln 4 / 3 \approx 0.462$ s⁻¹