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Damped and Driven Oscillations

Damping forces, resonance, and forced oscillations

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Damped and Driven Oscillations

Damped Oscillations

With damping force Fd=−bvF_d = -bv:

md2xdt2+bdxdt+kx=0m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0

Divide by mm:

d2xdt2+2γdxdt+ω02x=0\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2x = 0

where:

  • γ=b/(2m)\gamma = b/(2m) (damping coefficient)
  • ω0=k/m\omega_0 = \sqrt{k/m} (natural frequency)

Three Cases of Damping

Underdamped (γ<ω0\gamma < \omega_0):

x(t)=Ae−γtcos⁡(ω′t+ϕ)x(t) = Ae^{-\gamma t}\cos(\omega' t + \phi)

where ω′=ω02−γ2\omega' = \sqrt{\omega_0^2 - \gamma^2} (damped frequency)

Oscillates with decreasing amplitude.

Critically damped (γ=ω0\gamma = \omega_0):

x(t)=(A+Bt)e−γtx(t) = (A + Bt)e^{-\gamma t}

Returns to equilibrium fastest without oscillating.

Overdamped (γ>ω0\gamma > \omega_0):

x(t)=Ae−αt+Be−βtx(t) = Ae^{-\alpha t} + Be^{-\beta t}

where α,β=γ±γ2−ω02\alpha, \beta = \gamma \pm \sqrt{\gamma^2 - \omega_0^2}

Returns to equilibrium slowly without oscillating.

Quality Factor

Q=ω02γQ = \frac{\omega_0}{2\gamma}

High Q: Light damping, many oscillations before amplitude decays

Low Q: Heavy damping, few oscillations

Energy decay: E(t)=E0e−2γtE(t) = E_0 e^{-2\gamma t}

Time constant: τ=1/(2γ)\tau = 1/(2\gamma)

Driven Oscillations

With external driving force F(t)=F0cos⁡(ωt)F(t) = F_0\cos(\omega t):

md2xdt2+bdxdt+kx=F0cos⁡(ωt)m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = F_0\cos(\omega t)

d2xdt2+2γdxdt+ω02x=F0mcos⁡(ωt)\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2x = \frac{F_0}{m}\cos(\omega t)

Steady-State Solution

After transients die out:

x(t)=A(ω)cos⁡(ωt−δ)x(t) = A(\omega)\cos(\omega t - \delta)

Amplitude: A(ω)=F0/m(ω02−ω2)2+(2γω)2A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}

Phase lag: tan⁡δ=2γωω02−ω2\tan\delta = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}

Resonance

Amplitude is maximum when denominator is minimum.

Resonant frequency: ωr=ω02−2γ2\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}

For light damping (γ≪ω0\gamma \ll \omega_0): ωr≈ω0\omega_r \approx \omega_0

Maximum amplitude: Amax≈F02mγω0=F0Qmω02=F0QkA_{max} \approx \frac{F_0}{2m\gamma\omega_0} = \frac{F_0Q}{m\omega_0^2} = \frac{F_0Q}{k}

Power and Resonance

Average power absorbed: Pavg=12bω2A2P_{avg} = \frac{1}{2}b\omega^2 A^2

Maximum at: ω=ω0\omega = \omega_0 (exactly)

Power at resonance: Pmax=F024mγP_{max} = \frac{F_0^2}{4m\gamma}

Bandwidth and Q

Half-power points: where P=Pmax/2P = P_{max}/2

Occur at ω=ω0±γ\omega = \omega_0 \pm \gamma (for small γ\gamma)

Bandwidth: Δω=2γ\Delta\omega = 2\gamma

Q=ω0Δω=ω02γQ = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0}{2\gamma}

Sharp resonance: high Q, narrow bandwidth

Complex Representation

Using x~=x0eiωt\tilde{x} = x_0 e^{i\omega t}:

−ω2x~+2iγωx~+ω02x~=F0m-\omega^2\tilde{x} + 2i\gamma\omega\tilde{x} + \omega_0^2\tilde{x} = \frac{F_0}{m}

x~=F0/mω02−ω2+2iγω\tilde{x} = \frac{F_0/m}{\omega_0^2 - \omega^2 + 2i\gamma\omega}

Magnitude gives amplitude, argument gives phase.

Applications

Shock absorbers: Critical damping for quick settling

Musical instruments: High Q for pure tones

Bridges: Avoid driving at natural frequency

RLC circuits: Same equations as mechanical oscillators

📚 Practice Problems

1Problem 1hard

❓ Question:

A damped oscillator has mass m = 0.5 kg, spring constant k = 50 N/m, and damping coefficient b = 2.0 kg/s. Find: (a) the natural frequency ω₀, (b) the damping constant γ, (c) determine if the system is underdamped, critically damped, or overdamped, and (d) find the damped frequency ω_d.

💡 Show Solution

Given:

  • m = 0.5 kg
  • k = 50 N/m
  • b = 2.0 kg/s

(a) Natural frequency:

ω0=km=500.5=100\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.5}} = \sqrt{100}

ω0=10 rad/s\boxed{\omega_0 = 10 \text{ rad/s}}

(b) Damping constant:

γ=b2m=2.02(0.5)=2.01.0\gamma = \frac{b}{2m} = \frac{2.0}{2(0.5)} = \frac{2.0}{1.0}

γ=2.0 s−1\boxed{\gamma = 2.0 \text{ s}^{-1}}

(c) Type of damping:

Compare γ with ω₀:

  • Underdamped: γ < ω₀
  • Critically damped: γ = ω₀
  • Overdamped: γ > ω₀

Since γ=2.0<ω0=10\gamma = 2.0 < \omega_0 = 10:

System is UNDERDAMPED\boxed{\text{System is UNDERDAMPED}}

(d) Damped frequency:

ωd=ω02−γ2=100−4\omega_d = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{100 - 4}

ωd=96\omega_d = \sqrt{96}

ωd=9.80 rad/s\boxed{\omega_d = 9.80 \text{ rad/s}}

Motion: x(t)=Ae−γtcos⁡(ωdt+ϕ)x(t) = Ae^{-\gamma t}\cos(\omega_d t + \phi)

Period: Td=2πωd=0.641T_d = \frac{2\pi}{\omega_d} = 0.641 s

2Problem 2medium

❓ Question:

For the damped oscillator in the previous problem, if the initial amplitude is A₀ = 0.10 m, find: (a) the amplitude after one period, (b) the time for amplitude to decrease to 0.01 m (10% of original), and (c) the quality factor Q.

💡 Show Solution

From previous:

  • γ = 2.0 s⁻¹
  • ω_d = 9.80 rad/s
  • T_d = 2π/ω_d = 0.641 s
  • A₀ = 0.10 m

(a) Amplitude after one period:

Amplitude envelope: A(t)=A0e−γtA(t) = A_0 e^{-\gamma t}

A(Td)=A0e−γTd=0.10e−(2.0)(0.641)A(T_d) = A_0 e^{-\gamma T_d} = 0.10 e^{-(2.0)(0.641)}

A(Td)=0.10e−1.282=0.10(0.277)A(T_d) = 0.10 e^{-1.282} = 0.10(0.277)

A(Td)=0.0277 m\boxed{A(T_d) = 0.0277 \text{ m}}

Amplitude decreased to 28% in one period!

(b) Time to reach 10% of original:

0.01=0.10e−γt0.01 = 0.10 e^{-\gamma t}

0.1=e−2t0.1 = e^{-2t}

ln⁡(0.1)=−2t\ln(0.1) = -2t

t=−ln⁡(0.1)2=−−2.3032t = -\frac{\ln(0.1)}{2} = -\frac{-2.303}{2}

t=1.15 s\boxed{t = 1.15 \text{ s}}

(c) Quality factor:

Q=ω02γ=102(2.0)=104Q = \frac{\omega_0}{2\gamma} = \frac{10}{2(2.0)} = \frac{10}{4}

Q=2.5\boxed{Q = 2.5}

Low Q indicates significant damping. For high-Q systems (Q >> 1), oscillations persist many cycles.

Alternatively: Q=π×Q = \pi \times (number of oscillations to decay to 1/e)

3Problem 3hard

❓ Question:

A driven oscillator (m = 1.0 kg, k = 100 N/m, b = 2.0 kg/s) is subjected to driving force F(t) = F₀cos(ωt) where F₀ = 10 N. Find: (a) the resonance frequency, (b) the amplitude at resonance, and (c) the amplitude when driving frequency is ω = 5 rad/s.

💡 Show Solution

Given:

  • m = 1.0 kg
  • k = 100 N/m
  • b = 2.0 kg/s
  • F₀ = 10 N

(a) Resonance frequency:

Natural frequency: ω0=km=1001.0=10 rad/s\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1.0}} = 10 \text{ rad/s}

For lightly damped system, resonance occurs near ω₀:

ωres=ω02−2γ2\omega_{res} = \sqrt{\omega_0^2 - 2\gamma^2}

where γ=b2m=2.02=1.0\gamma = \frac{b}{2m} = \frac{2.0}{2} = 1.0 s⁻¹

ωres=100−2(1)=98\omega_{res} = \sqrt{100 - 2(1)} = \sqrt{98}

ωres=9.90 rad/s\boxed{\omega_{res} = 9.90 \text{ rad/s}}

(b) Amplitude at resonance:

Ares=F0bωres=10(2.0)(9.90)A_{res} = \frac{F_0}{b\omega_{res}} = \frac{10}{(2.0)(9.90)}

Ares=0.505 m\boxed{A_{res} = 0.505 \text{ m}}

(c) Amplitude at ω = 5 rad/s:

General formula: A(ω)=F0/m(ω02−ω2)2+(2γω)2A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}

A(5)=10/1.0(100−25)2+(2×1×5)2A(5) = \frac{10/1.0}{\sqrt{(100 - 25)^2 + (2 \times 1 \times 5)^2}}

A(5)=10752+102=105625+100A(5) = \frac{10}{\sqrt{75^2 + 10^2}} = \frac{10}{\sqrt{5625 + 100}}

A(5)=105725=1075.7A(5) = \frac{10}{\sqrt{5725}} = \frac{10}{75.7}

A(5)=0.132 m\boxed{A(5) = 0.132 \text{ m}}

Much smaller than at resonance!

Explain using:

📌 Related Topics in Oscillations

❓ Frequently Asked Questions

What is Damped and Driven Oscillations?▾
Damping forces, resonance, and forced oscillations
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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