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.
Damping forces, resonance, and forced oscillations
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mdt2d2x+
bdtdx+
kx=
0
Divide by m:
dt2d2x+2γdtdx+ω02x=0
where:
γ=b/(2m) (damping coefficient)
ω0=k/m (natural frequency)
Three Cases of Damping
Underdamped (γ<ω0):
x(t)=Ae−γtcos(ω′t+ϕ)
where ω′=ω02−γ2 (damped frequency)
Oscillates with decreasing amplitude.
Critically damped (γ=ω0):
x(t)=(A+Bt)e−γt
Returns to equilibrium fastest without oscillating.
Overdamped (γ>ω0):
x(t)=Ae−αt+Be−βt
where α,β=γ±γ2−ω02
Returns to equilibrium slowly without oscillating.
Quality Factor
Q=2γω0
High Q: Light damping, many oscillations before amplitude decays
Low Q: Heavy damping, few oscillations
Energy decay:E(t)=E0e−2γt
Time constant:τ=1/(2γ)
Driven Oscillations
With external driving force F(t)=F0cos(ωt):
mdt2d2x+bdtdx+kx=F0cos(ωt)
dt2d2x+2γdtdx+ω02x=mF0cos(ωt)
Steady-State Solution
After transients die out:
x(t)=A(ω)cos(ωt−δ)
Amplitude:A(ω)=(ω02−ω2)2+(2γω)2F0/m
Phase lag:tanδ=ω02−ω22γω
Resonance
Amplitude is maximum when denominator is minimum.
Resonant frequency:ωr=ω02−2γ2
For light damping (γ≪ω0): ωr≈ω0
Maximum amplitude:Amax≈2mγω0F0=mω02F0Q=kF0Q
Power and Resonance
Average power absorbed:Pavg=21bω2A2
Maximum at:ω=ω0 (exactly)
Power at resonance:Pmax=4mγF02
Bandwidth and Q
Half-power points: where P=Pmax/2
Occur at ω=ω0±γ (for small γ)
Bandwidth:Δω=2γ
Q=Δωω0=2γω0
Sharp resonance: high Q, narrow bandwidth
Complex Representation
Using x~=x0eiωt:
−ω2x~+2iγωx~+ω02x~=mF0
x~=ω02−ω2+2iγωF0/m
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
ω0=mk=0.550=100
ω0=10 rad/s
(b) Damping constant:
γ=2mb=2(0.5)2.0=1.02.0
γ=2.0 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:
System is UNDERDAMPED
(d) Damped frequency:
ωd=ω02−γ2=100−4
ωd=96
ωd=9.80 rad/s
Motion: x(t)=Ae−γtcos(ωdt+ϕ)
Period: Td=ωd2π=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−γt
A(Td)=A0e
A(Td)=0.10e−1.282=0.10(0.277)
A(Td)=0.0277 m
Amplitude decreased to 28% in one period!
(b) Time to reach 10% of original:
0.01=0.10e−γt
0.1=e−2t
ln(0.1)=−2t
t=−2ln(0.1)=−
t=1.15 s
(c) Quality factor:
Q=2γω0=
Q=2.5
Low Q indicates significant damping. For high-Q systems (Q >> 1), oscillations persist many cycles.
Alternatively: Q=π× (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=mk=1.0100=10 rad/s
For lightly damped system, resonance occurs near ω₀:
ωres=ω02−
where γ=2mb=2 s⁻¹
ωres=100−2(1)
ωres=9.90 rad/s
(b) Amplitude at resonance:
Ares=bωres
Ares=0.505 m
(c) Amplitude at ω = 5 rad/s:
General formula:
A(ω)=(ω0
A(5)=(100−25)2+(
A(5)=752+10
A(5)=5725
A(5)=0.132 m
Much smaller than at resonance!
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