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[parent] GRE Physics Companion: Damped Harmonic Motion

(Example)

GRE Physics Companion: Damped Harmonic Motion

The standard linear damped oscillator is

|-----------2-------|
¨x-+-2γ-˙x +-ω0x-=-0,--
(1)

with

                  ∘  ---
γ = -b-,     ω0 =    k-.
    2m               m
(2)

The characteristic roots are

|----------------------|
|           ∘   2    2 |
|r± = − γ ±   γ  − ω 0.|
-----------------------
(3)

PIC

Figure 1. A compact damping strategy: compute ω0 and γ, classify the roots, choose the correct solution form, then use energy or peak decay as an independent check.

1 High-value GRE facts

  1. γ = b∕(2m) for the convention ẍ + 2γẋ + ω02x = 0.
  2. ω0 = ∘  -----
   k∕m is the undamped natural angular frequency.
  3. Underdamped: γ < ω0.
  4. Critical: γ = ω0.
  5. Overdamped: γ > ω0.
  6. ωd = ∘  --------
   ω20 − γ2 in the underdamped regime.
  7. The underdamped envelope decays as e−γt.
  8. critical damping requires bc = 2√ ----
  mk.
  9. mechanical energy satisfies dE∕dt = −bv2.
  10. logarithmic decrement is Λ = 2πγ∕ωd.

Part I: Original GRE-style problems

Problem 1: damping rate

For

m x¨+ bx˙+ kx  = 0,
(4)

the damping rate in

x¨+  2γ ˙x + ω20x = 0
(5)

is

  1. b∕m
  2. b∕(2m)
  3. 2b∕m
  4. k∕b
  5. ∘ -----
  b∕m

Problem 2: regime classification

A system has

       −1
γ = 3 s  ,    ω0 =  5rad∕s.
(6)

It is

  1. undamped
  2. underdamped
  3. critically damped
  4. overdamped
  5. unstable

Problem 3: damped frequency

For the system in Problem 2,

ω
  d
(7)

is

  1. 2 rad/s
  2. 3 rad/s
  3. 4 rad/s
  4. 5 rad/s
  5. √34-- rad/s

Problem 4: amplitude decay

For an underdamped oscillator with

       −1
γ = 2 s  ,
(8)

the amplitude envelope after

0.50s
(9)

is multiplied by

  1. e−2
  2. e−1
  3. e−1∕2
  4. 2
  5. 1

Problem 5: critical coefficient

For

m  = 2 kg,     k = 50 N∕m,
(10)

critical damping requires

  1. 5 kg/s
  2. 10 kg/s
  3. 20 kg/s
  4. 25 kg/s
  5. 100 kg/s

Problem 6: critical solution

The general critically damped displacement has the form

  1. A cos ω0t
  2. Ae−γt cos ω dt
  3. (C + Dt)e−ω0t
  4. Ceω0t + De−ω0t
  5. C + Dt

Problem 7: energy loss rate

For viscous damping,

dE ∕dt
(11)

equals

  1. +bv2
  2. −bv2
  3. −kx2
  4. −mgv
  5. zero

Problem 8: phase plane

The underdamped phase-plane trajectory is approximately

  1. a closed ellipse
  2. an outward spiral
  3. an inward spiral
  4. a straight line
  5. a hyperbola

Problem 9: logarithmic decrement

Successive same-sign peaks are

An =  10,     An+1 = 5.
(12)

The logarithmic decrement is

  1. 1∕2
  2. ln 2
  3. 2
  4. e2
  5. π

Problem 10: overdamped roots

For an overdamped oscillator, the characteristic roots are

  1. purely imaginary
  2. complex conjugates with negative real part
  3. two distinct negative real numbers
  4. one positive and one negative number
  5. both zero

Problem 11: weak-damping quality factor

For weak viscous damping,

Q
(13)

is approximately

  1. 2γ∕ω0
  2. ω0∕(2γ)
  3. γω0
  4. ω02∕γ
  5. 1

Problem 12: damping and period

For an underdamped oscillator,

  1. ωd > ω0
  2. ωd = ω0 for all damping
  3. ωd < ω0
  4. ωd = 0 always
  5. ωd is imaginary

Part II: Complete worked solutions

Solution 1

Dividing by mass gives

     b     k
¨x + -- ˙x + --x = 0.
    m      m
(14)

Therefore

2 γ = -b,
      m
(15)

so

γ =  -b-.
     2m
(16)

Answer: (B).

Solution 2

Because

3 < 5,
(17)

γ < ω0.
(18)

The system is underdamped. Answer: (B).

Solution 3

ωd = ∘ --------
  ω20 − γ2 (19)
= √ -------
  25 − 9 (20)
= 4 rad∕s. (21)

Answer: (C).

Solution 4

The envelope factor is

e− γt = e− (2)(0.50) = e−1.
(22)

Answer: (B).

Solution 5

bc = 2√ ----
  mk (23)
= 2∘ -------
  (2)(50) (24)
= 20 kg∕s. (25)

Answer: (C).

Solution 6

A repeated root produces

x(t) = (C + Dt )e−ω0t.
(26)

Answer: (C).

Solution 7

The damping force power is

                   2
Pd =  (− bv)v = − bv .
(27)

Thus

             2
dE ∕dt = − bv .
(28)

Answer: (B).

Solution 8

Damping removes energy, so the closed undamped phase-plane orbit contracts toward the origin. In the underdamped case it spirals inward. Answer: (C).

Solution 9

      (    )
        10-
Λ = ln   5   =  ln 2.
(29)

Answer: (B).

Solution 10

For

γ > ω0,
(30)

∘ --------
  γ2 − ω2
        0
(31)

is real and smaller than

γ.
(32)

Thus both roots are real, distinct, and negative. Answer: (C).

Solution 11

For weak damping,

Q  ≈  ω0.
      2γ
(33)

Answer: (B).

Solution 12

     ∘  --------
ω  =    ω2 − γ2 < ω .
  d      0          0
(34)

Answer: (C).

2 GRE checklist

  1. Compute ω0 and γ before selecting a solution form.
  2. Compare γ directly with ω0.
  3. Keep the convention 2γ = b∕m consistent.
  4. For underdamping, use ωd = ∘ --------
  ω20 − γ2.
  5. For critical damping, remember the extra factor of t in the second independent solution.
  6. For overdamping, expect two decaying exponentials, not a sinusoid.
  7. Use dE∕dt = −bv2 as a sign and physics check.
  8. Use logarithmic decrement only with corresponding same-sign peaks.

References

References

[1]   PhysicsLibrary, M06-01, Simple Harmonic Motion.

[2]   A. P. French, Vibrations and Waves, W. W. Norton, 1971.

[3]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.


"GRE Physics Companion: Damped Harmonic Motion" is owned by bloftin.
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Other names:  M06-02G
Keywords:  GRE physics, damped harmonic motion, underdamped oscillator, critical damping, overdamped oscillator, damping ratio, damped frequency, logarithmic decrement, energy dissipation, phase plane

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Cross-references: power, force, mass, complex conjugates, viscous damping, displacement, amplitude, system, damping rate, logarithmic decrement, mechanical energy, critical damping, angular frequency, energy, oscillator

This is version 1 of GRE Physics Companion: Damped Harmonic Motion, born on 2026-10-18.
Object id is 1491, canonical name is GREPhysicsCompanionDampedHarmonicMotion.
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Classification:
Physics Classification: 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
 45.05.+x (General theory of classical mechanics of discrete systems)

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