Statics and Dynamics Quiz: Damping And Vibration Behavior

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Damping And Vibration BehaviorQuestion 1 of 15

Consider the free-vibration response of an underdamped system ( $\zeta = 0.1$ , $\omega_n = 10$ rad/s) that is given an initial velocity $v_0 > 0$ with zero initial displacement ( $x_0 = 0$ ).

Which statement correctly describes the qualitative shape of the displacement time history $x(t)$ for $t > 0$ , specifically addressing whether and when the displacement crosses zero and how this compares to the same system with initial displacement only ( $x_0 > 0$ , $v_0 = 0$ )?

With initial velocity only,

x(t)

rises from zero, reaches a positive maximum, then oscillates with exponentially decaying amplitude, crossing zero repeatedly. With initial displacement only,

x(t)

also oscillates and crosses zero repeatedly, but the first zero crossing occurs later for the displacement-excited case because that trajectory must travel a greater initial distance before reversing through equilibrium.

With initial velocity only,

x(t)

rises from zero, reaches a positive maximum, then oscillates with exponentially decaying amplitude, crossing zero repeatedly. With initial displacement only,

x(t)

moves toward zero immediately and crosses it sooner than the velocity-excited case, so the first zero crossing occurs earlier for the displacement-excited case.

With initial velocity only,

x(t)

rises monotonically from zero and never crosses zero again, because the initial kinetic energy is fully dissipated by damping before the restoring force can reverse the motion. This contrasts with initial displacement only, where the stored potential energy drives the mass through equilibrium, producing repeated zero crossings.

Both initial conditions produce identical zero-crossing patterns for this underdamped system because the superposition principle guarantees that any solution has the same qualitative behavior, and the spacing between zero crossings depends only on

\omega_d

, which is the same regardless of initial conditions.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Damping And Vibration Behavior

Practice Damping And Vibration Behavior in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Damping And Vibration Behavior, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Consider the free-vibration response of an underdamped system ( $\zeta = 0.1$ , $\omega_n = 10$ rad/s) that is given an initial velocity $v_0 > 0$ with zero initial displacement ( $x_0 = 0$ ).

With initial velocity only, $x(t)$ rises from zero, reaches a positive maximum, then oscillates with exponentially decaying amplitude, crossing zero repeatedly. With initial displacement only, $x(t)$ also oscillates and crosses zero repeatedly, but the first zero crossing occurs later for the displacement-excited case because that trajectory must travel a greater initial distance before reversing through equilibrium.
With initial velocity only, $x(t)$ rises from zero, reaches a positive maximum, then oscillates with exponentially decaying amplitude, crossing zero repeatedly. With initial displacement only, $x(t)$ moves toward zero immediately and crosses it sooner than the velocity-excited case, so the first zero crossing occurs earlier for the displacement-excited case. (correct answer)
With initial velocity only, $x(t)$ rises monotonically from zero and never crosses zero again, because the initial kinetic energy is fully dissipated by damping before the restoring force can reverse the motion. This contrasts with initial displacement only, where the stored potential energy drives the mass through equilibrium, producing repeated zero crossings.
Both initial conditions produce identical zero-crossing patterns for this underdamped system because the superposition principle guarantees that any solution has the same qualitative behavior, and the spacing between zero crossings depends only on $\omega_d$ , which is the same regardless of initial conditions.

Explanation: When analyzing free-vibration initial conditions, your first tool should be the general underdamped solution:

x(t) = e^{-\zeta\omega_n t}\left[A\cos(\omega_d t) + B\sin(\omega_d t)\right]

, where coefficients A and B are set by initial conditions. For initial velocity only (

x_0 = 0

v_0 > 0

): applying initial conditions gives

A = 0

and

B = v_0/\omega_d

, so

x(t) = \frac{v_0}{\omega_d}e^{-\zeta\omega_n t}\sin(\omega_d t)

. This starts at zero, rises to a positive peak, then oscillates with decaying amplitude — crossing zero each time

\sin(\omega_d t) = 0

, first at

t = \pi/\omega_d

. For initial displacement only (

x_0 > 0

v_0 = 0

): you get a cosine-dominated response. The displacement begins at

x_0

, immediately heads toward zero, and crosses it near

t = \pi/(2\omega_d)

— roughly half the time it takes the velocity-excited case to reach its first zero crossing. This confirms B: the displacement-excited trajectory crosses zero sooner. A is wrong because it reverses the comparison — it incorrectly claims the displacement-excited case takes longer to first cross zero, when in fact it crosses sooner (starting partway through a cosine cycle rather than at the base of a sine cycle). C is wrong because

\zeta = 0.1

is underdamped, meaning oscillation always occurs; the restoring force absolutely reverses the motion. D is wrong because while

\omega_d

governs the spacing between subsequent zero crossings, the timing of the first crossing depends critically on initial conditions. Your strategy: always sketch which trig function governs the response — sine (velocity IC) or cosine (displacement IC) — since this immediately tells you where the first zero crossing falls.

Question 2

A single-degree-of-freedom spring-mass-damper system has mass $m$ , spring constant $k$ , and viscous damping coefficient $c$ . An engineer increases the damping coefficient from $c_1$ to $c_2 = 4c_1$ while holding $m$ and $k$ fixed. Originally the system was underdamped with $\zeta_1 = 0.3$ .

After the damping coefficient is quadrupled, which of the following correctly describes both the new damping ratio $\zeta_2$ and the qualitative change in the system's free-vibration response?

$\zeta_2 = 1.2$ , and the system is overdamped: it returns to equilibrium without oscillating, but more slowly than the critically damped case because the dominant characteristic root moves closer to zero as $\zeta$ increases beyond 1. (correct answer)
$\zeta_2 = 0.6$ , and the system remains underdamped: it still oscillates but with a lower damped natural frequency and faster amplitude decay than before the change, since doubling $\zeta$ doubles the decay rate $\zeta\omega_n$ .
$\zeta_2 = 1.2$ , and the system is critically damped: it returns to equilibrium in the shortest possible time because $\zeta_2$ just exceeds 1 and the response sits at the boundary of oscillatory behavior.
$\zeta_2 = 0.75$ , and the system is underdamped: the damped natural frequency increases because higher damping stiffens the effective restoring force, shifting energy toward higher frequencies.

Explanation: Whenever you see a question about a spring-mass-damper system, your first instinct should be to work with the damping ratio

\zeta = \frac{c}{2\sqrt{mk}}

. Notice that

\zeta

is directly proportional to

c

, so if

c

quadruples while

m

and

k

stay fixed, the damping ratio also quadruples:

\zeta_2 = 4\zeta_1 = 4(0.3) = 1.2

. Since

\zeta_2 > 1

, the system crosses into the overdamped regime, meaning it returns to equilibrium exponentially without oscillating. Answer A correctly identifies both results. The overdamped free response contains two real, negative characteristic roots:

s_{1,2} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)

. As

\zeta

grows beyond 1, one root approaches zero while the other becomes more negative. The root near zero dominates the long-term response, making return to equilibrium slower than the critically damped case (

\zeta = 1

), which is the fastest non-oscillatory return possible. A is correct. Answer B makes two errors: it miscalculates

\zeta_2

as 0.6 (confusing quadrupling

c

with doubling

\zeta

, perhaps imagining a square-root relationship) and incorrectly keeps the system underdamped. Answer C gets the math right (

\zeta_2 = 1.2

) but misidentifies the regime — critical damping is exactly

\zeta = 1

, not

\zeta > 1

, and

\zeta_2 = 1.2

is overdamped. Answer D invents an incorrect value of

\zeta_2 = 0.75

and fabricates a physical mechanism; higher damping never increases the damped natural frequency. As a study tip: always distinguish the three regimes by the threshold

\zeta = 1

, and remember that overdamped is slower than critically damped — a counterintuitive but frequently tested fact.

Question 3

A free response crosses equilibrium repeatedly. The damping cannot be

Underdamped, ζ just below 1
Overdamped or critical (correct answer)
Lightly damped, ζ small
Underdamped, ζ near 1

Explanation: Crossing equilibrium repeatedly means the response oscillates, so the system must be underdamped, with zeta below 1. Overdamped and critically damped responses decay without oscillation; they can cross zero at most once, never repeatedly. The tempting wrong idea is that zeta just below 1 is too damped to oscillate, but any zeta below 1 still gives repeated zero crossings, just with slower oscillation.

Question 4

For an underdamped oscillator, increasing damping while staying underdamped makes it oscillate with

Lower frequency, faster decay (correct answer)
Higher frequency, faster decay
Same frequency, slower decay
Lower frequency, slower decay

Explanation: Increasing damping removes energy faster, so the amplitude decays more quickly. It also opposes the motion more strongly, which slows the oscillation, so the frequency drops. The tempting mistake is thinking damping only shrinks amplitude while leaving frequency unchanged, but the damped frequency decreases as damping increases toward critical damping.

Question 5

Released from rest, a critically damped system compared with ζ = 1.05 returns

Faster, with one overshoot
Slower, with no overshoot
Faster, with no overshoot (correct answer)
Slower, with one overshoot

Explanation: Critical damping gives the fastest return to equilibrium without oscillating. A system with zeta = 1.05 is overdamped, so it also has no overshoot but settles more slowly. The tempting wrong answer is slower with no overshoot because both cases avoid overshoot, but critical damping is the quickest non-oscillatory response.

Question 6

In a linear underdamped free vibration, successive peak amplitudes

Decrease linearly to zero
Drop by a constant amount
Remain nearly constant
Drop by a constant ratio (correct answer)

Explanation: In an underdamped free vibration, the displacement envelope decays exponentially. Because an exponential curve shrinks by the same fraction over each equal time interval, each successive peak is a fixed percentage of the previous one. The tempting wrong answer is a constant amount, but that would describe linear decay, not the exponential decay of underdamped motion.

Question 7

An underdamped oscillator's amplitude decays because the damping force

Does negative work each cycle (correct answer)
Does positive work each cycle
Keeps total energy constant
Stores energy in the damper

Explanation: The damping force always opposes the motion, so as the oscillator moves through each cycle the force and displacement are opposite in direction. That makes the work done by damping negative, removing mechanical energy and shrinking the amplitude. The tempting wrong answer is that the damper stores energy, but a damper dissipates energy as heat rather than storing it.

Question 8

Two identical spring-mass systems (same $m$ and $k$ ) are given different damping coefficients: System 1 has $\zeta = 0.5$ and System 2 has $\zeta = 2.0$ . Both are released from the same initial displacement with zero initial velocity. Which statement most accurately compares their responses at a very large time $t \to \infty$ ?

Both displacements approach zero, but System 1 reaches a practically negligible amplitude sooner because its oscillatory decay envelope falls off faster than the dominant slow exponential in the overdamped System 2. (correct answer)
System 2 reaches zero displacement in finite time because overdamped systems have no oscillatory component, whereas System 1 oscillates indefinitely at a constant amplitude since the damping only shifts the frequency.
Both displacements approach zero at the same rate because, for large $t$ , all linear viscous systems decay with the same time constant determined solely by $\omega_n$ , regardless of $\zeta$ .
System 1 reaches negligible amplitude sooner only if its damped period is shorter than the decay time of System 2, which cannot be determined without knowing the actual values of $m$ and $k$ .

Explanation: Whenever you see a question comparing underdamped and overdamped systems at large times, your focus should shift from oscillation behavior to exponential decay rates. The long-term behavior is governed entirely by the slowest-decaying exponential term. For a spring-mass-damper system, the general decay involves

e^{-\zeta\omega_n t}

as the characteristic envelope scale. For the underdamped System 1 (

\zeta = 0.5

), the solution is

x(t) = Ae^{-0.5\omega_n t}\cos(\omega_d t + \phi)

, decaying with time constant

\tau_1 = 1/(0.5\omega_n) = 2/\omega_n

. For the overdamped System 2 (

\zeta = 2.0

), the solution is a sum of two pure exponentials:

x(t) = C_1 e^{-s_1 t} + C_2 e^{-s_2 t}

, where

s_{1,2} = (\zeta \mp \sqrt{\zeta^2-1})\omega_n

. The slower root gives

s_{\min} = (2 - \sqrt{3})\omega_n \approx 0.27\omega_n

, meaning the dominant time constant is

\tau_2 \approx 3.7/\omega_n

. Since

\tau_1 < \tau_2

, System 1 decays to negligible amplitude faster. Answer A correctly captures this. B is wrong on two counts: overdamped systems never reach zero in finite time (exponentials are asymptotic), and underdamped systems absolutely do decay — the amplitude envelope shrinks to zero. C is wrong because decay rates depend explicitly on

\zeta

; different damping ratios produce different time constants even with identical

\omega_n

. D is a trap — since both systems share the same

m

and

k

(hence the same

\omega_n

), the comparison is fully determined by

\zeta

alone. Study tip: For long-time behavior, always identify the slowest exponential root — that term dominates, and overdamped systems often surprise students by decaying more slowly than underdamped ones.

Question 9

A viscously damped system is described by $\ddot{x} + 2\zeta\omega_n \dot{x} + \omega_n^2 x = 0$ . An engineer claims: 'Increasing $\zeta$ from 0.9 to 1.1 always makes the system settle faster because the damping force is larger.' Which of the following best evaluates this claim?

The claim is true only if $\omega_n$ is simultaneously increased to maintain a constant decay rate $\sigma = \zeta\omega_n$ ; otherwise, increasing $\zeta$ alone at fixed $\omega_n$ has an indeterminate effect on settling time near $\zeta = 1$ .
The claim is true. Any increase in $\zeta$ increases the real part of the characteristic roots, which always accelerates the exponential decay regardless of whether the system transitions from underdamped to overdamped.
The claim is false. Increasing $\zeta$ beyond 0.9 reduces the damped natural frequency $\omega_d$ , which lengthens the period of oscillation and therefore increases the time required for the amplitude envelope to decay to a negligible level.
The claim is false. The settling time is minimized at $\zeta = 1$ (critical damping). Moving from $\zeta = 0.9$ to $\zeta = 1.1$ crosses through the minimum, so the system at $\zeta = 1.1$ settles more slowly than at $\zeta = 1$ even though the damping force is larger than at $\zeta = 0.9$ . (correct answer)

Explanation: When analyzing damped systems, don't just ask "is there more damping?" — ask "how does the damping change the roots of the characteristic equation, and what does that do to settling time?" For

\ddot{x} + 2\zeta\omega_n\dot{x} + \omega_n^2 x = 0

, the characteristic roots are

s = -\zeta\omega_n \pm \omega_n\sqrt{\zeta^2 - 1}

. When

\zeta < 1

(underdamped), both roots are complex with real part

-\zeta\omega_n

, and settling time is governed by

\tau = 1/(\zeta\omega_n)

. As

\zeta \to 1

, this decay rate improves. But once

\zeta > 1

(overdamped), the roots become two distinct negative reals:

s_1 = -\zeta\omega_n + \omega_n\sqrt{\zeta^2-1}

and

s_2 = -\zeta\omega_n - \omega_n\sqrt{\zeta^2-1}

. The slower root

s_1

now dominates settling, and its magnitude decreases as

\zeta

increases beyond 1. So at

\zeta = 1.1

, the system actually settles more slowly than at

\zeta = 1

, confirming that D is correct: critical damping minimizes settling time, and crossing through it in either direction makes settling slower. Choice A is wrong because this isn't about simultaneously tuning

\omega_n

— the question holds

\omega_n

fixed, and the effect of increasing

\zeta

past 1 is entirely determinate, not ambiguous. Choice B is wrong because beyond critical damping, increasing

\zeta

actually reduces the magnitude of the dominant real root, slowing — not accelerating — decay. Choice C identifies a real phenomenon (reduced

\omega_d

) but draws the wrong conclusion; the oscillation frequency is irrelevant once the system is overdamped, and the envelope argument doesn't properly account for the root behavior near

\zeta = 1

. Your study tip: always sketch or compute the dominant characteristic root as a function of

\zeta

. The "more damping = faster settling" intuition breaks down past critical damping, and exam questions love to exploit exactly that transition.

Question 10

A structural engineer is comparing two building models with the same undamped natural frequency $\omega_n$ . Building X has $\zeta_X = 0.02$ (lightly damped steel frame) and Building Y has $\zeta_Y = 0.15$ (heavily damped base-isolated structure). Both are subjected to free vibration after an impulsive load. Which statement correctly compares both the number of observable oscillations before the amplitude drops to 1% of its initial value and the damped natural frequencies of the two buildings?

Building Y undergoes more oscillations than Building X before reaching 1% amplitude because heavier damping dissipates less energy per cycle relative to the stored energy, allowing more cycles before decay, and Building Y's damped frequency is higher than Building X's due to the stiffening effect of the isolation bearings.
Building X undergoes approximately 7.5 times as many oscillations as Building Y before reaching 1% amplitude, and both buildings have essentially the same damped natural frequency because the difference between $\omega_n\sqrt{1-0.02^2}$ and $\omega_n\sqrt{1-0.15^2}$ is less than 2% and therefore negligible for practical purposes.
Building X undergoes approximately 7.5 times as many oscillations as Building Y before reaching 1% amplitude, and Building Y's damped natural frequency $\omega_{dY}$ is measurably lower than $\omega_{dX}$ because $\omega_d = \omega_n\sqrt{1-\zeta^2}$ and the $\zeta^2$ term is significantly larger for Building Y. (correct answer)
Both buildings undergo the same number of oscillations before reaching 1% amplitude because the number of cycles to decay depends only on $\omega_n$ , which is identical for both buildings, and the 1% threshold is reached at the same time since the decay envelope $e^{-\zeta\omega_n t}$ is evaluated at the same $\omega_n$ .

Explanation: When analyzing damped free vibration, you need to track two separate quantities: how quickly the amplitude decays (governed by the exponential envelope) and how fast the system oscillates (governed by the damped natural frequency). For the amplitude decay, the envelope is

e^{-\zeta\omega_n t}

. Setting this equal to 0.01 and solving gives

t_{1\%} = \frac{\ln(100)}{\zeta\omega_n} \approx \frac{4.605}{\zeta\omega_n}

. The number of oscillations before reaching 1% is this time divided by the damped period, which simplifies to approximately

N \approx \frac{4.605}{2\pi\zeta} \cdot \sqrt{1-\zeta^2}

. For small

\zeta

, this reduces to roughly

\frac{0.733}{\zeta}

. Plugging in: Building X gives

N_X \approx 36.6

cycles and Building Y gives

N_Y \approx 4.9

cycles — a ratio of approximately 7.5. So Building X clearly undergoes far more oscillations. For the damped frequency,

\omega_d = \omega_n\sqrt{1-\zeta^2}

. For Building X:

\sqrt{1-0.0004} \approx 0.9998\omega_n

, essentially unchanged. For Building Y:

\sqrt{1-0.0225} \approx 0.9887\omega_n

, about 1.1% lower — small but measurably different, especially when observing cycle timing over many seconds. This confirms C is correct. A is wrong because it reverses the logic entirely — heavier damping dissipates more energy per cycle, causing faster decay, not slower. The claim about stiffening from isolation bearings is also fabricated physics. B is wrong because it correctly identifies the 7.5× ratio but then incorrectly dismisses the frequency difference as negligible — 1.1% is measurable in structural monitoring contexts, and the question specifically asks about it. D is wrong because it conflates the time to decay with the number of cycles; identical

\omega_n

does not mean identical cycle counts when

\zeta

differs. Your study tip: always separate the decay rate (

\zeta\omega_n

) from the oscillation rate (

\omega_d

) — exam questions frequently mix these two distinct effects to create plausible-sounding distractors.

Question 11

The characteristic equation for a viscously damped SDOF system is $ms^2 + cs + k = 0$ . For a specific system, the two roots are found to be $s_{1,2} = -4 \pm 3j$ (where $j = \sqrt{-1}$ ). Which of the following statements correctly identifies the damping condition and interprets the physical meaning of the imaginary part of the roots?

The system is underdamped ( $\zeta < 1$ ); the imaginary part $3$ rad/s equals the damped natural frequency $\omega_d$ , which is always less than the undamped natural frequency $\omega_n = \sqrt{16+9} = 5$ rad/s. (correct answer)
The system is underdamped ( $\zeta < 1$ ); the imaginary part $3$ rad/s equals the undamped natural frequency $\omega_n$ , confirming that damping does not alter the fundamental resonant frequency of the system.
The system is critically damped ( $\zeta = 1$ ); the imaginary part $3$ rad/s represents a phase angle in radians rather than a frequency, since critically damped systems have repeated real roots with a small oscillatory correction term.
The system is overdamped ( $\zeta > 1$ ); complex conjugate roots with a nonzero imaginary part indicate that the exponential decay rate exceeds the natural frequency, which is the defining condition for overdamping.

Explanation: When you see complex roots for a damped SDOF system, your first task is to decode what each part of the root tells you physically. The roots

s_{1,2} = \sigma \pm j\omega_d

always take this form for an underdamped system: the real part

\sigma = -\zeta\omega_n

governs exponential decay, and the imaginary part equals the damped natural frequency

\omega_d

, not the undamped frequency. For the roots

s_{1,2} = -4 \pm 3j

, the presence of a nonzero imaginary part immediately tells you the system is underdamped — it oscillates while decaying. The undamped natural frequency is recovered from both parts together:

\omega_n = \sqrt{\sigma^2 + \omega_d^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{25} = 5

rad/s. The damping ratio follows as

\zeta = 4/5 = 0.8 < 1

, confirming underdamping. Since

\omega_d = \omega_n\sqrt{1-\zeta^2} < \omega_n

, the imaginary part

3

rad/s is always less than

5

rad/s — exactly what A states. B is wrong because it misidentifies

\omega_d = 3

as the undamped frequency

\omega_n

, and falsely claims damping doesn't shift the oscillation frequency — it always does. C is wrong because critically damped systems produce repeated real roots with zero imaginary part; the imaginary part here is definitely a frequency, not a phase angle. D is wrong because overdamped systems produce two distinct real roots — complex conjugate roots with a nonzero imaginary part are the hallmark of underdamping, not overdamping. A reliable memory aid: complex roots → underdamped; repeated real roots → critically damped; two distinct real roots → overdamped. Knowing this table cold will let you classify any SDOF system instantly.

Question 12

A damped SDOF system has its damping provided entirely by Coulomb (dry) friction rather than viscous damping. The system is released from an initial displacement of 50 mm.

A student applies the standard viscous damping classification (underdamped, critically damped, overdamped) to this Coulomb-damped system. Which of the following best describes the fundamental error in this approach and the actual qualitative behavior of the Coulomb-damped system?

The error is that Coulomb friction introduces a nonlinear term that causes the system to oscillate at a frequency lower than $\omega_n$ , similar to the underdamped viscous case but with an exponential decay envelope whose rate changes each half-cycle, making the standard classification approximately correct but quantitatively inaccurate.
The error is that viscous damping classifications apply only to systems with more than one degree of freedom; for SDOF systems, Coulomb friction can be equivalently modeled as viscous damping using an effective damping coefficient, so the underdamped/overdamped framework is still applicable without loss of accuracy.
The error is that viscous damping classifications assume the damping force is proportional to velocity, whereas Coulomb friction provides a constant-magnitude force opposing motion. Coulomb-damped free vibration oscillates at the undamped natural frequency with a linearly decreasing amplitude envelope — not an exponential one — and the motion arrests permanently in finite time when the spring restoring force can no longer overcome static friction. (correct answer)
The error is that the damping ratio $\zeta$ cannot be defined for Coulomb friction because the system has no characteristic equation, and therefore no classification is possible; the Coulomb-damped system behaves chaotically for large initial displacements and only approaches periodic motion near equilibrium.

Explanation: Whenever you see a question mixing different damping types, your first instinct should be to ask: what does the damping force actually depend on? Viscous damping force scales with velocity (

F_d = c\dot{x}

), which is what makes the underdamped/critically damped/overdamped classification meaningful — it emerges directly from the characteristic equation of a linear ODE. Coulomb friction, by contrast, produces a constant-magnitude force

F_f = \mu N

that simply reverses direction with motion. This fundamental difference invalidates the entire viscous framework. For Coulomb-damped free vibration, the equation of motion is piecewise linear, not truly linear. Each half-cycle, the system oscillates at exactly the undamped natural frequency

\omega_n

(the friction term shifts the equilibrium point but doesn't alter the frequency). Crucially, the amplitude decreases by a fixed amount each half-cycle — producing a linear decay envelope, not the exponential envelope of viscous damping. Motion stops permanently in finite time once the restoring force

kx

falls below the static friction limit. This makes C the correct answer. Choice A is tempting but wrong on two counts: the frequency remains

\omega_n

(not lower), and the envelope is linear, not exponential with a changing rate. Choice B is doubly false — equivalent viscous damping is an approximation used for forced response analysis, not an exact equivalence, and the SDOF/MDOF distinction is irrelevant here. Choice D incorrectly claims no characteristic equation exists and invents chaotic behavior; Coulomb systems are well-defined and deterministic, just nonlinear. Your study tip: memorize that Coulomb damping → linear amplitude decay + finite stop time, while viscous damping → exponential decay + asymptotic approach to rest. Exams love testing this contrast.

Question 13

A mechanical engineer is designing a door-closing mechanism modeled as a rotational spring-mass-damper. The specification requires that the door return to the closed position as quickly as possible without overshooting (i.e., without the door bouncing back open). The engineer has freedom to choose the damping coefficient $c$ .

Which damping condition satisfies the design specification, and what is the physical consequence of choosing a damping coefficient that is even slightly below this target value?

Critical damping ( $\zeta = 1$ ) satisfies the specification because it provides the fastest return to equilibrium without oscillation. If $c$ is slightly below critical, the system becomes underdamped and the door will overshoot the closed position at least once, violating the no-bounce requirement. (correct answer)
Overdamping ( $\zeta > 1$ ) satisfies the specification because any non-oscillatory response meets the no-bounce criterion. If $c$ is slightly below the chosen overdamped value, the system moves closer to critical damping and closes faster, so the specification continues to be met or exceeded.
Critical damping ( $\zeta = 1$ ) satisfies the specification because no oscillation occurs at this condition. If $c$ is slightly below critical, the time constant of the response increases but the door still returns without overshoot, so the no-bounce requirement is partially satisfied even in the underdamped regime.
Overdamping ( $\zeta > 1$ ) satisfies the specification because the two distinct real characteristic roots guarantee a monotonic return with no oscillation. If $c$ drops below the overdamped threshold, the system becomes underdamped and the door overshoots, so the engineer must keep $\zeta$ safely above 1 to guarantee no bounce at the cost of a slower closure.

Explanation: When analyzing vibration control problems, always start by identifying the two competing requirements: speed of response and absence of oscillation. These goals pull in opposite directions, and the damping ratio

\zeta

is the parameter that balances them. Critical damping (

\zeta = 1

) sits exactly at the boundary between oscillatory and non-oscillatory behavior. Mathematically, the characteristic equation has a repeated real root, producing a response of the form

(A + Bt)e^{-\omega_n t}

. This decays to zero faster than any overdamped case while still being strictly non-oscillatory — making it the optimal solution when you need the fastest return without any overshoot. Answer A captures this precisely. The critical insight is what happens just below: when

c

drops even slightly below the critical value,

\zeta < 1

, the roots become complex conjugates, and the response includes a sinusoidal component. This means the door swings past the closed position — exactly the bounce the specification forbids. Answer B is wrong because overdamping does satisfy no-bounce, but it is not the fastest response. The specification demands both conditions simultaneously, and critical damping uniquely achieves this. B also incorrectly implies that moving toward critical damping always helps — it does increase speed, but crossing below

\zeta = 1

introduces oscillation. Answer C contains a dangerous error: underdamped systems do overshoot. Claiming the no-bounce requirement is "partially satisfied" misrepresents the physics — any oscillation means the door bounces. Answer D wrongly prescribes overdamping as the solution, sacrificing the speed requirement without necessity. Study tip: Remember that critical damping is the unique sweet spot — it's not a range but a single value, making it both the target and a boundary you cannot cross downward safely.

Question 14

A student measures the free vibration of a system and records the following three successive positive displacement peaks: $x_1 = 20$ mm, $x_2 = 12$ mm, $x_3 = 7.2$ mm.

Using the logarithmic decrement method, what is the damping ratio $\zeta$ , and which statement about the computation is correct?

The logarithmic decrement is $\delta = \ln(x_1/x_2) = \ln(20/12) \approx 0.511$ , giving $\zeta = \delta/(2\pi) \approx 0.081$ . However, the method requires that consecutive peaks be separated by exactly one damped period $T_d$ , and since the problem does not explicitly confirm the time between recorded peaks equals $T_d$ , the damping ratio cannot be determined from displacement magnitudes alone.
The logarithmic decrement is $\delta = \ln(x_1/x_3) = \ln(20/7.2) \approx 1.022$ , giving $\zeta = \delta/\sqrt{4\pi^2 + \delta^2} \approx 0.161$ . This two-cycle value is more accurate than the single-cycle estimate because averaging over more cycles reduces the effect of measurement noise on the result.
The logarithmic decrement is $\delta = \ln(x_1/x_2) = \ln(20/12) \approx 0.511$ , and the small- $\delta$ approximation $\zeta \approx \delta/(2\pi) \approx 0.081$ is used here. The exact formula $\zeta = \delta/\sqrt{4\pi^2+\delta^2}$ yields a meaningfully different result only when $\delta > 1$ , so the approximation and exact formula are considered interchangeable for this problem.
The logarithmic decrement is $\delta = \ln(x_1/x_2) = \ln(20/12) \approx 0.511$ , giving $\zeta = \delta/\sqrt{4\pi^2 + \delta^2} \approx 0.081$ . Using two cycles gives $\delta = \frac{1}{2}\ln(x_1/x_3) = \frac{1}{2}\ln(20/7.2) \approx 0.511$ , confirming consistency with the single-cycle result. (correct answer)

Explanation: Whenever you see a free-vibration problem with recorded displacement peaks, your first instinct should be the logarithmic decrement method: the natural log of the ratio of successive same-direction peaks, separated by exactly one damped period

T_d

. The single-cycle logarithmic decrement is

\delta = \ln(x_1/x_2) = \ln(20/12) \approx 0.511

. Plugging into the exact formula gives

\zeta = \delta/\sqrt{4\pi^2 + \delta^2} = 0.511/\sqrt{4\pi^2 + 0.511^2} \approx 0.081

. To verify consistency, you can average over two cycles:

\delta = \frac{1}{2}\ln(x_1/x_3) = \frac{1}{2}\ln(20/7.2) \approx 0.511

. The identical result confirms the data is internally consistent and your single-cycle answer is reliable. That's exactly what D demonstrates. A is a trap that invents a problem — it claims the timing of peaks must be explicitly confirmed in the problem statement. In practice, "successive positive peaks" implies one damped period between them by definition; no separate timing confirmation is needed. B correctly uses two cycles but makes a critical algebra error: it computes

\delta = \ln(x_1/x_3) \approx 1.022

without dividing by the number of cycles (2), inflating

\delta

by a factor of two and doubling the final

\zeta

. C correctly computes

\delta \approx 0.511

but makes a false claim: it says the approximation

\zeta \approx \delta/(2\pi)

and the exact formula are "interchangeable" only when

\delta > 1

. In fact they diverge for any nonzero

\delta

; you should always prefer the exact formula. Study tip: When multiple cycles are available, use the averaged form

\delta = \frac{1}{n}\ln(x_1/x_{n+1})

as a cross-check — matching results between one-cycle and two-cycle calculations is a reliable consistency signal on exam problems.

Question 15

A researcher observes two freely vibrating systems released from rest at the same initial displacement $x_0$ . System A shows successive peaks at $x_0 = 10$ mm, $6.7$ mm, and $4.5$ mm. System B returns smoothly to zero without crossing it, reaching $x_0/2$ in time $T$ and $x_0/10$ in time $3T$ .

Based on the observed responses, which classification is correct, and what key feature of System B's response rules out the possibility that System B is critically damped rather than overdamped?

System A is underdamped and System B is critically damped. No feature of System B's description rules out overdamping, because both critically damped and overdamped systems return to zero without oscillating, and the displacement values given are consistent with a single decaying exponential of the form $(A + Bt)e^{-\omega_n t}$ .
System A is underdamped and System B is overdamped. The feature ruling out critical damping is the non-constant effective decay rate between times $T$ and $3T$ : a critically damped system has a single effective time constant, whereas the slowing decay observed here indicates two distinct exponential time constants, which is characteristic of overdamping. (correct answer)
System A is underdamped and System B is overdamped. The definitive feature ruling out critical damping is that System B never crosses zero displacement; a critically damped system released from rest always crosses the equilibrium position exactly once before settling, while overdamped systems do not cross zero.
System A is underdamped and System B is overdamped. The feature ruling out critical damping is that System B reaches $x_0/10$ at time $3T$ rather than $2T$ ; critical damping always causes displacement to drop by one decade per two time constants, whereas overdamping requires three or more time constants for the same reduction due to the slower dominant mode.

Explanation: When analyzing damped vibration problems, your first job is to classify each system by its response signature, then carefully distinguish why one classification fits over another — especially between critically damped and overdamped, which can look similar at first glance. System A oscillates through successive peaks (10 mm → 6.7 mm → 4.5 mm), confirming underdamping — energy is lost each cycle, but the system still oscillates. System B never crosses zero and decays smoothly, which eliminates underdamping. The real challenge is distinguishing overdamped from critically damped. Here's the key insight that makes B correct: examine System B's effective decay rate. From rest to time

T

, displacement drops from

x_0

x_0/2

— a 50% reduction. From

T

3T

(twice as long), displacement drops from

x_0/2

x_0/10

— an 80% reduction over double the interval. The rate of decay is slowing down over time. A critically damped system has the form

(A + Bt)e^{-\omega_n t}

, which produces a single characteristic decay rate that doesn't slow in this way. An overdamped system, however, contains two distinct exponential terms with different time constants — the slower one dominates at large

t

, causing the apparent decay rate to decrease. That non-constant effective decay rate is the fingerprint of overdamping. A is wrong because it claims System B could be critically damped — the changing decay rate rules this out. C contains a factual error: a critically damped system released from rest does not cross zero; neither does an overdamped one, so this distinction doesn't help you separate them. D fabricates a rule about "one decade per two time constants" that has no basis in vibration theory. Your study tip: when distinguishing overdamped from critically damped, don't just check whether the system crosses zero — check whether the decay rate remains consistent or slows over time. Slowing decay signals two competing exponentials, which means overdamping.