FE Mechanical · knowledge area 7 · 10–15 questions
FE dynamics review, motion, energy, vibration
Dynamics, kinematics and vibrations is one of the heaviest areas of the outline at 10–15 questions. It is also the one where units do the most damage: rpm, rad/s and Hz all describe the same shaft, and only one of them goes into the formula.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam10–15
worked problems here18
of 14 knowledge areas7
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What the area covers
The specification splits the area into particles, rigid bodies and vibrations:
Particles: kinematics in rectangular, normal-tangential and polar coordinates; Newton's second law; work-energy; impulse-momentum, including impact.
Kinetic friction: sliding bodies where the friction force is μk·N and does negative work.
Rigid bodies: kinematics of translation, fixed-axis rotation and general plane motion; Newton's second law with the mass moment of inertia; work-energy and impulse-momentum for rotating parts.
Mechanisms: linkages, gear trains and the velocity and acceleration of points on them.
Vibrations: free and forced response of single-degree-of-freedom systems, with and without damping.
18 dynamics problems, every option explained
0 / 18 answered · 0 correct
Question 1 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A structural resonance test produces a frequency response curve with a distinct peak. The maximum amplitude occurs at exactly 30 Hz. The amplitude drops to 0.707 times the peak maximum value at frequencies of 28.5 Hz and 31.5 Hz. What is most nearly the estimated damping ratio for this system?
Given
Peak frequency fn = 30 Hz
Lower half-power frequency f1 = 28.5 Hz
Upper half-power frequency f2 = 31.5 Hz
Find
Damping ratio ζ
Solution
ζ ≈ (f2 − f1)/(2fn)
ζ ≈ (31.5 − 28.5)/(2 × 30) = 0.050
Dividing the bandwidth by 4fn halves the estimate; the half-power formula has 2fn in the denominator.
This divides the correct 0.050 by 0.707; the 0.707 ratio only identifies the half-power points and does not enter the formula.
Δf/fn = 0.100 equals 2ζ (the loss factor); the factor of 2 in the denominator was dropped.
Multiplying Δf/fn by √2 mixes the amplitude ratio into the result; the bandwidth alone sets ζ.
Question 2 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
An undamped spring-mass system has a measurable stiffness of 500 lbf/in and an oscillating weight of 161 lbf. A harmonic external force is applied at an operational frequency of 600 RPM. What is most nearly the dimensionless frequency ratio r of the system? (Assume g = 32.2 ft/s2)
FE Reference Handbook, Dynamics: free and forced vibration.
Using 600 rpm = 10 Hz as if it were rad/s gives 10/34.6; the forcing speed must be converted to 62.8 rad/s.
This is ωn/ω, the ratio inverted; the frequency ratio is forcing over natural frequency.
m = 161/(32.2 × 12) = 0.4167 lbf·s2/in, ωn = √(500/0.4167) = 34.6 rad/s, ω = 62.8 rad/s, so r = 1.81.
This is r2; the frequency ratio itself is not squared.
Taking mass as 161/32.2 = 5 slug with k still in lbf/in mixes inches and feet; k must be 6000 lbf/ft or m must carry the factor 12.
Question 3 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A uniform heavy spring with a mass of 3.0 kg and a stiffness of 400 N/m supports a lumped tip mass of 8.0 kg. Calculate the fundamental natural frequency of this single-degree-of-freedom system.
Given
Spring mass ms = 3.0 kg
Spring stiffness k = 400 N/m
Tip mass m = 8.0 kg
Find
Natural frequency ωn (rad/s)
Solution
meq = m + ms/3 = 8.0 + 1.0 = 9.0 kg
ωn = √(k/meq) = √(400/9.0) = 6.667 rad/s
This ignores the spring's own mass, which a heavy spring adds to the vibrating mass.
Rayleigh's correction adds one third of the spring mass: ωn = √(400/(8 + 1)) = 6.667 rad/s.
Adding half of the spring mass overstates its share; the equivalent mass of a uniform spring is ms/3.
Adding the whole 3.0 kg spring mass overstates the moving mass; only ms/3 participates.
This is 6.667/2π, the frequency in hertz, printed with rad/s units.
Question 4 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A 5 kg mass is fired from a spring gun with spring constant k = 500 N/m. If the spring was initially compressed 0.2 m, what is the initial velocity of the mass when it leaves the gun (assuming no friction)?
Given
Mass m = 5 kg
Spring constant k = 500 N/m
Spring compression x = 0.2 m
Find
Launch velocity v (m/s)
Solution
½kx2 = ½mv2
v = x√(k/m) = (0.2 m)√(500/5) = 2.0 m/s
FE Reference Handbook, Dynamics: work-energy principle.
Writing ½kx2 = mv2 drops the ½ from the kinetic energy and gives v = √(kx2/(2m)) = 1.4 m/s.
½kx2 = ½mv2 → v = x√(k/m) = 0.2 × √100 = 2.0 m/s.
Writing kx2 = ½mv2 drops the ½ from the spring energy and gives v = √(2kx2/m) = 2.8 m/s; both energy terms carry ½.
This is kx2/m = 4 without the square root; it has units of m2/s2, not m/s.
Question 5 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A 500 g ball traveling at 8 m/s undergoes a perfectly elastic head-on collision with a 1 kg ball initially at rest. What is the velocity of the 500 g ball after the collision (positive in its original direction of travel)?
Given
Ball 1 mass m1 = 500 g
Ball 1 velocity v1 = 8 m/s
Ball 2 mass m2 = 1 kg (at rest)
Find
Velocity of the 500 g ball after collision v1′ (m/s)
The size is right, but a lighter ball striking a heavier one at rest rebounds, so the velocity is negative.
Question 6 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A uniform disk with radius 0.3 m and mass 5 kg rolls without slipping down a 20° incline. What is the disk's linear acceleration?
Given
Disk radius r = 0.3 m
Disk mass m = 5 kg
Incline angle θ = 20°
Find
Linear acceleration a (m/s²)
Solution
a = g sin θ/(1 + I/(mr2)) with I = ½mr2
a = (9.81)(sin 20°)/1.5 = 2.24 m/s2
FE Reference Handbook, Dynamics: plane motion of a rigid body (mass moment of inertia of a cylinder).
Keeping only the rotational term, a = g sin θ/(I/(mr2)) = 2g sin θ = 6.71 m/s2, drops the 1 that accounts for translation.
g sin θ is the acceleration of a frictionless sliding block; rolling ties up energy in rotation.
This uses I = 2/5·mr2 for a solid sphere, g sin θ/1.4; a disk has I = ½mr2.
For a solid disk I = ½mr2, so a = g sin θ/(1 + ½) = 3.36/1.5 = 2.24 m/s2; mass and radius cancel.
Question 7 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A vibrating machine is mounted on a base isolator system designed such that the frequency ratio is r = 4.0. If the damping ratio of the isolator is subsequently increased from 0.05 to 0.20, what is the effect on the steady-state transmissibility of the system?
FE Reference Handbook, Dynamics: forced vibration (transmissibility).
Resonance depends on the frequency ratio; damping barely changes the natural frequency and r stays at 4.
A damper transmits force through c·ẋ, so adding damping can only add a transmitted component.
The 2ζr term appears in the transmissibility formula, so damping changes TR at every r except √2.
Damping lowers transmissibility only for r < √2; in the isolation range it raises it.
Above r = √2 damping works against isolation: at r = 4, TR rises from about 0.07 at ζ = 0.05 to about 0.13 at ζ = 0.20.
Question 8 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A vibrating machine with a mass of 100 kg is mounted on an isolator with a stiffness of 40,000 N/m and a damping ratio of 0.20. An external harmonic force defined by F(t) = 500 sin(30t) N is applied. Determine the steady-state amplitude of the machine.
X = (F0/k)/√((1 − r2)2 + (2ζr)2) = 12.5 mm/√(1.5625 + 0.36) = 9.02 mm
FE Reference Handbook, Dynamics: forced vibration.
ωn = 20 rad/s, r = 1.5, and X = (F0/k)/√((1 − r2)2 + (2ζr)2) = 12.5/1.386 = 9.02 mm.
This uses ζr instead of 2ζr in the denominator, understating the damping term.
This is the undamped result 12.5/|1 − r2|; the 2ζr term is left out.
F0/k is the static deflection; above resonance the dynamic amplitude is smaller than that.
δst/(2ζ) is the amplitude at resonance (r = 1); the machine runs at r = 1.5.
Question 9 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
An engineer calculates the front and rear normal forces on a transfer cart accelerating along a straight, level track. The free-body diagram is solved with ΣFy = 0 and ΣM = 0 about the rear wheel contact point, exactly as in a statics problem. Which dynamic effect was omitted from this calculation?
Solution
ΣMrear contact = (m·a)·h (not 0) → front/rear normal forces shift by m·a·h/L
FE Reference Handbook, Dynamics: plane motion of a rigid body (D'Alembert).
The cart runs in a straight line, so there is no normal acceleration and no centrifugal effect.
Drag at shop speeds is negligible, and it is an external force any free-body diagram can include; it is not the dynamic term.
Rolling resistance is small and acts the same at constant speed; it is not what distinguishes the accelerating case.
Gyroscopic moments need a change in the direction of the wheel spin axes, which a straight track does not produce.
About a point other than the CG, ΣM equals the moment of m·a at height h, which shifts load between the axles; static equilibrium leaves it out.
Question 10 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A discrete point on the rim of a mechanical flywheel with a radius of 0.5 m is analyzed. At a specific instant, the flywheel has an instantaneous angular velocity of 4.0 rad/s and an angular acceleration of 6.0 rad/s2. What is most nearly the total absolute acceleration magnitude of the point?
Given
Radius r = 0.5 m
Angular velocity ω = 4.0 rad/s
Angular acceleration α = 6.0 rad/s2
Find
Total acceleration a (m/s2)
Solution
at = αr = 6.0 × 0.5 = 3.0 m/s2, an = ω2r = 16 × 0.5 = 8.0 m/s2
a = √(at2 + an2) = √73 ≈ 8.5 m/s2
FE Reference Handbook, Dynamics: plane circular motion.
This is only the tangential component αr; the centripetal component ω2r is larger here.
Adding αr = 3 to ωr = 2 mixes a velocity (m/s) into an acceleration; the normal component is ω2r.
This is only the centripetal component ω2r; the angular acceleration adds a tangential part.
at = αr = 3.0 m/s2 and an = ω2r = 8.0 m/s2 are perpendicular, so a = √(32 + 82) ≈ 8.5 m/s2.
Adding 3 + 8 algebraically ignores that the two components are at right angles.
Question 11 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A linear single-degree-of-freedom mechanical system consists of a 50 kg mass attached to a spring with a stiffness of 20 kN/m. During free vibration, the maximum kinetic energy recorded is 10 J. Determine the amplitude of vibration.
Given
Mass m = 50 kg
Spring stiffness k = 20 kN/m
Maximum kinetic energy KEmax = 10 J
Find
Amplitude A (mm)
Solution
KEmax = ½kA2
A = √(2KEmax/k) = √(2 × 10/20,000) = 0.03162 m = 31.62 mm
FE Reference Handbook, Dynamics: free vibration.
KE/k = 10/20,000 m skips both the factor 2 and the square root.
2·KE/k = 0.001 m gives A2 in m2, not A; the square root is missing.
√(KE/k) drops the ½ in ½kA2, understating the amplitude by √2.
Maximum kinetic energy equals the spring energy at full amplitude, ½kA2 = 10 J, so A = √(20/20,000) m = 31.62 mm.
This is the peak speed √(2KE/m) = 0.632 m/s read as a length; the amplitude follows from the spring stiffness.
Question 12 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A flywheel with moment of inertia I = 5 kg·m² rotates at 300 rpm. What torque is required to bring it to rest in 15 seconds?
Given
Moment of inertia I = 5 kg·m²
Initial speed n = 300 rpm
Stopping time t = 15 s
Find
Required torque τ (N·m)
Solution
ω0 = 300 × 2π/60 = 31.42 rad/s
T = I·ω0/t = 5 × 31.42/15 = 10.47 N·m
FE Reference Handbook, Dynamics: angular impulse and momentum.
Dividing I × 300 by 15 s uses rpm directly; the speed must first be converted to rad/s.
This is Iω = 157 kg·m2/s, the angular momentum; dividing by the 15 s stopping time is missing.
This converts 300 rpm to 1885 rad/min but never to rad/s: 5 × 1885/15 = 628.32; the factor 60 is missing.
Question 13 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
Two gears are in mesh with pitch radii of 4 cm and 12 cm. If the smaller gear rotates at 1200 rpm clockwise, what is the angular velocity of the larger gear in rad/s?
The speed is right, but two external gears in mesh always rotate in opposite senses.
Multiplying by r2/r1 instead of r1/r2 inverts the ratio; the larger gear is the slower one.
ω2 = ω1r1/r2 = 125.66 × 4/12 = 41.89 rad/s, and meshing external gears turn in opposite directions.
This is the smaller gear's own speed (1200 rpm) in rad/s; the larger gear turns three times slower.
Question 14 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A particle moves along a path described by x(t) = t3 - 6t2 + 9t (in meters). Determine the instantaneous acceleration at t = 4 s.
Given
Position x(t) = t3 − 6t2 + 9t m
Time t = 4 s
Find
Acceleration a (m/s2)
Solution
v = dx/dt = 3t2 − 12t + 9
a = dv/dt = 6t − 12 = 6(4) − 12 = 12 m/s2
This is the average acceleration over 0–4 s, [v(4) − v(0)]/4 = (9 − 9)/4 = 0; the question asks for the instantaneous value.
This is the position x(4) = 64 − 96 + 36 = 4 m; the function has to be differentiated twice.
This is the velocity v(4) = 3(16) − 12(4) + 9 = 9 m/s, one derivative short.
a = d2x/dt2 = 6t − 12 = 6(4) − 12 = 12 m/s2.
Question 15 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A projectile is launched from ground level with an initial speed of 30 m/s at an angle of 40 degrees above the horizontal. Neglecting air resistance, determine the total time of flight and the horizontal range.
Given
Initial speed v0 = 30 m/s
Launch angle θ = 40°
Find
Time of flight T (s) and horizontal range R (m)
Solution
t = 2v0 sin θ/g = 2(30)(sin 40°)/9.81 = 3.93 s
R = v0 cos θ · t = (30)(cos 40°)(3.93) = 90.3 m
FE Reference Handbook, Dynamics: projectile motion.
t = 2v0 sin θ/g = 2(30)(0.643)/9.81 = 3.93 s and R = v0 cos θ · t = 23.0 × 3.93 = 90.3 m.
1.97 s is the time to the top of the arc; the projectile needs as long again to come down.
Using cos θ in the flight time gives t = 2v0 cos θ/g = 4.69 s; multiplying that by v0 cos θ = 23.0 m/s gives 107.7 m. The flight time depends on the vertical component v0 sin θ.
The time is right, but the range uses the vertical component v0 sin θ instead of v0 cos θ.
Question 16 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A damped oscillator with m = 1 kg and k = 16 N/m has a damping ratio of 0.5. Determine the time required for its amplitude to decay to 20% of its initial value.
Given
Mass m = 1 kg
Stiffness k = 16 N/m
Damping ratio ζ = 0.5
Amplitude ratio = 20 %
Find
Decay time t (s)
Solution
ωn = √(k/m) = 4 rad/s, x(t)/x0 = e−ζωnt = 0.20
t = ln 5/(ζωn) = 1.609/2 = 0.80 s
The envelope decays as e−ζωnt with ζωn = 0.5 × 4 = 2 s−1, so t = ln 5/2 = 0.80 s.
ln 5/ωn leaves ζ out of the decay rate ζωn.
This uses log10 5 = 0.699 instead of ln 5 = 1.609: 0.699/2 = 0.35 s.
ln(1/0.8)/2 is the time to lose 20 % of the amplitude, not to fall to 20 %.
Question 17 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A mass-spring-damper system has m = 2 kg, k = 32 N/m, and damping constant c = 8 Ns/m. Determine the damped natural frequency and the damping ratio of the system.
Given
Mass m = 2 kg
Spring constant k = 32 N/m
Damping constant c = 8 N·s/m
Find
Damped natural frequency ωd (rad/s) and damping ratio ζ
ωn(1 − ζ2) = 3.00 forgets the square root on (1 − ζ2).
Taking the critical damping as 2√(k/m) = 8 makes the system look critically damped; cc = 2√(km) = 16 N·s/m.
4.00 rad/s is the undamped natural frequency; damping lowers the oscillation frequency.
Question 18 of 18 · Dynamics, Kinematics, and Vibrations · 10–15 items
A rigid link AB undergoes general planar motion. At a given instant, the absolute velocity of point A is purely horizontal and the absolute velocity of point B is purely vertical. Point A is at (0, 2) m and point B is at (3, 0) m. What are the coordinates of the instantaneous center of rotation of the link at this instant?
Given
Point A = (0, 2) m, velocity horizontal
Point B = (3, 0) m, velocity vertical
Find
Coordinates of the instantaneous center
Solution
Perpendicular to vA (horizontal) through A: x = 0. Perpendicular to vB (vertical) through B: y = 0.
IC = (0, 0) m
FE Reference Handbook, Dynamics: instantaneous center of rotation.
The midpoint of AB is not special; the center is set by the velocity directions, not by the link geometry alone.
This is point A itself; A is moving, so it cannot be the center of zero velocity.
This intersects lines drawn along the velocities; the instantaneous center lies on lines perpendicular to them.
This is point B itself, which has a nonzero velocity.
The instantaneous center lies on the perpendicular to each velocity: the vertical line x = 0 through A and the horizontal line y = 0 through B meet at the origin.
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Common traps
The mistakes below come up again and again in Meriam and Kraige's Engineering Mechanics: Dynamics and in the Dynamics chapter of the FE Reference Handbook, which holds the particle, rigid-body and vibration relations:
Three units for one speed. Circular frequency ω is in rad/s, cyclic frequency f is in Hz, and shaft speed is in rpm. Convert rpm with 2π/60 before any formula that expects radians.
The wrong moment of inertia. Rotation needs the mass moment of inertia (kg·m2), not the area moment from statics. For rotation about an axis away from the center of mass, add m·d2.
Kinetic energy with a piece missing. A body that rolls or swings has both translational and rotational kinetic energy. Dropping the ½·I·ω2 term inflates the speed.
Momentum is a vector. In a collision, keep the sign of each velocity along the line of impact; the coefficient of restitution compares relative speeds of separation and approach along that line.
Springs in series and in parallel. Parallel stiffnesses add; series stiffnesses combine like resistors in parallel. Swapping the two moves the natural frequency by a large factor.
Undamped versus damped frequency. A question about the oscillation you would actually observe wants the damped value; a question about resonance or the frequency ratio usually starts from the undamped natural frequency.
Static deflection shortcut
For a mass hanging on a spring, the static deflection under its own weight already contains k/m. That is why vibration problems sometimes give a deflection instead of a stiffness: the natural frequency follows from g and the deflection alone.
Where dynamics shows up again
Vibration response reappears as the dynamic response of physical systems in measurements, instrumentation and controls, where a sensor or a mass-spring-damper is treated as a first- or second-order system. Gear trains and shaft speeds feed power transmission in mechanical design and analysis.
Dynamics is explained on this page for the whole site: the civil exam's smaller dynamics area links here. Differential equations behind free vibration belong to the shared mathematics section.
No. The natural frequency, damping ratio and forced-response relations are in the Dynamics chapter of the on-screen Handbook. What saves time is knowing which quantity a question wants and which unit it is in.
How many dynamics questions are on the FE Mechanical exam?
The specification gives 10–15 questions for dynamics, kinematics and vibrations together. Only fluid mechanics, thermodynamics and mechanical design share that range; no area is larger.
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