Mechanical vibrations: practice problems with step-by-step solutions
Free Mechanical vibrations practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: natural frequency, damping ratio, springs in series and parallel.
Free practice problemEasy
A machine rests on spring mounts, which compress by 15 mm under its weight. Treat it as an undamped single-degree-of-freedom system and use g = 9.81 m/s². Find the natural frequency fn of vertical vibration in Hz.
Hang a mass on a spring, pull it down a little and let go. It bounces up and down around its resting position. That repeating motion is a vibration.
Every such system has its own favourite rhythm, the natural frequency. It depends only on how stiff the spring is and how heavy the mass is, not on how far you pulled.
Engineers care because a machine that is shaken at its natural frequency can vibrate violently. This is called resonance. Engine mounts, washing machines and bridges are all designed with it in mind.
A stiffer spring makes the vibration faster. A heavier mass makes it slower.
Self-check You replace the mass on a spring with one that is four times heavier. What happens to the natural frequency?
It becomes four times lower
It doubles
It halves
It stays the same
Answer: It halves. The frequency depends on √(k/m). Four times the mass divides the square root by √4 = 2, so the frequency halves.
2Natural frequency and spring combinations
For a mass m (kg) on a spring of stiffness k (N/m), the natural circular frequency in rad/s is:
ωn = √(k / m)
One full cycle is 2π radians. So the natural frequency in hertz (cycles per second) and the period (seconds per cycle) are:
fn = ωn / (2π) T = 1 / fn
Sometimes you only know how far the spring sags under the weight, the static deflection δst = m·g/k. Then ωn = √(g / δst), with δst in metres.
When there are several springs, first replace them by one equivalent spring:
Side by side (parallel), the springs share the load and the stiffnesses add.
End to end (series), each spring takes the full load, and the equivalent spring is softer than either one.
Self-check Two springs of 10 kN/m each are connected end to end (in series). What is the equivalent stiffness?
10 kN/m
100 kN/m
20 kN/m
5 kN/m
Answer: 5 kN/m. 1/k = 1/10 + 1/10 = 2/10, so k = 5 kN/m. Adding them to 20 kN/m is the rule for springs in parallel.
3Damping
Real vibrations die out. A damper (a piston moving through oil, like a car’s shock absorber) takes energy out. Its strength is the damping coefficient c in N·s/m.
The critical damping is the amount of damping that just stops the mass from bouncing at all:
cc = 2 · √(k · m) = 2 · m · ωn
The damping ratio compares the real damping with that critical value:
ζ = c / cc
ζ < 1: underdamped. The mass still vibrates, but the swings shrink.
ζ = 1: critically damped. It returns to rest as fast as possible without overshooting.
ζ > 1: overdamped. It creeps back slowly without vibrating.
An underdamped system vibrates a little slower than the undamped one, at the damped natural frequency:
ωd = ωn · √(1 − ζ²)
Self-check A 10 kg mass sits on a spring of 1000 N/m with a damper of c = 50 N·s/m. What is the damping ratio?
0.25
0.5
2.5
4
Answer: 0.25. cc = 2·√(1000 · 10) = 200 N·s/m, so ζ = 50/200 = 0.25. Forgetting the factor 2 gives 0.5, and dividing the other way round gives 4.
4The method step by step
Convert everything to SI units: kN/m to N/m (× 1000), mm to m, grams to kg.
If there are several springs, combine them into one k: add them in parallel, add their reciprocals in series.
Find ωn = √(k/m) in rad/s.
If the question asks for hertz, divide by 2π: fn = ωn/(2π). For the period, take T = 1/fn.
With damping, find cc = 2·√(k·m), then ζ = c/cc and, if asked, ωd = ωn·√(1 − ζ²).
Keep ω (rad/s) and f (Hz) apart. They differ by a factor 2π ≈ 6.28, and mixing them up is the most common way to lose the marks.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A 50 kg mass hangs from a spring with a stiffness of 20 kN/m. Ignore damping. Find the natural frequency fn in Hz and the period of vibration.
1
Draw and convert
The stiffness must be in N/m: k = 20 kN/m = 20 000 N/m. The mass is already in kg.
2
Natural circular frequency
ωn = √(k / m) = √(20 000 / 50) = √400 = 20 rad/s
3
Convert to hertz
Divide by 2π to count full cycles per second:
fn = 20 / (2π) = 3.18 Hz
4
The period
T = 1 / fn = 1 / 3.183 = 0.314 s
The mass makes a little more than three full up-and-down cycles every second.
5
Check the answer
The static deflection is δst = m·g/k = 50 · 9.81 / 20 000 = 0.0245 m.
ωn = √(9.81 / 0.0245) = 20 rad/s
The same 20 rad/s, found in a different way. The answer is right.
6
Going further: add a damper
With a damper of c = 400 N·s/m: cc = 2·√(20 000 · 50) = 2000 N·s/m, so ζ = 400/2000 = 0.2.
ωd = 20 · √(1 − 0.2²) = 19.6 rad/s, fd = 3.12 Hz
Answer
fn ≈ 3.18 Hz (ωn = 20 rad/s), and the period is T ≈ 0.314 s.
6Common mistakes
Giving ωn in rad/s when the question asks for fn in Hz, or the other way round.
Leaving the stiffness in kN/m. The formula needs N/m.
Adding springs in series as if they were in parallel.
Forgetting the 2 in cc = 2·√(k·m).
Using the weight m·g instead of the mass m in ωn = √(k/m).
Self-check A 4 kg mass hangs from a spring of 400 N/m. What is its natural frequency in Hz?
0.05 Hz
1.59 Hz
10 Hz
62.8 Hz
Answer: 1.59 Hz. ωn = √(400/4) = 10 rad/s, and fn = 10/(2π) = 1.59 Hz. Stopping at 10 gives rad/s, not Hz; multiplying by 2π gives 62.8; using 0.4 (kN/m) gives 0.05 Hz.
7How exam questions look
Easy: natural frequency. A mass on a spring: find ωn in rad/s or fn in Hz. Some questions give only the static deflection of a machine on its mounts.
Medium: damping ratio. Find ζ from m, k and c, or find the damped frequency fd from a given ζ. Others give two peaks from a free-vibration test.
Hard: springs in series and parallel, sometimes both (two springs side by side holding a bar, with a third spring below). Find keq first, then fn.
For two peaks x0 and xn that are n cycles apart, use the logarithmic decrement:
A 40 kg mass hangs from a spring with a stiffness of 15 kN/m. Ignore damping. Find the natural frequency fn in Hz.
m = 40 kgk = 15 kN/m
Find fn(Hz)
Show hints
Work in SI units: k in N/m.
ωn = √(k/m), fn = ωn/(2π).
k = 15000 N/m.
Show the answer and the step-by-step solution
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Natural circular frequency
ωn = √(k/m) = √(15000/40) = 19.36 rad/s
In hertz
fn = ωn/(2π) = 19.36/(2π) = 3.082 Hz
Answer
fn = 3.082 Hz
Example 2Medium
A 10 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
m = 10 kgk = 50 kN/mζ = 0.4
Find fd(Hz)
Show hints
Find the undamped natural frequency first.
ωn = √(k/m), fn = ωn/(2π).
fd = fn·√(1 − ζ²).
Show the answer and the step-by-step solution
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Undamped natural frequency
ωn = √(50000/10) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 11.25·√(1 − 0.4²) = 10.31 Hz
Answer
fd = 10.31 Hz
Example 3Hard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 20 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 80 kN/mk2 = 20 kN/m
Find keq(kN/m)
Show hints
Parallel springs share the deflection: stiffnesses add. Series springs share the force: compliances 1/k add.
1/keq = 1/k₁ + 1/k₂.
keq < 20 kN/m.
Show the answer and the step-by-step solution
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
Hang a mass on a spring, pull it down a little and let go. It bounces up and down around its resting position. That repeating motion is a vibration.
Every such system has its own favourite rhythm, the natural frequency. It depends only on how stiff the spring is and how heavy the mass is, not on how far you pulled.
How do I solve Mechanical vibrations problems step by step?
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
What are common mistakes in Mechanical vibrations?
That is ωn in rad/s. Divide by 2π to get the frequency in Hz.
Convert kN/m to N/m (multiply by 1000) before using ωn = √(k/m).
Divide ωn by 2π to get Hz; do not multiply.
What do the Mechanical vibrations exercises cover?
Easy: natural frequency
Medium: damping ratio
Hard: springs in series and parallel
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.