Vibrations: Free, Damped and Forced Response of Single-Degree-of-Freedom Systems

This chapter answers Section XE3.3, Vibrations, of XE3 Solid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The syllabus text is: “Free and forced vibration of single-degree-of-freedom systems; effect of damping; base excitation.” It teaches the equation of motion and natural frequency, equivalent stiffness and mass with the energy method, viscous damping with the damping ratio and logarithmic decrement, forced harmonic response with resonance and rotating unbalance, and base excitation with transmissibility and isolation. The numericals are a cantilever with a tip mass, a damping ratio from decaying peaks, amplitudes of a forced and an unbalanced machine, and the stiffness of a vibration isolator. Take g = 9.81 m/s², linear springs, viscous (force proportional to velocity) damping, and masses as rigid and concentrated, unless the stem says otherwise.

1. Free undamped vibration of a single-degree-of-freedom system

A system with one degree of freedom has its configuration fixed by a single coordinate x. For a mass m on a spring of stiffness k, Newton’s second law about the static equilibrium position (where the spring force already balances the weight, so gravity drops out) gives m ẍ + k x = 0. Its solution is x = A cos ω_n t + B sin ω_n t = X cos(ω_n t − φ), with natural frequency ω_n = √(k/m) rad/s, f_n = ω_n/2π Hz and period T = 2π/ω_n. The initial conditions x₀, v₀ fix the amplitude X = √(x₀² + (v₀/ω_n)²). Frequency depends on k and m alone, not on the amplitude. Doubling the mass lowers ω_n by √2; making the spring four times stiffer doubles it.

If the spring deflects statically by δ_st under the weight, k = mg/δ_st and so ω_n = √(g/δ_st); a measured static sag gives the frequency at once. For rotation, J θ̈ + k_t θ = 0 gives ω_n = √(k_t/J), with torsional stiffness k_t = GJ_p/L for a uniform circular shaft (J_p is the polar second moment of area, not the inertia J of the disc). Worked example. A 2 kg mass on a 800 N/m spring: ω_n = √(800/2) = 20 rad/s, f_n = 3.18 Hz, T = 0.314 s. Released with x₀ = 10 mm and v₀ = 0.3 m/s, X = √(0.01² + (0.3/20)²) = √(1.0 × 10⁻⁴ + 2.25 × 10⁻⁴) = 18.0 mm.

⚠️ Gravity and units
Measure x from the static equilibrium position, otherwise a constant mg appears and misleads. The result ω_n is in rad/s; f = ω/2π is in Hz, and a speed quoted in rpm must be converted with ω = 2πN/60. Mixing these gives answers wrong by a factor of 2π or 60.

2. Equivalent stiffness and mass, and the energy method

Springs in parallel share the same deflection and their stiffnesses add: k_eq = k₁ + k₂. Springs in series carry the same force and their flexibilities add: 1/k_eq = 1/k₁ + 1/k₂. Elastic members act as springs, with k = force per unit deflection at the mass: a cantilever with a tip load has k = 3EI/L³, a simply supported beam loaded at mid-span has k = 48EI/L³, a beam fixed at both ends and loaded at mid-span has k = 192EI/L³, and an axial rod has k = AE/L. A mass carried by a flexible beam on a spring is a series arrangement, so it is softer than either.

The energy method avoids the free-body diagram. For a conservative system with one coordinate q, the kinetic energy is T = ½ m_eq q̇² and the potential energy is V = ½ k_eq q². Conservation of T + V gives m_eq q̈ + k_eq q = 0, so ω_n = √(k_eq/m_eq); equivalently, Rayleigh’s method equates the largest kinetic energy (at the equilibrium position) to the largest potential energy (at the extreme): T_max = V_max, with q = Q cos ω_n t. Write every velocity in terms of q̇ by the kinematic constraint, which is how rolling and pulley masses are included.

Worked example (rolling cylinder). A uniform solid cylinder of mass m and radius R rolls without slipping, its axle joined to a wall by a spring k. With x of the centre, ω = ẋ/R: T = ½mẋ² + ½(½mR²)(ẋ/R)² = ¾mẋ², so m_eq = 3m/2 and V = ½kx², giving ω_n = √(2k/3m), lower than √(k/m) because of the rotational inertia. Worked example (cantilever). A 12 kg tip mass on a steel cantilever of L = 0.5 m, E = 200 GPa, I = 1.0 × 10⁻⁸ m⁴ (beam mass neglected): k = 3EI/L³ = 3(200 × 10⁹)(10⁻⁸)/0.125 = 48 000 N/m, and ω_n = √(48 000/12) = 63.2 rad/s. Pendulums. A simple pendulum has ω_n = √(g/L); a uniform rod of length L pivoted at its end has ω_n = √(3g/2L), from (mgL/2)/(mL²/3), for small angles.

3. Damped free vibration: damping ratio, logarithmic decrement

With a viscous damper of coefficient c (force c ẋ opposing velocity), m ẍ + c ẋ + k x = 0. The critical damping coefficient is c_c = 2√(km) = 2mω_n, and the damping ratio is ζ = c/c_c, which gives ẍ + 2ζω_n ẋ + ω_n² x = 0. The roots are s = ω_n(−ζ ± √(ζ² − 1)). Three cases follow.

Free response of a damped system
CaseConditionMotion
Under-dampedζ < 1Decaying oscillation x = X e^(−ζω_n t) cos(ω_d t − φ), ω_d = ω_n√(1 − ζ²)
Critically dampedζ = 1No oscillation; the fastest return to rest without overshoot
Over-dampedζ > 1No oscillation; a slower creep back to rest than at critical damping

The logarithmic decrement for two successive peaks one period T_d = 2π/ω_d apart is δ = ln(x_n/xn+1) = 2πζ/√(1 − ζ²), and over N cycles δ = (1/N) ln(x₀/x_N). For small ζ, δ ≈ 2πζ. Inverting, ζ = δ/√(4π² + δ²). Each cycle multiplies the amplitude by e^(−δ). Worked example. m = 10 kg, k = 4000 N/m, c = 80 N·s/m: ω_n = 20 rad/s, c_c = 2(10)(20) = 400 N·s/m, ζ = 0.2, ω_d = 20√0.96 = 19.6 rad/s, δ = 2π(0.2)/√0.96 = 1.283, so each peak is e^(−1.283) = 0.277 of the previous one.

⚠️ Damped frequency and critical damping
ω_d is always below ω_n, but for ζ = 0.1 the difference is only 0.5%, so ω_d ≈ ω_n for light damping. The ratio of successive peaks depends on ζ only, not on the amplitude (the decay is geometric), and the peaks fall by the same ratio whatever the starting amplitude. Critical damping is the dividing line, not a quantity that stops the motion: an over-damped system returns more slowly.

4. Forced harmonic vibration, resonance and rotating unbalance

Under a force F₀ sin ωt, m ẍ + c ẋ + k x = F₀ sin ωt. After the transient dies out, the steady-state response is x = X sin(ωt − φ) at the driving frequency. With frequency ratio r = ω/ω_n: X = (F₀/k)/√((1 − r²)² + (2ζr)²) and tan φ = 2ζr/(1 − r²). The magnification factor is M = X/(F₀/k) = 1/√((1 − r²)² + (2ζr)²). For r ≪ 1, M ≈ 1 (the system follows the force like a static load) and φ ≈ 0. At resonance r = 1, X = F₀/(2ζk), M = 1/(2ζ) and φ = 90° (displacement lags force). For r ≫ 1, M ≈ 1/r² (the mass cannot follow; X ≈ F₀/mω²) and φ → 180°.

The peak of M occurs at r = √(1 − 2ζ²), slightly below 1, with M_max = 1/(2ζ√(1 − ζ²)); a peak exists only for ζ < 1/√2. For small ζ the peak is at r ≈ 1 and M_max ≈ 1/(2ζ). Worked example. F₀ = 50 N, k = 5000 N/m (so F₀/k = 10 mm), ζ = 0.1, r = 2: M = 1/√((1 − 4)² + (0.4)²) = 1/√9.16 = 0.330, so X = 3.30 mm. At r = 1 the same system would give X = 10/(0.2) = 50 mm.

Rotating unbalance. A machine of total mass M has an unbalanced rotor mass m₀ at eccentricity e turning at ω. The vertical exciting force is m₀eω² sin ωt, which grows with speed, so M ẍ + c ẋ + k x = m₀eω² sin ωt. Then MX/(m₀e) = r²/√((1 − r²)² + (2ζr)²). It is zero at r = 0, equals 1/(2ζ) at r = 1 and tends to 1 for r ≫ 1, so X → m₀e/M: the machine ends up vibrating about its centre of mass, with an amplitude set by the unbalance alone. Worked example. M = 100 kg, m₀e = 0.05 kg·m, r = 2, ζ = 0.1: MX/(m₀e) = 4/√9.16 = 1.32, so X = 1.32 × 0.05/100 = 0.66 mm.

⚠️ Where the peak is, and which force
The exact peak is not at r = 1 unless damping is negligible, but at resonance the phase is exactly 90° whatever ζ is. For a constant-amplitude force the amplitude falls as 1/r² at high speed; for a rotating unbalance the force grows as r², so the amplitude levels off at m₀e/M instead of falling to zero.

5. Base excitation, transmissibility and vibration isolation

When the support moves as y = Y sin ωt and the mass is joined to it by a spring k and damper c, the absolute displacement x of the mass satisfies m ẍ + c(ẋ − ẏ) + k(x − y) = 0. The steady amplitude ratio, the displacement transmissibility, is TR = X/Y = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²). The force transmissibility F_T/F₀ for a machine with exciting force F₀ mounted on spring and damper has the same expression: the force passed to the floor is X√(k² + c²ω²). The motion of the mass relative to the base, z = x − y, has Z/Y = r²/√((1 − r²)² + (2ζr)²).

Transmissibility regions
Frequency ratioTransmissibilityEffect of more damping
r < √2TR > 1, peak near r = 1Lowers TR, especially near resonance
r = √2TR = 1 for every ζNo effect
r > √2TR < 1: isolationRaises TR, which worsens isolation

Isolation therefore needs ω > √2 ω_n, so the isolator must be soft: ω_n well below the disturbing frequency. For an undamped isolator TR = 1/(r² − 1) (taking the magnitude, r > √2). Since ω_n = √(g/δ_st), a large static deflection means a low natural frequency. Worked example. A 200 kg machine runs at 1500 rpm, so ω = 2π(1500)/60 = 157.1 rad/s. For TR = 0.2 on undamped mounts, r² − 1 = 5, r² = 6, ω_n² = ω²/6 = 4112 rad²/s², so k = mω_n² = 200 × 4112 = 822 kN/m and δ_st = g/ω_n² = 2.39 mm. Heavier damping would give TR above 0.2 at this speed. A lightly damped isolator is still passed through resonance at start-up, which is why a little damping is kept.

⚠️ Soft is good, and damping is a compromise
A stiffer isolator raises ω_n and r falls, which moves the system towards resonance. Above r = √2, adding damping makes TR worse (r = 3, ζ = 0 gives 0.125; ζ = 0.2 gives 0.193), while below √2 it helps. Do not mix the two ratios: TR compares the output with the base motion, whereas Z/Y compares relative motion with it.

Key takeaways

  • A one-degree-of-freedom spring-mass system has ω_n = √(k/m) = √(g/δ_st), f = ω_n/2π, T = 2π/ω_n, independent of amplitude. Measure x from static equilibrium so gravity drops out, and convert rpm with ω = 2πN/60.
  • Parallel springs add, series springs add flexibilities; a cantilever has 3EI/L³. By the energy method, ω_n = √(k_eq/m_eq), with every velocity written in the coordinate; a rolling cylinder on a spring has m_eq = 3m/2.
  • With ζ = c/(2√(km)): under-damped if ζ < 1, with ω_d = ω_n√(1 − ζ²); critical at ζ = 1; over-damped above it. Logarithmic decrement δ = 2πζ/√(1 − ζ²), and ζ = δ/√(4π² + δ²).
  • For forced vibration X = (F₀/k)M with M = 1/√((1 − r²)² + (2ζr)²); at r = 1, M = 1/(2ζ) and the phase is 90°. For r ≫ 1, M falls as 1/r². Rotating unbalance has MX/(m₀e) = r²M, which tends to 1 at high speed.
  • Transmissibility TR = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²) equals 1 at r = √2 for every ζ. Isolation needs r > √2, where more damping worsens it; for undamped mounts TR = 1/(r² − 1).
  • Exam traps: using the damping ratio in place of c, ω_d for ω_n, rpm in place of rad/s, putting the peak exactly at r = 1, forgetting the static-deflection form, and using more damping to improve isolation above √2.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A mass on a linear spring vibrates freely without damping. If the mass is made four times larger and the spring is unchanged, the natural frequency becomes:

    1. Four times smaller, since the frequency is inversely proportional to the mass
    2. Half of its original value, since the frequency varies as 1/√m
    3. Unchanged, since the natural frequency does not depend on the mass but only on the amplitude
    4. Two times larger, since a heavier mass stores more energy and vibrates faster
    Show answer

    Answer: B — Half of its original value, since the frequency varies as 1/√m

    ω_n = √(k/m), so m → 4m gives ω_n → ω_n/2. The inverse-proportion answer forgets the square root. The natural frequency of a linear system is independent of the amplitude, and a larger mass lowers, not raises, the frequency because more inertia slows the response.

  2. A mass is supported by two springs in series, of stiffness k and 2k. The equivalent stiffness seen by the mass is:

    1. 3k, because the stiffnesses add when the springs carry the same force
    2. 2k, because the stiffer spring controls the deflection
    3. 2k/3, because the flexibilities 1/k and 1/(2k) add
    4. 3k/2, the arithmetic mean of the two stiffnesses
    Show answer

    Answer: C — 2k/3, because the flexibilities 1/k and 1/(2k) add

    Springs in series carry the same force, so the deflections and hence the flexibilities add: 1/k_eq = 1/k + 1/(2k) = 3/(2k), giving k_eq = 2k/3, softer than either spring. Adding the stiffnesses to 3k is the parallel result. The equivalent can never exceed the softer spring, so 2k and 3k/2 are wrong.

  3. A uniform solid cylinder of mass m rolls without slipping on a horizontal surface, and its axle is connected to a fixed wall by a horizontal spring of stiffness k. For small oscillations the natural frequency is:

    1. √(k/m), since only the mass of the cylinder matters
    2. √(k/2m), since rolling halves the effective stiffness
    3. √(3k/2m), since the rotation makes the effective mass smaller than m
    4. √(2k/3m), since the kinetic energy is (3/4)m ẋ², so the effective mass is 3m/2
    Show answer

    Answer: D — √(2k/3m), since the kinetic energy is (3/4)m ẋ², so the effective mass is 3m/2

    Translation gives ½mẋ² and rotation gives ½(½mR²)(ẋ/R)² = ¼mẋ², so T = ¾mẋ² = ½(3m/2)ẋ² and m_eq = 3m/2. With V = ½kx², ω_n = √(k/m_eq) = √(2k/3m). Using √(k/m) ignores the rotational energy, and rotation adds inertia, so the effective mass is larger than m, not smaller.

  4. A uniform rod of length 1.5 m is pivoted at one end on a frictionless pin and swings in a vertical plane through small angles. Take g = 9.81 m/s². Its natural frequency is:

    1. 2.56 rad/s, √(g/L), as for a simple pendulum of the same length
    2. 4.43 rad/s, √(3g/L), from a moment arm of L and inertia mL²/3
    3. 3.13 rad/s, √(3g/2L), from the weight moment mgL/2 and inertia mL²/3
    4. 2.09 rad/s, √(2g/3L), from inertia 3mL²/2 and a moment arm of L/2
    Show answer

    Answer: C — 3.13 rad/s, √(3g/2L), from the weight moment mgL/2 and inertia mL²/3

    For small θ the restoring moment is (mgL/2)θ and the inertia about the pivot is mL²/3, so ω_n² = (mgL/2)/(mL²/3) = 3g/(2L) = 3(9.81)/3 = 9.81 and ω_n = 3.13 rad/s. The simple-pendulum value treats the mass as concentrated at the end, and the others use a wrong arm or a wrong inertia; the rod’s centre of mass is at L/2, not at L.

  5. A 12 kg mass is attached to the free end of a steel cantilever of length 0.5 m, with E = 200 GPa and second moment of area I = 1.0 × 10⁻⁸ m⁴. The mass of the cantilever is neglected, and the tip stiffness is 3EI/L³. The natural frequency of the system is ____ rad/s, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 63.25

    k = 3EI/L³ = 3(200 × 10⁹)(1.0 × 10⁻⁸)/0.5³ = 6000/0.125 = 48 000 N/m. Then ω_n = √(k/m) = √(48 000/12) = √4000 = 63.25 rad/s.

  6. A single-degree-of-freedom system has a damping ratio ζ = 1.5. Following a displacement from rest, it:

    1. Returns to equilibrium without oscillating, more slowly than with critical damping
    2. Oscillates with an amplitude that decays at a rate set by ζ = 1.5 and then stops
    3. Oscillates indefinitely at the damped frequency, because ζ > 1 adds energy
    4. Returns to equilibrium faster than at critical damping, because more damping always means a faster return
    Show answer

    Answer: A — Returns to equilibrium without oscillating, more slowly than with critical damping

    For ζ > 1 both roots are real and negative, so the motion is non-oscillatory and the slower root controls the return, which is slower than at ζ = 1. Oscillation requires ζ < 1. Damping never adds energy, and the fastest non-oscillatory return occurs at critical damping, not above it.

  7. In a free-decay test of an under-damped single-degree-of-freedom system, the ratio of the amplitudes of two successive peaks is 1.8. The damping ratio ζ of the system is ____, to three decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.093

    δ = ln 1.8 = 0.5878. From δ = 2πζ/√(1 − ζ²), ζ = δ/√(4π² + δ²) = 0.5878/√(39.478 + 0.3455) = 0.5878/6.311 = 0.0931, i.e. 0.093. The small-damping estimate δ/2π = 0.0935 is close but not the exact value.

  8. For an under-damped single-degree-of-freedom system with damping ratio ζ (0 < ζ < 1), which of the following are correct?

    1. The damped natural frequency is ω_d = ω_n√(1 − ζ²), which is less than ω_n
    2. The logarithmic decrement is δ = 2πζ/√(1 − ζ²), and the ratio of successive peaks is e^δ at every cycle
    3. The critical damping coefficient is c_c = 2√(km), and ζ = c/c_c
    4. The damped frequency ω_d exceeds ω_n, because damping pushes the mass back faster
    Show answer

    Answer: A — The damped natural frequency is ω_d = ω_n√(1 − ζ²), which is less than ω_n; B — The logarithmic decrement is δ = 2πζ/√(1 − ζ²), and the ratio of successive peaks is e^δ at every cycle; C — The critical damping coefficient is c_c = 2√(km), and ζ = c/c_c

    The first three are standard results: the oscillation of an under-damped system is slower than the undamped one, successive peaks fall by the constant factor e^(−δ), and c_c = 2√(km). The damped frequency is always below ω_n, because the damper adds a retarding force and lengthens the period.

  9. A lightly damped single-degree-of-freedom system is driven by a sinusoidal force at exactly its undamped natural frequency (r = 1). In the steady state, the displacement:

    1. Is in phase with the force, because resonance means the force and the motion peak together
    2. Lags the force by 45°, which is the average of the in-phase and the opposite-phase limits
    3. Lags the force by 180°, because the mass is moving against the spring force
    4. Lags the force by 90°, and its amplitude is F₀/(2ζk)
    Show answer

    Answer: D — Lags the force by 90°, and its amplitude is F₀/(2ζk)

    At r = 1 the spring and inertia forces cancel, the damper force balances the applied force, and tan φ = 2ζr/(1 − r²) → ∞, so φ = 90° (the velocity is in phase with the force); the amplitude is F₀/(cω_n) = F₀/(2ζk). In phase is the low-frequency limit and 180° the high-frequency limit; 45° is not reached at r = 1 for any damping.

  10. A machine of total mass M has an unbalanced rotor (mass m₀ at eccentricity e) and is mounted on a spring and damper. As the running speed is raised very far above the natural frequency (r ≫ 1), the steady vibration amplitude of the machine tends to:

    1. Zero, since the amplitude of a forced system falls as 1/r² at high frequency
    2. Infinity, since the exciting force m₀eω² grows without limit with speed
    3. m₀e/M, because the force growing as ω² cancels the 1/ω² fall of the response
    4. m₀eω²/k, the static deflection under the instantaneous exciting force
    Show answer

    Answer: C — m₀e/M, because the force growing as ω² cancels the 1/ω² fall of the response

    MX/(m₀e) = r²/√((1 − r²)² + (2ζr)²) → 1 as r → ∞, so X → m₀e/M: the exciting force grows as ω² while the response factor falls as 1/ω², and the product is constant. The fall to zero holds only for a constant-amplitude force, and infinity occurs only at resonance in an undamped system. The static-deflection expression ignores the inertia that dominates at high speed.

  11. A single-degree-of-freedom system has k = 5000 N/m, damping ratio ζ = 0.1 and is driven by a harmonic force of amplitude 50 N at a frequency twice its natural frequency (r = 2). The steady-state displacement amplitude is ____ mm, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 3.30

    The static deflection is F₀/k = 50/5000 = 10 mm. M = 1/√((1 − r²)² + (2ζr)²) = 1/√((−3)² + 0.4²) = 1/√9.16 = 0.3304. So X = 10 × 0.3304 = 3.30 mm.

  12. A machine of total mass 100 kg (including the rotor) has an unbalance m₀e = 0.05 kg·m. It is mounted on a spring-damper support with damping ratio ζ = 0.1 and runs at a speed equal to twice the undamped natural frequency of the support (r = 2). The steady vertical vibration amplitude is ____ mm, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.66

    MX/(m₀e) = r²/√((1 − r²)² + (2ζr)²) = 4/√9.16 = 1.3217. So X = 1.3217 × m₀e/M = 1.3217 × 0.05/100 = 6.608 × 10⁻⁴ m = 0.66 mm. Forgetting the r² factor (using M = 0.3304 alone) would give 0.17 mm, which is wrong because the exciting force grows as ω².

  13. A single-degree-of-freedom system with damping ratio ζ (0 < ζ < 0.5) is driven by a harmonic force of constant amplitude F₀ at frequency ratio r. Which of the following are correct?

    1. At low frequency (r ≪ 1) the amplitude is close to the static deflection F₀/k
    2. At r = 1 the amplitude is F₀/(2ζk), and it grows without bound as ζ → 0
    3. At high frequency (r ≫ 1) the amplitude is approximately F₀/(mω²), which is small
    4. For every value of ζ, the maximum of the amplitude occurs exactly at r = 1
    Show answer

    Answer: A — At low frequency (r ≪ 1) the amplitude is close to the static deflection F₀/k; B — At r = 1 the amplitude is F₀/(2ζk), and it grows without bound as ζ → 0; C — At high frequency (r ≫ 1) the amplitude is approximately F₀/(mω²), which is small

    For r ≪ 1, M ≈ 1; at r = 1, M = 1/(2ζ); for r ≫ 1, M ≈ 1/r², so X ≈ F₀/(k r²) = F₀/(mω²). The peak of the displacement amplitude lies at r = √(1 − 2ζ²), which is below 1 whenever ζ > 0, so it is not exactly at r = 1 (it is the velocity amplitude that peaks exactly at r = 1).

  14. A machine is mounted on a spring-damper isolator, and the disturbance at the base is harmonic. For any value of the damping ratio ζ, the displacement transmissibility at frequency ratio r = √2 is:

    1. Zero, because isolation begins at that frequency ratio
    2. Equal to 1, independent of ζ
    3. Equal to √2, the same as the frequency ratio, whatever ζ is
    4. Equal to 1/(2ζ), the value at resonance, so it depends strongly on ζ
    Show answer

    Answer: B — Equal to 1, independent of ζ

    At r² = 2: numerator √(1 + 8ζ²) and denominator √((1 − 2)² + 8ζ²) = √(1 + 8ζ²), so TR = 1 for every ζ. Isolation (TR < 1) begins only above this ratio, so TR is not zero there. The 1/(2ζ) value belongs to resonance in the forced-response magnification, not to base excitation at √2.

  15. An engineer reduces the transmitted vibration of a machine by using isolators that operate at a frequency ratio r = 3, where the transmissibility is 0.125 with no damping. If viscous damping with ζ = 0.2 is now added to the isolators, the transmissibility at the same running speed becomes about:

    1. 0.19, higher than before, because above r = √2 added damping lets more vibration through
    2. 0.06, lower than before, because damping always dissipates energy and reduces transmission
    3. 0.125, unchanged, because the transmissibility depends on r alone and not on ζ
    4. 0.25, doubled, because transmissibility is proportional to 1 + ζ
    Show answer

    Answer: A — 0.19, higher than before, because above r = √2 added damping lets more vibration through

    TR = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²) = √(1 + 1.44)/√(64 + 1.44) = 1.562/8.089 = 0.193. Above r = √2 the damper transmits force directly across the isolator, so TR rises from 0.125 to about 0.19. Damping does help below √2 (near resonance), which is the tempting generalisation; TR does depend on ζ, and there is no 1 + ζ rule.

  16. A 200 kg machine runs at 1500 rpm on undamped isolators. To limit the transmissibility to 0.2, the isolator stiffness (total, in the vertical direction) must be ____ kN/m, to the nearest integer. Take the machine as a single rigid mass and use TR = 1/(r² − 1) for r > √2.

    Numerical answer — type the value.

    Show answer

    Answer: 822

    ω = 2π(1500)/60 = 157.08 rad/s. TR = 1/(r² − 1) = 0.2 gives r² = 6. Then ω_n² = ω²/6 = 24 674/6 = 4112.3 rad²/s², and k = mω_n² = 200 × 4112.3 = 822 467 N/m = 822 kN/m (nearest integer).

  17. Which of the following statements about transmissibility TR = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²) and vibration isolation are correct?

    1. TR = 1 at r = √2 for every damping ratio, and TR < 1 only for r > √2
    2. For r > √2, increasing the damping ratio increases the transmissibility
    3. For an undamped isolator, a lower natural frequency (a softer mount) gives a lower transmissibility at a fixed running speed above √2 ω_n
    4. For ζ > 0.5, the transmissibility is below 1 at every frequency ratio
    Show answer

    Answer: A — TR = 1 at r = √2 for every damping ratio, and TR < 1 only for r > √2; B — For r > √2, increasing the damping ratio increases the transmissibility; C — For an undamped isolator, a lower natural frequency (a softer mount) gives a lower transmissibility at a fixed running speed above √2 ω_n

    TR passes through 1 at r = √2 for all ζ and is below 1 beyond; above √2 the damper’s direct force path makes TR rise with ζ; with ζ = 0 and fixed ω, lowering ω_n raises r and so lowers 1/(r² − 1). Transmissibility exceeds 1 for r < √2 whatever the damping, since the TR = 1 crossing at √2 is independent of ζ, so a larger ζ does not remove the amplification region.