Interactive Lecture
Natural Frequency, Damping, and Frequency Response Functions
by Dr Arnaldo Delli Carri
Single Degree of Freedom (SDOF) vibration is a fundamental concept in mechanical engineering that describes the oscillatory motion of a system with one principal direction of movement. These systems are characterised by:
A single degree of freedom system can be represented by the following differential equation:
\[m\ddot{x}+c\dot{x}+kx=F_0\cos(\omega t)\]
Where:
The behaviour of a SDOF system depends critically on its damping ratio \(\zeta\), which is defined as:
\[\zeta = \frac{c}{c_{c}} = \frac{c}{2\sqrt{km}}\]
where \(c_c=2\sqrt{km}\) is the critical damping coefficient [Ns/m].
There are three damping conditions that lead to distinctly different system behaviours, depending on whether the damping coefficient is smaller, equal or greater than the critical damping coefficient:
Damping has several important effects on SDOF systems:
Natural frequency (pulsation): \(\omega_n = \sqrt{\frac{k}{m}}\) [rad/s]
Natural frequency: \(f_n = \frac{1}{2\pi}\omega_n\) [Hz]
Damped/transient natural pulsation: \(\omega_d = \omega_n\sqrt{1-\zeta^2}\) [rad/s]
Damped/transient natural frequency: \(f_d = \frac{1}{2\pi}\omega_d\) [Hz]
For an underdamped system (\(\zeta < 1\)) released from rest (\(\dot{x}(0) = 0\)) at an initial displacement \(x_0\), the free vibration response is:
\[x(t) = x_0 e^{-\zeta \omega_n t}\left[\cos(\omega_d t) + \frac{\zeta\omega_n}{\omega_d}\sin(\omega_d t)\right]\]
For light damping (\(\zeta \ll 1\)) the sine term is small, so the response is close to \(x(t) \approx x_0 e^{-\zeta\omega_n t}\cos(\omega_d t)\): a cosine bounded by the decaying envelope \(\pm x_0 e^{-\zeta\omega_n t}\).
The decay rate can be read straight off the response. The logarithmic decrement \(\delta\) is the natural log of the ratio of two successive displacement peaks one period apart:
\[\delta = \ln\frac{x_n}{x_{n+1}} = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\]
Inverting this recovers the damping ratio from a measured decrement:
\[\zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}}\]
When a harmonic force \(F(t) = F_0 \cos(\omega t)\) is applied to the system, the steady-state response is:
\[x(t) = X \cos(\omega t - \theta)\]
Where:
\[X = \frac{F_0/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\]
\[\theta = \tan^{-1}\left(\frac{2\zeta r}{1-r^2}\right)\]
\[r=\omega/\omega_n\]
Resonance: When the forcing frequency approaches the natural frequency (\(\omega \approx \omega_n\)) the amplitude of vibration becomes very large, especially for lightly damped systems. This phenomenon is called resonance and can lead to catastrophic failures in engineering structures if not properly managed.
The resonance frequency \(\omega_R\) for a SDOF system is:
\[\omega_R=\omega_n\sqrt{1-2\zeta^2}\]
The Frequency Response Function is a fundamental tool in vibration analysis that shows how a system responds to different forcing frequencies:
\[H(\omega) = \frac{X(\omega)}{F(\omega)} = \frac{1/k}{\sqrt{[1-(\omega/\omega_n)^2]^2 + [2\zeta(\omega/\omega_n)]^2}}\]
The FRF has two components:
Experiment with different parameters to see how they affect the behaviour of a SDOF system:
| Concept | Equation | Description |
|---|---|---|
| Natural Frequency | \(\omega_n = \sqrt{\frac{k}{m}}\) | Undamped natural frequency |
| Damping Ratio | \(\zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{km}}\) | Fraction of critical damping |
| Damped Frequency | \(\omega_d = \omega_n\sqrt{1-\zeta^2}\) | Frequency with damping |
| Free Response | \(x(t) = x_0 e^{-\zeta\omega_n t}\cos(\omega_d t)\) | Underdamped free vibration (light damping) |
| Forced Response | \(X = \frac{F_0/k}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}\) | Steady-state amplitude |
| Frequency Ratio | \(r = \frac{\omega}{\omega_n}\) | Excitation to natural frequency |
A machine component has a mass of 5 kg and is supported by a spring with stiffness 8000 N/m. Calculate:
Show Solution
Solution:
a) The natural frequency in rad/s:
\(\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{8000}{5}} = \sqrt{1600} = 40\) rad/s
b) The natural frequency in Hz:
\(f_n = \frac{\omega_n}{2\pi} = \frac{40}{2\pi} = 6.37\) Hz
c) The period of oscillation:
\(T = \frac{1}{f_n} = \frac{1}{6.37} = 0.157\) seconds
A vibrating system with mass 2 kg and spring stiffness 1800 N/m has a logarithmic decrement of 0.5. Calculate:
Show Solution
Solution:
a) The damping ratio:
The logarithmic decrement \(\delta = 0.5\)
\(\zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} = \frac{0.5}{\sqrt{4\pi^2 + 0.5^2}} = \frac{0.5}{\sqrt{39.74}} = \frac{0.5}{6.3} = 0.079\)
b) The damping coefficient:
\(\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1800}{2}} = 30\) rad/s
\(c = 2\zeta\omega_n m = 2 \times 0.079 \times 30 \times 2 = 9.48\) Ns/m
c) The critical damping coefficient:
\(c_{critical} = 2m\omega_n = 2 \times 2 \times 30 = 120\) Ns/m
d) Since \(\zeta = 0.079 < 1\), the system is underdamped.
A machine with mass 100 kg is mounted on a spring with stiffness \(k = 4 \times 10^5\) N/m and a damper with damping coefficient \(c = 1000\) Ns/m. The machine is subjected to a harmonic force with amplitude \(F_0 = 5000\) N. Calculate:
Show Solution
Solution:
a) The natural frequency:
\(\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{4 \times 10^5}{100}} = \sqrt{4000} = 63.25\) rad/s
\(f_n = \frac{\omega_n}{2\pi} = \frac{63.25}{2\pi} = 10.07\) Hz
b) The amplitude of vibration at resonance:
First, calculate the damping ratio:
\(c_{critical} = 2m\omega_n = 2 \times 100 \times 63.25 = 12650\) Ns/m
\(\zeta = \frac{c}{c_{critical}} = \frac{1000}{12650} = 0.079\)
At resonance \(\omega = \omega_R\)
\(\omega_R=\omega_n\sqrt{1-2\zeta^2}=62.85\) rad/s, the amplitude is:
\(X = \frac{F_0/k}{2\zeta\sqrt{1-\zeta^2}} = \frac{5000/(4 \times 10^5)}{2 \times 0.079 \times \sqrt{1-0.079^2}} = \frac{0.0125}{0.158} = 0.079\) m (or 79 mm)
c) The amplitude of vibration when the forcing frequency is 10 Hz:
\(\omega = 2\pi f = 2\pi \times 10 = 62.83\) rad/s
\(r = \frac{\omega}{\omega_n} = \frac{62.83}{63.25} = 0.993\)
\(X = \frac{F_0/k}{\sqrt{[1-r^2]^2 + [2\zeta r]^2}} = \frac{0.0125}{\sqrt{[1-0.993^2]^2 + [2 \times 0.079 \times 0.993]^2}}\)
\(X = \frac{0.0125}{\sqrt{[0.014]^2 + [0.157]^2}} = \frac{0.0125}{0.158} = 0.079\) m (or 79 mm)
Note: The 10 Hz excitation sits almost exactly on the resonance peak (\(f_R = \frac{\omega_R}{2\pi} \approx 10.00\) Hz, just below the natural frequency \(f_n = 10.07\) Hz), so the amplitude is nearly the maximum.