Machine Dynamics and Vibrations: free practice, theory and problems

1. Free vibrations

What is it about?

Free vibration is the motion a system performs when it is disturbed from equilibrium and then left to oscillate without any further external force. A mass–spring system, a beam that shudders after being struck, or a shaft with a flywheel are typical examples. As an engineer you need to know a structure's natural frequency, because if it is hit by a load at the same frequency the vibrations can grow dangerously large (resonance, covered in the next unit). This unit gives you the tools to find the natural frequency and describe the vibration itself.

Concepts and formulas

The equation of motion for an undamped mass–spring system is

• Natural angular frequency: $\omega_n = \sqrt{k/m}$ (rad/s). Natural frequency in Hz: $f_n = \omega_n/(2\pi)$. Period: $T = 1/f_n = 2\pi/\omega_n$.
• The solution is harmonic: $x(t) = X\cos(\omega_nt - \varphi)$, where $X$ is the amplitude and $\varphi$ the phase angle, both set by the initial conditions $x(0)$ and $\dot x(0)$.
• Maximum velocity and acceleration: $\dot x_{max} = X\omega_n$ and $\ddot x_{max} = X\omega_n^2$.
• Springs in parallel: $k_{tot} = k_1 + k_2$. Springs in series: $1/k_{tot} = 1/k_1 + 1/k_2$.
• Torsional vibration: $\omega_n = \sqrt{k_t/J}$, where $k_t$ is the torsional stiffness (Nm/rad) and $J$ the mass moment of inertia (kgm²) about the axis.
• From the static deflection $\delta$ (the spring stretched by the weight): $k\delta = mg$, so $\omega_n = \sqrt{g/\delta}$.
• The period is independent of the amplitude in a linear (harmonic) vibration – that is the whole point of calling it "harmonic".

How to solve the problems

1. Find the equivalent stiffness $k$ (add parallel springs, combine series springs) or the torsional stiffness $k_t$.
2. Set up $\omega_n = \sqrt{k/m}$ (or $\sqrt{k_t/J}$, or $\sqrt{g/\delta}$ from the static deflection).
3. Convert to $f_n$ or $T$ as the problem asks.
4. If you need velocity or acceleration in the vibration, use $X\omega_n$ and $X\omega_n^2$ with the given amplitude.
5. Check the units: $k$ in N/m and $m$ in kg give $\omega_n$ in rad/s.

Example

A machine of 80 kg rests on four springs in parallel, each with a stiffness of 3000 N/m. It vibrates with an amplitude of 5 mm. Find the natural frequency in Hz and the maximum acceleration.

1. Total stiffness: $k = 4\cdot 3000 = 12000$ N/m.
2. $\omega_n = \sqrt{12000/80} = \sqrt{150} \approx 12.25$ rad/s.
3. $f_n = \omega_n/(2\pi) \approx 1.95$ Hz.
4. $\ddot x_{max} = X\omega_n^2 = 0.005\cdot 150 = 0.75$ m/s².

Answer: $f_n \approx 1.95$ Hz and $\ddot x_{max} \approx 0.75$ m/s².

Common mistakes

• Mixing up $\omega_n$ (rad/s) and $f_n$ (Hz) – remember the factor $2\pi$.
• Assuming series springs follow the same formula as parallel springs (it is the opposite of electrical resistors).
• Believing the amplitude affects the natural frequency. It does not, in a linear vibration.
• Using the mass in kg when the stiffness is given in kN/m without converting.
• Forgetting that $\delta$ (static deflection) must be measured along the direction the spring force acts.

Concepts in this part

Practise free vibrations in the app →

2. Damping

What is it about?

Most real vibrating systems have damping: friction, air resistance or a shock absorber removes energy from the system, so the vibrations die out. Damping decides whether a car's suspension keeps oscillating after a bump in the road or settles down smoothly, and whether an instrument needle settles quickly or swings back and forth for a long time. This unit is about describing and measuring damping, and understanding how it changes the vibration.

Concepts and formulas

The equation of motion with viscous damping is

• Damping ratio: $\zeta = c/c_c$, where the critical damping coefficient is $c_c = 2\sqrt{km} = 2m\omega_n$.
• Three cases: $\zeta < 1$ underdamped (oscillates, decaying amplitude), $\zeta = 1$ critically damped (fastest return without overshoot), $\zeta > 1$ overdamped (returns slowly, without oscillating).
• Damped natural frequency (only for an underdamped system): $\omega_d = \omega_n\sqrt{1-\zeta^2}$, and damped period $T_d = 2\pi/\omega_d$.
• The amplitude of a free, underdamped vibration decays as $X(t) = X_0e^{-\zeta\omega_nt}$ – an exponential envelope.
• Logarithmic decrement between two successive peaks: $\delta = \ln(x_1/x_2)$. Over $n$ periods: $\delta = \dfrac{1}{n}\ln(x_0/x_n)$. Relation to $\zeta$: $\delta = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}}$, and for small $\zeta$: $\zeta \approx \delta/(2\pi)$.
• Time until the amplitude is reduced to a given fraction: solve $e^{-\zeta\omega_nt} = X/X_0$ for $t$.

How to solve the problems

1. Find $\omega_n$ from $k$ and $m$ if it is not given directly.
2. Compute $\zeta$ from $c$, $c_c$, or from amplitude measurements (logarithmic decrement).
3. Decide the type of damping by comparing $\zeta$ with 1.
4. Use $\omega_d$ and $T_d$ when you need the damped vibration frequency, and the exponential formula when you need the amplitude or the time.
5. Check that $\zeta \geq 0$ and that the underdamped formulas are only used when $\zeta < 1$.

Example

A system has $m = 10$ kg, $k = 4000$ N/m and $c = 80$ Ns/m. Find the damping ratio and the damped natural frequency.

1. $c_c = 2\sqrt{km} = 2\sqrt{4000\cdot 10} = 2\sqrt{40000} \approx 400$ Ns/m.
2. $\zeta = c/c_c = 80/400 = 0.2$.
3. $\omega_n = \sqrt{k/m} = \sqrt{400} = 20$ rad/s.
4. $\omega_d = \omega_n\sqrt{1-\zeta^2} = 20\sqrt{1-0.04} \approx 19.6$ rad/s.

Answer: $\zeta = 0.2$ (underdamped) and $\omega_d \approx 19.6$ rad/s.

Common mistakes

• Mixing up $\omega_n$ and $\omega_d$ – they are close only when $\zeta$ is very small.
• Using the damped formula $\omega_d = \omega_n\sqrt{1-\zeta^2}$ when $\zeta \geq 1$ (the expression is then meaningless – the system does not oscillate).
• Believing critical damping means the system does not move at all. It simply returns as fast as possible without overshooting equilibrium.
• Forgetting the exponential term and using only the undamped amplitude formula.
• Confusing the logarithmic decrement (between two peaks) with the damping ratio $\zeta$ itself – they are related through the formula above.

Concepts in this part

Practise damping in the app →

3. Forced vibrations and isolation

What is it about?

When a vibrating system is driven by a periodic force or motion – from an unbalanced motor, waves, or an uneven road – the result is called forced vibration. The system eventually vibrates at the same frequency as the excitation, but how large the response becomes depends strongly on how close the excitation frequency is to the natural frequency. Near resonance even a small force can produce huge amplitudes, and that is exactly what an engineer must either avoid or exploit – by adding damping, by shifting the natural frequency away from the operating frequency, or by isolating the machine from its foundation.

Concepts and formulas

• Frequency ratio: $r = \omega/\omega_n$, where $\omega$ is the angular frequency of the excitation.

The magnification factor (dynamic response divided by the static deflection $F_0/k$) is

• At resonance ($r = 1$), $M \approx 1/(2\zeta)$ for lightly damped systems – the smaller the damping, the larger the response.
• Phase angle between the force and the response: $\tan\varphi = \dfrac{2\zeta r}{1-r^2}$. Below resonance the force and the response are nearly in phase ($\varphi \approx 0$), at resonance $\varphi = 90°$, and well above resonance they are nearly out of phase ($\varphi \approx 180°$).
• Limiting behavior: for $r \ll 1$ the response approaches the static deflection $F_0/k$ (the mass follows the force quasi-statically). For $r \gg 1$ the response goes to zero (the mass cannot keep up).
• Transmissibility (force or motion transmitted to the foundation), undamped: $TR = 1/|1-r^2|$. Good isolation requires $r > \sqrt2$, meaning the natural frequency must be low compared with the operating frequency (soft isolators).
• Unbalance force from a rotating mass $m_e$ with eccentricity $e$: $F_0 = m_ee\omega^2$, which grows with the square of the rotational speed.

How to solve the problems

1. Find $\omega_n$ for the system (as in the previous units) and the excitation frequency $\omega$ (convert rpm to rad/s: $\omega = 2\pi n/60$).
2. Compute the frequency ratio $r = \omega/\omega_n$.
3. Substitute into $M(r,\zeta)$ or $TR = 1/|1-r^2|$, whichever the problem asks for.
4. For isolation: check whether $r > \sqrt2$. If not, the force is amplified rather than reduced.
5. Check the limiting cases: is $r$ close to 1 (risk of resonance), close to 0 (quasi-static), or large (isolated)?

Example

A 150 kg machine with an unbalance rotates at 900 rpm and rests on springs with a total stiffness of 200 kN/m. Damping is negligible. Is the isolation effective?

1. $\omega_n = \sqrt{k/m} = \sqrt{200000/150} \approx 36.5$ rad/s.
2. $\omega = 2\pi\cdot 900/60 \approx 94.2$ rad/s.
3. $r = \omega/\omega_n \approx 2.58$.
4. Since $r > \sqrt2 \approx 1.41$, the isolation is effective: $TR = 1/|1-r^2| = 1/|1-6.66| \approx 0.177$, so only 17.7% of the force is transmitted.

Answer: Yes, the isolation works well ($TR \approx 0.18$).

Common mistakes

• Assuming the amplification is always largest when $r$ is large. It is largest near $r = 1$ (resonance); for $r \gg 1$ the response decreases.
• Forgetting to convert rpm to rad/s before computing $r$.
• Believing isolation works for any $r < 1$. Isolation (reduced force transmission) requires $r > \sqrt2$; for $r < \sqrt2$ the force is amplified.
• Using the undamped transmissibility formula when damping is significant – then $\zeta$ must be included.
• Forgetting the square in the unbalance force $F_0 = m_ee\omega^2$: doubling the speed quadruples the force.

Concepts in this part

Practise forced vibrations and isolation in the app →